Get longest continuous sequence of 1s - arrays

I recently encountered a problem statement it says:
Given an array of 0s and 1s, find the position of 0 to be
replaced with 1 to get longest continuous sequence of 1s.
For example : Array- 1,1,0,0,1,0,1,1,1,0,1,1,1
Output - index 9
I tried a brute force approach replacing every encountered 0 with 1 and after each such replacement, i counted the largest continuous repetitive sequence of 1 and updated it every time.
Is there a better approach/algorithm to this problem?

There should be a one-pass solution to this. The overall idea is to count the ones and to add up the lengths for each zero as you go. Well, not each zero, just the last encountered one and the longest.
You need to keep track of two things:
The longest chain so far.
The previous zero value, along with the length of the preceding ones.
The process then goes as following:
Starting walking through the string until you encounter a zero. Keep track of the number of ones as you go.
When you hit the zero, remember the position of the zero along with the number of preceding 1s.
Count the 1s to the next zero.
Go back to the previous zero and add the new "ones" to the previous "ones". If this is longer than the longest chain, then replace the longest chain.
Remember this zero along with the preceding 1s.
Repeat until you have reached the end of the string.
At then end of the string, go back and add the length to the previous zero and replace the longest chain if appropriate.

You can imagine you have to maintain a set of 1 allowing only one 0 among them,
so
1) walk over the array,
2) if you are getting a 1,
check a flag if you are already in a set, if no,
then you start one and keep track of the start,
else if yes, you just update the end point of set
3) if you get a 0, then check if it can be included in the set,
(i.e. if only one 0 surrounded by 1 "lonely zero" )
if no, reset that flag which tells you you are in a set
else
is this first time ? (store this 0 pos, initialized to -1)
yes, then just update the zero position
else okk, then previous set, of one..zero..one gets finished here,
now the new set's first half i.e. first consecutive ones are the previous set's last portion,
so set the beginning of the set marker to last zero pos +1, update the zero position.
So when to get check if the current set is having highest length? See , we update the end point only in 2 -> else portion, so just check with max start, max end etc etc at that point and it should be enough

Here is my solution. It is clean, takes O(n) time and O(1) memory.
public class Q1 {
public Q1() {
}
public static void doit(int[] data) {
int state = 0;
int left, right, max_seq, max_i, last_zero;
left = right = 0;
max_seq = -1;
max_i = -1;
// initialization
right = data[0];
last_zero = (data[0]==0) ? 0 : -1;
for (int i = 1; i < data.length; i++) {
state = data[i - 1] * 10 + data[i];
switch (state) {
case 00: //reset run
left = right = 0;
last_zero = i;
break;
case 01: // beginning of a run
right++;
break;
case 10:// ending of a run
if(left+right+1>max_seq){
max_seq = left+right+1;
max_i = last_zero;
}
last_zero = i; //saving zero position
left = right; // assigning left
right = 0; // resetting right
break;
case 11: // always good
right++;
break;
}
}
//wrapping up
if(left+right+1>max_seq){
max_seq = left+right+1;
max_i = last_zero;
}
System.out.println("seq:" + max_seq + " index:" + max_i);
}
public static void main(String[] args) {
//Q1.doit(new int[] { 1,1,0,0,1,0,1,1,1,0,1,1,1 });
Q1.doit(new int[] { 1,1,0,0,1,0,1,1,1,0,1,1,1 });
}
}

Using Dynamic programming you can solve this code.
Time complexity is O(n) and space complexity is O(n).
public static int Flipindex(String mystring){
String[] arr = mystring.split(",");
String [] arrays= new String[arr.length];
for(int i=0;i<arr.length;i++){
arrays[i]="1";
}
int lastsum = 0;
int[] sumarray =new int[arr.length];
for(int i=0;i<arr.length;i++){
if(!arr[i].equals(arrays[i])){
++lastsum;
}
sumarray[i]=lastsum;
}
int [] consecsum = new int [sumarray[sumarray.length-1]+1];
for(int i: sumarray){
consecsum[i]+=1;
}
int maxconsecsum=0,startindex=0;
for(int i=0;i<consecsum.length-1;i++){
if((consecsum[i]+consecsum[i+1])>maxconsecsum){
maxconsecsum=(consecsum[i]+consecsum[i+1]);
startindex=i;
}
}
int flipindex=0;
for(int i=0;i<=startindex;i++){
flipindex+=consecsum[i];
}
return flipindex;
}
public static void main(String[] args) {
String s= "1,1,0,0,1,0,1,1,1,0,1,1,1";
System.out.println(Flipindex(s));
}

Playing around with console yielded me this, touch up and cover edge case then you are good to go
function getIndices(arr, val) {
var indexes = [], i = -1;
while ((i = arr.indexOf(val, i+1)) != -1){
indexes.push(i);
}
return indexes;
}
var a = [1,1,1,1,1,0,0,1,0,0,1,1,1,0,1,1,1,1,1,1,0];
var z = getIndices(a, 0);
z.unshift(0);
var longestchain = 0;
var target = 0;
for(var i=0;i<z.length;i++) {
if(i == 0) { //first element
longestchain = z[i] + z[i+1];
target = i;
} else if (i == z.length-1) { //last element
var lastDistance = Math.abs(z[i] - z[i-1]);
if(lastDistance > longestchain) {
longestchain = lastDistance;
target = i;
}
} else {
if(Math.abs(z[i] - z[i+1]) > 1) { //consecutive 0s
//look before and ahead
var distance = Math.abs(z[i-1] - z[i]) + Math.abs(z[i] - z[i+1]);
if(distance > longestchain) {
longestchain = distance;
target = i;
}
}
}
}
console.log("change this: " + z[target]);
I first search for zeroes in the array and stored the position in another array, so in my e.g. you will get something like this [0,5,6,8,9,13,20], then i just run a single loop to find the greatest distance from each element with their adjacent ones, and storing the distance in the "longestchain", everytime i find a longer chain, i take note of the index, in this case "13".

This C code implementation is based on the algorithm provided by #gordon-linoff above.
int maxOnesIndex1(bool arr[], int n)
{
int prevZeroPos = 0;
int oldOneCnt = 0;
int newOneCnt = 0;
int longestChainOfOnes = 0;
int longestChainPos = 0;
int i;
for(i=0; i<n; i++)
{
if(arr[i]!=0)
{
oldOneCnt++;
}
else // arr[i] == 0
{
prevZeroPos = i;
newOneCnt = 0;
// move by one to find next sequence of 1's
i++;
while(i<n && arr[i] == 1)
{
i++;
newOneCnt++;
}
if((oldOneCnt+newOneCnt) > longestChainOfOnes)
{
longestChainOfOnes = oldOneCnt+newOneCnt+1;
longestChainPos = prevZeroPos;
}
oldOneCnt = 0;
i = prevZeroPos;
}
}
if((oldOneCnt+newOneCnt) > longestChainOfOnes)
{
longestChainOfOnes = oldOneCnt+newOneCnt+1;
longestChainPos = prevZeroPos;
}
return longestChainPos;
}

Space Complexity - O(1)
Time Complexity - O(n)
A = map(int, raw_input().strip().split(' '))
left = 0 #Numbers of 1 on left of current index.
right = 0 #Number of 1 on right of current index.
longest = 0 #Longest sequence so far
index = 0
final_index = 0 # index of zero to get the longest sequence
i = 0
while i < A.__len__():
if A[i] == 0:
left = right
index = i
i += 1
right = 0
while i < A.__len__() and A[i] != 0:
right += 1
i += 1
if left + right + 1 > longest:
final_index = index
longest = left + right + 1
else:
right += 1
i += 1
print final_index, longest

Here is little different algorithm
public static int zeroIndexToGetMaxOnes(int[] binArray) {
int prevPrevIndex = -1, prevIndex = -1,currentLenght= -1, maxLenght = -1, requiredIndex = -1;
for (int currentIndex = 0; currentIndex < binArray.length; currentIndex++) {
if (binArray[currentIndex] == 0) {
if (prevPrevIndex != -1) {
currentLenght = currentIndex - (prevPrevIndex + 1);
if (currentLenght > maxLenght) {
maxLenght = currentLenght;
requiredIndex = prevIndex;
}
}
prevPrevIndex = prevIndex;
prevIndex = currentIndex;
} else {// case when last element is not zero, and input contains more than 3 zeros
if (prevIndex != -1 && prevPrevIndex != -1) {
currentLenght = currentIndex - (prevPrevIndex + 1);
if (currentLenght > maxLenght) {
maxLenght = currentLenght;
requiredIndex = prevIndex;
}
}
}
}
if (maxLenght == -1) { // less than three zeros
if (prevPrevIndex != -1) { // 2 zeros
if (prevIndex > (binArray.length - prevPrevIndex - 1)) {
requiredIndex = prevPrevIndex;
} else {
requiredIndex = prevIndex;
}
} else { // one zero
requiredIndex = prevIndex;
}
}
return requiredIndex;
}
Here is the unit tests
#Test
public void replace0ToGetMaxOnesTest() {
int[] binArray = {1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1};
int index = ArrayUtils.zeroIndexToGetMaxOnes(binArray);
assertThat(index, is(9));
binArray = new int[]{1,0,1,1,1,0};
index = ArrayUtils.zeroIndexToGetMaxOnes(binArray);
assertThat(index, is(1));
binArray = new int[]{0,1,1,1,0,1};
index = ArrayUtils.zeroIndexToGetMaxOnes(binArray);
assertThat(index, is(4));
binArray = new int[]{1,1,1,0,1,0};
index = ArrayUtils.zeroIndexToGetMaxOnes(binArray);
assertThat(index, is(3));
binArray = new int[]{0,1,1,1,0};
index = ArrayUtils.zeroIndexToGetMaxOnes(binArray);
assertThat(index, is(4));
binArray = new int[]{1,1,1,1,0};
index = ArrayUtils.zeroIndexToGetMaxOnes(binArray);
assertThat(index, is(4));
binArray = new int[]{0,1,1,1,1};
index = ArrayUtils.zeroIndexToGetMaxOnes(binArray);
assertThat(index, is(0));
}

def sol(arr):
zeros = [idx for idx, val in enumerate(arr) if val == 0]
if len(arr) == 0 or len(zeros) == 0:
return None
if len(arr) - 1 > zeros[-1]:
zeros.append(len(arr))
if len(zeros) == 1:
return zeros[0]
if len(zeros) == 2:
return max(zeros)
max_idx = None
diff = 0
for i in range(len(zeros) - 2):
# Calculating the difference of i+2 and i, since i+1 should be filled with 1 to find the max index
if zeros[i+2] - zeros[i] > diff:
diff = zeros[i + 2] - zeros[i] - 1
max_idx = zeros[i+1]
return max_idx
arr = [1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1]
print(sol(arr))

Related

Given an array of 0 and 1, find minimum no. of swaps to bring all 1s together (only adjacent swaps allowed)

If given an array of 1's and 0's, what's good algorithm to show the minimum number of adjacent swaps needed to group all of the 1's together. The 1's don't need to be grouped at any specific place in the array. They just need to be grouped in whatever place provides for the minimum number of adjacent swaps.
For example, if the array looks like this...
1,0,0,1,1,0,1
...the minimum number of adjacent swaps would be 3, because you'd center on index 4 and do the following swaps:
Swap indices 0 and 1, resulting in:
0,1,0,1,1,0,1
Swap indices 1 and 2, resulting in:
0,0,1,1,1,0,1
Swap indices 5 and 6, resulting in:
0,0,1,1,1,1,0
Anyone have a good algorithm for finding the minimum number of adjacent swaps for any array of 1's and 0's?
UPDATED:
The algorithm determines center by just getting an array of all indices of 1's. The center of that array will always hold the center index. Much faster.
oneIndices = array of indices of all 1's in the input
middleOfOnesIndices = round(oneIndices.length/2)-1 // index to the center index
minimumSwaps = 0
foreach index i of oneIndices
minimumSwaps += aboluteValue(oneIndices[middleOfOneIndices]-oneIndices[i])-absoluteValue(middleOfOneIndices-i);
Here's a fiddle to see it in action:
https://jsfiddle.net/3pmwrk0d/6/
This was a fun one. Thanks for the question.
Hi, firstly I would like to suggest that the minimum number of adjacent swaps would be 2 for your given example instead of 3. As just swap index 0 with index 2. So 1 swap from left and 1 swap from right.
Here is my way to find minimum of swaps to bring the array in consecutive 1's form -
Step 1 : First find the centre index for maximum number of consecutive 1's
Step 2 : Parse the left side of array to swap it and count the number of swap in a efficient manner(Do not swap unnecessarily)
Step 3 : Do the same for the right side array
Step 4 : Plus the counts of both side.
Please have a look at my java program based on same strategy :
`public class MinimumSwap
{
//function to find consecutive number index
public static int[] getMaxConsecutiveIndex(List<Integer> array)
{
int desiredIndex = -1;
int count = 0;
int dupDesiredIndex = -1;
int dupCount = 0;
int i = 0;
while(i < array.size())
{
if(array.get(i) == 0)
{
//pass duplcateIndex value to desiredIndex if count is more
if(dupCount > count)
{
desiredIndex = dupDesiredIndex;
count = dupCount;
}
dupDesiredIndex = -1;
dupCount = 0;
}
else
{
if(dupDesiredIndex == -1)
{
dupDesiredIndex = i;
dupCount = 1;
}
else
{
dupCount++;
}
}
i++;
}
return new int[]{desiredIndex,count};
}
public static int swapCount(List<Integer> array,int startIndex, int endIndex, boolean side)
{
// side == false means 0 at the left
// side == true means 1 at the left
System.out.println("startIndex "+startIndex+" endIndex "+endIndex+" side "+side);
int swapCount = 0;
if(side == false)
{
while(startIndex <= endIndex)
{
if(array.get(endIndex) == 0) // swap from the end only if it is 0
{
//check for first 1 from left to swap
while(array.get(startIndex) == 0 && (startIndex != endIndex))
startIndex++;
if(array.get(startIndex) == 1)
{
// now swap
int temp = array.get(startIndex);
array.set(startIndex, array.get(endIndex));
array.set(endIndex,temp);
swapCount++;
endIndex--;
}
}
endIndex--;
}
}
else
{
while(startIndex <= endIndex)
{
if(array.get(startIndex) == 0) // swap from the starting only if it is 0
{
//check for first 1 from right to swap
while(array.get(endIndex) == 0 && (startIndex != endIndex))
endIndex--;
if(array.get(endIndex) == 1)
{
// now swap
int temp = array.get(startIndex);
array.set(startIndex, array.get(endIndex));
array.set(endIndex,temp);
swapCount++;
startIndex++;
}
}
startIndex++;
}
}
return swapCount;
}
public static void main(String...strings)
{
List<Integer> arr = new ArrayList<Integer>();
int temp[] = {0,1,1,0,0,0,1,1,1,0,1,1,1,0,1,1,1,1,0,1};
//int temp[] = {1,0,0,1,1,0,1};
for(int i=0; i<temp.length; i++)
arr.add(temp[i]);
int centerIndex = getMaxConsecutiveIndex(arr)[0];
int consequtivecount = getMaxConsecutiveIndex(arr)[1];
System.out.println("centerIndex "+centerIndex+" consequtivecount "+consequtivecount);
int swapCountLeft = swapCount(arr,0, centerIndex-1, false);
int swapCountRight = swapCount(arr,centerIndex+consequtivecount, arr.size()-1, true);
System.out.println("total swap count "+swapCountLeft+" :: "+swapCountRight);
System.out.println("array after swapping "+arr);
}
}
`
I am not very sure about performance. But as per my knowledge it should not be inefficient. If anyone finds any performance issue please do let me know :)
Approach :
This can be done by finding number of zeroes to the right side of every 1 and add them. In order to sort the array every one always has to perform a swap operation with every zero on its right side.
So the total number of swap operations for a particular 1 in array is the number of zeroes on its right hand side. Find the number of zeroes on right side for every one i.e. the number of swaps and add them all to obtain the total number of swaps.
// Java code to find minimum number of swaps to sort a binary array
class MinimumNumberOfSwapsNeeded {
static int findMinSwaps(int arr[], int n)
{
// Array to store count of zeroes
int noOfZeroes[] = new int[n];
int i, count = 0;
// Count number of zeroes
// on right side of every one.
noOfZeroes[n - 1] = 1 - arr[n - 1];
for (i = n - 2; i >= 0; i--)
{
noOfZeroes[i] = noOfZeroes[i + 1];
if (arr[i] == 0)
noOfZeroes[i]++;
}
// Count total number of swaps by adding number
// of zeroes on right side of every one.
for (i = 0; i < n; i++)
{
if (arr[i] == 1)
count += noOfZeroes[i];
}
return count;
}
// Driver Code
public static void main(String args[])
{
int ar[] = { 0, 0, 1, 0, 1, 0, 1, 1 };
System.out.println(findMinSwaps(ar, ar.length));
}
}
**
Grouping the array of 0's and 1's such that minimum swaps can be calculated in O(2*n) ~ O(n) complexity.**
package com.segregate.array;
import java.util.ArrayList;
import java.util.List;
public class ArraySegregation {
public static void main(String[] args) {
List<Integer> arr = new ArrayList<>();
/*
*
* List -> low high [1 1 0 0 1 0] -> [ 000111] or [111000]
*
* 1 1 0 0 1 0 -> 000111
*/
arr.add(0);
arr.add(0);
arr.add(0);
arr.add(1);
arr.add(1);
arr.add(0);
arr.add(1);
arr.add(0);
arr.add(0);
List<Integer> arr1 = new ArrayList<>(arr);
int low = 0, high = arr.size() - 1;
int counter1 = 0, counter2 = 0;
// case for swaps such that all 0 in the left side.
while (low < high) {
switch (arr.get(low)) {
case 0:
while (arr.get(low) == 0)
low++;
break;
case 1:
while (arr.get(high) == 1)
high--;
swap(low, high, arr);
counter1++;
high--;
low++;
break;
}
}
// case for swaps such that all 0 in the right side.
/*
* [1 1 0 0 1 0] -> 11 1 0 0 0
*
*
*/
low=0;high = arr1.size() - 1;
while (low < high) {
switch (arr1.get(low)) {
case 0:
while (arr1.get(high) == 0)
high--;
swap(low, high, arr1);
counter2++;
high--;
low++;
break;
case 1:
while (arr1.get(low) == 1)
low++;
break;
}
}
int count = (counter1 > counter2) ? counter2 : counter1;
System.out.println(count);
}
private static void swap(int low, int high, List<Integer> arr) {
int temp1 = 0;
temp1 = arr.get(low);// 1
arr.remove(low);
arr.add(low, arr.get(high-1));
arr.remove(high-1);
arr.add(high, temp1);
}
}
Here is a simple, but not very clever algorithm that will perform an exhaustive search for any input in the range [0, 255].
Input:
binary string
Output:
optimal number of steps
number of optimal solutions
one detailed example
var transition = [],
isSolution = [];
function init() {
var msk = [ 3, 6, 12, 24, 48, 96, 192 ],
i, j, n, x, cnt, lsb, msb, sz = [];
for(i = 0; i < 0x100; i++) {
for(n = cnt = msb = 0, lsb = 8; n < 8; n++) {
if(i & (1 << n)) {
cnt++;
lsb = Math.min(lsb, n);
msb = Math.max(msb, n);
}
}
sz[i] = msb - lsb;
isSolution[i] = (sz[i] == cnt - 1);
}
for(i = 0; i < 0x100; i++) {
for(j = 0, transition[i] = []; j < 0x100; j++) {
x = i ^ j;
if(msk.indexOf(x) != -1 && (x & i) != x && (x & j) != x && sz[j] <= sz[i]) {
transition[i].push(j);
}
}
}
}
function solve() {
var x = parseInt(document.getElementById('bin').value, 2),
path = [ x ],
list = [],
i, min, sol = [], res = [];
recurse(x, path, list);
for(i in list) {
if(min === undefined || list[i].length <= min) {
min = list[i].length;
(sol[min] = (sol[min] || [])).push(list[i]);
}
}
console.log('Optimal length: ' + (min - 1) + ' step(s)');
console.log('Number of optimal solutions: ' + sol[min].length);
console.log('Example:');
for(i in sol[min][0]) {
res.push(('0000000' + sol[min][0][i].toString(2)).substr(-8, 8));
}
console.log(res.join(' -> '));
}
function recurse(x, path, list) {
if(isSolution[x]) {
list.push(path);
return;
}
for(i in transition[x]) {
if(path.indexOf(y = transition[x][i]) == -1) {
recurse(y, path.slice().concat(y), list);
}
}
}
init();
<input id="bin" maxlength="8" placeholder="enter binary string">
<button onclick="solve()">solve</button>

How to optimally divide an array into two subarrays so that sum of elements in both are same, otherwise give an error?

How to optimally divide an array into two subarrays so that sum of elements in both subarrays is same, otherwise give an error?
Example 1
Given the array
10, 20 , 30 , 5 , 40 , 50 , 40 , 15
It can be divided as
10, 20, 30, 5, 40
and
50, 40, 15
Each subarray sums up to 105.
Example 2
10, 20, 30, 5, 40, 50, 40, 10
The array cannot be divided into 2 arrays of an equal sum.
There exists a solution, which involves dynamic programming, that runs in O(n*TotalSum), where n is the number of elements in the array and TotalSum is their total sum.
The first part consists in calculating the set of all numbers that can be created by adding elements to the array.
For an array of size n, we will call this T(n),
T(n) = T(n-1) UNION { Array[n]+k | k is in T(n-1) }
(The proof of correctness is by induction, as in most cases of recursive functions.)
Also, remember for each cell in the dynamic matrix, the elements that were added in order to create it.
Simple complexity analysis will show that this is done in O(n*TotalSum).
After calculating T(n), search the set for an element exactly the size of TotalSum / 2.
If such an item exists, then the elements that created it, added together, equal TotalSum / 2, and the elements that were not part of its creation also equal TotalSum / 2 (TotalSum - TotalSum / 2 = TotalSum / 2).
This is a pseudo-polynomial solution. AFAIK, this problem is not known to be in P.
This is called partition problem. There are optimal solutions for some special cases. However, in general, it is an NP-complete problem.
In its common variant, this problem imposes 2 constraints and it can be done in an easier way.
If the partition can only be done somewhere along the length of the array (we do not consider elements out of order)
There are no negative numbers.
The algorithm that then works could be:
Have 2 variables, leftSum and rightSum
Start incrementing leftSum from the left, and rightSum from the right of the array.
Try to correct any imbalance in it.
The following code does the above:
public boolean canBalance(int[] nums) {
int leftSum = 0, rightSum = 0, i, j;
if(nums.length == 1)
return false;
for(i=0, j=nums.length-1; i<=j ;){
if(leftSum <= rightSum){
leftSum+=nums[i];
i++;
}else{
rightSum+=nums[j];
j--;
}
}
return (rightSum == leftSum);
}
The output:
canBalance({1, 1, 1, 2, 1}) → true OK
canBalance({2, 1, 1, 2, 1}) → false OK
canBalance({10, 10}) → true OK
canBalance({1, 1, 1, 1, 4}) → true OK
canBalance({2, 1, 1, 1, 4}) → false OK
canBalance({2, 3, 4, 1, 2}) → false OK
canBalance({1, 2, 3, 1, 0, 2, 3}) → true OK
canBalance({1, 2, 3, 1, 0, 1, 3}) → false OK
canBalance({1}) → false OK
canBalance({1, 1, 1, 2, 1}) → true OK
Ofcourse, if the elements can be combined out-of-order, it does turn into the partition problem with all its complexity.
a=[int(g) for g in input().split()] #for taking the array as input in a
single line
leftsum=0
n=len(a)
for i in range(n):
leftsum+=a[i] #calculates the sum of first subarray
rightsum=0
for j in range(i+1):
rightsum+=a[j] #calculates the sum of other subarray
if leftsum==rightsum:
pos=i+1 #if the sum of subarrays are equal,
break set position where the condition
gets satisfied and exit the loop
else:
pos=-1 #if the sum of subarrays is not
equal, set position to -1
if pos=-1 or pos=n:
print('It is not possible.')
else: #printing the sub arrays`
for k in range(n):
if pos=k:
print('')
print(str(a[k]),end='')
This Problem says that if an array can have two subarrays with their sum of elements as same.
So a boolean value should be returned.
I have found an efficient algorithm :
Algo: Procedure
Step 1: Take an empty array as a container , sort the initial array and keep in the empty one.
Step 2: now take two dynamically allocatable arrays and take out highest and 2nd highest from the auxilliary array and keep it in the two subarrays respectively , and delete from the auxiliary array.
Step 3: Compare the sum of elements in the subarrays , the smaller sum one will have chance to fetch highest remaining element in the array and then delete from the container.
Step 4: Loop thru Step 3 until the container is empty.
Step 5: Compare the sum of two subarrays , if they are same return true else false.
// The complexity with this problem is that there may be many combinations possible but this algo has one unique way .
Tried a different solution . other than Wiki solutions (Partition Problem).
static void subSet(int array[]) {
System.out.println("Input elements :" + Arrays.toString(array));
int sum = 0;
for (int element : array) {
sum = sum + element;
}
if (sum % 2 == 1) {
System.out.println("Invalid Pair");
return;
}
Arrays.sort(array);
System.out.println("Sorted elements :" + Arrays.toString(array));
int subSum = sum / 2;
int[] subSet = new int[array.length];
int tmpSum = 0;
boolean isFastpath = true;
int lastStopIndex = 0;
for (int j = array.length - 1; j >= 0; j--) {
tmpSum = tmpSum + array[j];
if (tmpSum == subSum) { // if Match found
if (isFastpath) { // if no skip required and straight forward
// method
System.out.println("Found SubSets 0..." + (j - 1) + " and "
+ j + "..." + (array.length - 1));
} else {
subSet[j] = array[j];
array[j] = 0;
System.out.println("Found..");
System.out.println("Set 1" + Arrays.toString(subSet));
System.out.println("Set 2" + Arrays.toString(array));
}
return;
} else {
// Either the tmpSum greater than subSum or less .
// if less , just look for next item
if (tmpSum < subSum && ((subSum - tmpSum) >= array[0])) {
if (lastStopIndex > j && subSet[lastStopIndex] == 0) {
subSet[lastStopIndex] = array[lastStopIndex];
array[lastStopIndex] = 0;
}
lastStopIndex = j;
continue;
}
isFastpath = false;
if (subSet[lastStopIndex] == 0) {
subSet[lastStopIndex] = array[lastStopIndex];
array[lastStopIndex] = 0;
}
tmpSum = tmpSum - array[j];
}
}
}
I have tested. ( It works well with positive number greater than 0) please let me know if any one face issue.
This is a recursive solution to the problem, one non recursive solution could use a helper method to get the sum of indexes 0 to a current index in a for loop and another one could get the sum of all the elements from the same current index to the end, which works. Now if you wanted to get the elements into an array and compare the sum, first find the point (index) which marks the spilt where both side's sum are equal, then get a list and add the values before that index and another list to go after that index.
Here's mine (recursion), which only determines if there is a place to split the array so that the sum of the numbers on one side is equal to the sum of the numbers on the other side. Worry about indexOutOfBounds, which can easily happen in recursion, a slight mistake could prove fatal and yield a lot of exceptions and errors.
public boolean canBalance(int[] nums) {
return (nums.length <= 1) ? false : canBalanceRecur(nums, 0);
}
public boolean canBalanceRecur(int[] nums, int index){ //recursive version
if(index == nums.length - 1 && recurSumBeforeIndex(nums, 0, index)
!= sumAfterIndex(nums, index)){ //if we get here and its still bad
return false;
}
if(recurSumBeforeIndex(nums, 0, index + 1) == sumAfterIndex(nums, index + 1)){
return true;
}
return canBalanceRecur(nums, index + 1); //move the index up
}
public int recurSumBeforeIndex(int[] nums, int start, int index){
return (start == index - 1 && start < nums.length)
? nums[start]
: nums[start] + recurSumBeforeIndex(nums, start + 1, index);
}
public int sumAfterIndex(int[] nums, int startIndex){
return (startIndex == nums.length - 1)
? nums[nums.length - 1]
: nums[startIndex] + sumAfterIndex(nums, startIndex + 1);
}
Found solution here
package sort;
import java.util.ArrayList;
import java.util.List;
public class ArraySumSplit {
public static void main (String[] args) throws Exception {
int arr[] = {1 , 2 , 3 , 4 , 5 , 5, 1, 1, 3, 2, 1};
split(arr);
}
static void split(int[] array) throws Exception {
int sum = 0;
for(int n : array) sum += n;
if(sum % 2 == 1) throw new Exception(); //impossible to split evenly
List<Integer> firstPart = new ArrayList<Integer>();
List<Integer> secondPart = new ArrayList<Integer>();
if(!dfs(0, sum / 2, array, firstPart, secondPart)) throw new Exception(); // impossible to split evenly;
//firstPart and secondPart have the grouped elements, print or return them if necessary.
System.out.print(firstPart.toString());
int sum1 = 0;
for (Integer val : firstPart) {
sum1 += val;
}
System.out.println(" = " + sum1);
System.out.print(secondPart.toString());
int sum2 = 0;
for (Integer val : secondPart) {
sum2 += val;
}
System.out.println(" = " + sum2);
}
static boolean dfs(int i, int limit, int[] array, List<Integer> firstPart, List<Integer> secondPart) {
if( limit == 0) {
for(int j = i; j < array.length; j++) {
secondPart.add(array[j]);
}
return true;
}
if(limit < 0 || i == array.length) {
return false;
}
firstPart.add(array[i]);
if(dfs(i + 1, limit - array[i], array, firstPart, secondPart)) return true;
firstPart.remove(firstPart.size() - 1);
secondPart.add(array[i]);
if(dfs(i + 1, limit, array, firstPart, secondPart)) return true;
secondPart.remove(secondPart.size() - 1);
return false;
}
}
def listSegmentation(theList):
newList = [[],[]]
print(theList)
wt1 = 0
wt2 = 0
dWt = 0
for idx in range(len(theList)):
wt = theList[idx]
if (wt > (wt1 + wt2) and wt1 > 0 and wt2 > 0):
newList[0] = newList[0] + newList[1]
newList[1] = []
newList[1].append(wt)
wt1 += wt2
wt2 = wt
elif ((wt2 + wt) >= (wt1 + wt)):
wt1 += wt
newList[0].append(wt)
elif ((wt2 + wt) < (wt1 + wt)):
wt2 += wt
newList[1].append(wt)
#Balancing
if(wt1 > wt2):
wtDiff = sum(newList[0]) - sum(newList[1])
ls1 = list(filter(lambda x: x <= wtDiff, newList[0]))
ls2 = list(filter(lambda x: x <= (wtDiff/2) , newList[1]))
while len(ls1) > 0 or len(ls2) > 0:
if len(ls1) > 0:
elDif1 = max(ls1)
newList[0].remove(elDif1)
newList[1].append(elDif1)
if len(ls2) > 0:
elDif2 = max(ls2)
newList[0].append(elDif2)
newList[1].remove(elDif2)
wtDiff = sum(newList[0]) - sum(newList[1])
ls1 = list(filter(lambda x: x <= wtDiff, newList[0]))
ls2 = list(filter(lambda x: x <= (wtDiff/2) , newList[1]))
if(wt2 > wt1):
wtDiff = sum(newList[1]) - sum(newList[0])
ls2 = list(filter(lambda x: x <= wtDiff, newList[1]))
ls1 = list(filter(lambda x: x <= (wtDiff/2) , newList[0]))
while len(ls1) > 0 or len(ls2) > 0:
if len(ls1) > 0:
elDif1 = max(ls1)
newList[0].remove(elDif1)
newList[1].append(elDif1)
if len(ls2) > 0:
elDif2 = max(ls2)
newList[0].append(elDif2)
newList[1].remove(elDif2)
wtDiff = sum(newList[1]) - sum(newList[0])
ls2 = list(filter(lambda x: x <= wtDiff, newList[1]))
ls1 = list(filter(lambda x: x <= (wtDiff/2) , newList[0]))
print(ls1, ls2)
print(sum(newList[0]),sum(newList[1]))
return newList
#Test cases
lst1 = [4,9,8,3,11,6,13,7,2,25,28,60,19,196]
lst2 = [7,16,5,11,4,9,15,2,1,13]
lst3 = [8,17,14,9,3,5,19,11,4,6,2]
print(listSegmentation(lst1))
print(listSegmentation(lst2))
print(listSegmentation(lst3))
This Python3 function will split and balance a list of numbers to two separate lists equal in sum, if the sum is even.
Python3 solution:
def can_partition(a):
mylist1 = []
mylist2 = []
sum1 = 0
sum2 = 0
for items in a:
# Take total and divide by 2.
total = sum(a)
if total % 2 == 0:
half = total//2
else:
return("Exiting, sum has fractions, total %s half %s" % (total, total/2))
mylist1.append(items)
print('Total is %s and half is %s' %(total, total/2))
for i in a:
sum1 = sum(mylist1)
sum2 = sum(mylist2)
if sum2 < half:
mypop = mylist1.pop(0)
mylist2.append(mypop)
# Function to swtich numbers between the lists if sums are uneven.
def switchNumbers(list1, list2,switch_diff):
for val in list1:
if val == switch_diff:
val_index = list1.index(val)
new_pop = list1.pop(val_index)
list2.append(new_pop)
#Count so while do not get out of hand
count = len(a)
while count != 0:
sum1 = sum(mylist1)
sum2 = sum(mylist2)
if sum1 > sum2:
diff = sum1 -half
switchNumbers(mylist1, mylist2, diff)
count -= 1
elif sum2 > sum1:
diff = sum2 - half
switchNumbers(mylist2, mylist1, diff)
count -= 1
else:
if sum1 == sum2:
print('Values of half, sum1, sum2 are:',half, sum1,sum2)
break
count -= 1
return (mylist1, mylist2)
b = [ 2, 3, 4, 2, 3, 1, 2, 5, 4, 4, 2, 2, 3, 3, 2 ]
can_partition(b)
Output:
Total is 42 total, half is 21.0
Values of half, sum1 & sum2 are : 21 21 21
([4, 4, 2, 2, 3, 3, 2, 1], [2, 3, 4, 2, 3, 2, 5])
A non optimal solution in python,
from itertools import permutations
def get_splitted_array(a):
for perm in permutations(a):
l1 = len(perm)
for i in range(1, l1):
if sum(perm[0:i]) == sum(perm[i:l1]):
return perm[0:i], perm[i:l1]
>>> a = [6,1,3,8]
>>> get_splitted_array(a)
((6, 3), (1, 8))
>>> a = [5,9,20,1,5]
>>>
>>> get_splitted_array(a)
((5, 9, 1, 5), (20,))
>>>
Its O(n) time and O(n) space
def equal_subarr(arr):
n=len(arr)
post_sum = [0] * (n- 1) + [arr[-1]]
for i in range(n - 2, -1, -1):
post_sum[i] = arr[i] + post_sum[i + 1]
prefix_sum = [arr[0]] + [0] * (n - 1)
for i in range(1, n):
prefix_sum[i] = prefix_sum[i - 1] + arr[i]
for i in range(n - 1):
if prefix_sum[i] == post_sum[i + 1]:
return [arr[:i+1],arr[i+1:]]
return -1
arr=[10, 20 , 30 , 5 , 40 , 50 , 40 , 15]
print(equal_subarr(arr))
>>> [[10, 20, 30, 5, 40], [50, 40, 15]]
arr=[10, 20, 30, 5, 40, 50, 40, 10]
print(equal_subarr(arr))
>>> -1
First, if the elements are integers, check that the total is evenly divisible by two- if it isn't success isn't possible.
I would set up the problem as a binary tree, with level 0 deciding which set element 0 goes into, level 1 deciding which set element 1 goes into, etc. At any time if the sum of one set is half the total, you're done- success. At any time if the sum of one set is more than half the total, that sub-tree is a failure and you have to back up. At that point it is a tree traversal problem.
public class Problem1 {
public static void main(String[] args) throws IOException{
Scanner scanner=new Scanner(System.in);
ArrayList<Integer> array=new ArrayList<Integer>();
int cases;
System.out.println("Enter the test cases");
cases=scanner.nextInt();
for(int i=0;i<cases;i++){
int size;
size=scanner.nextInt();
System.out.println("Enter the Initial array size : ");
for(int j=0;j<size;j++){
System.out.println("Enter elements in the array");
int element;
element=scanner.nextInt();
array.add(element);
}
}
if(validate(array)){
System.out.println("Array can be Partitioned");}
else{
System.out.println("Error");}
}
public static boolean validate(ArrayList<Integer> array){
boolean flag=false;
Collections.sort(array);
System.out.println(array);
int index=array.size();
ArrayList<Integer> sub1=new ArrayList<Integer>();
ArrayList<Integer> sub2=new ArrayList<Integer>();
sub1.add(array.get(index-1));
array.remove(index-1);
index=array.size();
sub2.add(array.get(index-1));
array.remove(index-1);
while(!array.isEmpty()){
if(compareSum(sub1,sub2)){
index=array.size();
sub2.add(array.get(index-1));
array.remove(index-1);
}
else{
index=array.size();
sub1.add(array.get(index-1));
array.remove(index-1);
}
}
if(sumOfArray(sub1).equals(sumOfArray(sub2)))
flag=true;
else
flag=false;
return flag;
}
public static Integer sumOfArray(ArrayList<Integer> array){
Iterator<Integer> it=array.iterator();
Integer sum=0;
while(it.hasNext()){
sum +=it.next();
}
return sum;
}
public static boolean compareSum(ArrayList<Integer> sub1,ArrayList<Integer> sub2){
boolean flag=false;
int sum1=sumOfArray(sub1);
int sum2=sumOfArray(sub2);
if(sum1>sum2)
flag=true;
else
flag=false;
return flag;
}
}
// The Greedy approach //
I was asked this question in an interview, and I gave below simple solution, as I had NOT seen this problem in any websiteS earlier.
Lets say Array A = {45,10,10,10,10,5}
Then, the split will be at index = 1 (0-based index) so that we have two equal sum set {45} and {10,10,10,10,5}
int leftSum = A[0], rightSum = A[A.length - 1];
int currentLeftIndex = 0; currentRightIndex = A.length - 1
/*
Move the two index pointers towards mid of the array untill currentRightIndex != currentLeftIndex. Increase leftIndex if sum of left elements is still less than or equal to sum of elements in right of 'rightIndex'.At the end,check if leftSum == rightSum. If true, we got the index as currentLeftIndex+1(or simply currentRightIndex, as currentRightIndex will be equal to currentLeftIndex+1 in this case).
*/
while (currentLeftIndex < currentRightIndex)
{
if ( currentLeftIndex+1 != currentRightIndex && (leftSum + A[currentLeftIndex + 1) <=currentRightSum )
{
currentLeftIndex ++;
leftSum = leftSum + A[currentLeftIndex];
}
if ( currentRightIndex - 1 != currentLeftIndex && (rightSum + A[currentRightIndex - 1] <= currentLeftSum)
{
currentRightIndex --;
rightSum = rightSum + A[currentRightIndex];
}
}
if (CurrentLeftIndex == currentRightIndex - 1 && leftSum == rightSum)
PRINT("got split point at index "+currentRightIndex);
#Gal Subset-Sum problem is NP-Complete and has a O(n*TotalSum) pseudo-polynomial Dynamic Programming algorithm. But this problem is not NP-Complete. This is a special case and in fact this can be solved in linear time.
Here we are looking for an index where we can split the array into two parts with same sum.
Check following code.
Analysis: O(n), as the algorithm only iterates through the array and does not use TotalSum.
public class EqualSumSplit {
public static int solution( int[] A ) {
int[] B = new int[A.length];
int[] C = new int[A.length];
int sum = 0;
for (int i=0; i< A.length; i++) {
sum += A[i];
B[i] = sum;
// System.out.print(B[i]+" ");
}
// System.out.println();
sum = 0;
for (int i=A.length-1; i>=0; i--) {
sum += A[i];
C[i] = sum;
// System.out.print(C[i]+" ");
}
// System.out.println();
for (int i=0; i< A.length-1; i++) {
if (B[i] == C[i+1]) {
System.out.println(i+" "+B[i]);
return i;
}
}
return -1;
}
public static void main(String args[] ) {
int[] A = {-7, 1, 2, 3, -4, 3, 0};
int[] B = {10, 20 , 30 , 5 , 40 , 50 , 40 , 15};
solution(A);
solution(B);
}
}
Algorithm:
Step 1) Split the array into two
Step 2) If the sum is equal, split is complete
Step 3) Swap one element from array1 with array2, guided by the four rules:
IF the sum of elements in array1 is less than sum of elements in array2
Rule1:
Find a number in array1 that is smaller than a number in array2 in such a way that swapping of
these elements, do not increase the sum of array1 beyond the expected sum. If found, swap the
elements and return.
Rule2:
If Rule1 is not is not satisfied, Find a number in array1 that is bigger than a number in array2 in
such a way that the difference between any two numbers in array1 and array2 is not smaller than
the difference between these two numbers.
ELSE
Rule3:
Find a number in array1 that is bigger than a number in array2 in such a way that swapping these
elements, do not decrease the sum of array1 beyond the expected sum. If found, swap the elements and return.
Rule4:
If Rule3 is not is not satisfied, Find a number in array1 that is smaller than a number in array2 in
such a way that the difference between any two numbers in array1 and array2 is not smaller than
the difference between these two numbers.
Step 5) Go to Step2 until the swap results in an array with the same set of elements encountered already
Setp 6) If a repetition occurs, this array cannot be split into two halves with equal sum. The current set of arrays OR the set that was formed just before this repetition should be the best split of the array.
Note: The approach taken is to swap element from one array to another in such a way that the resultant sum is as close to the expected sum.
The java program is available at Java Code
Please try this and let me know if not working. Hope it will helps you.
static ArrayList<Integer> array = null;
public static void main(String[] args) throws IOException {
ArrayList<Integer> inputArray = getinputArray();
System.out.println("inputArray is " + inputArray);
Collections.sort(inputArray);
int totalSum = 0;
Iterator<Integer> inputArrayIterator = inputArray.iterator();
while (inputArrayIterator.hasNext()) {
totalSum = totalSum + inputArrayIterator.next();
}
if (totalSum % 2 != 0) {
System.out.println("Not Possible");
return;
}
int leftSum = inputArray.get(0);
int rightSum = inputArray.get(inputArray.size() - 1);
int currentLeftIndex = 0;
int currentRightIndex = inputArray.size() - 1;
while (leftSum <= (totalSum / 2)) {
if ((currentLeftIndex + 1 != currentRightIndex)
&& leftSum != (totalSum / 2)) {
currentLeftIndex++;
leftSum = leftSum + inputArray.get(currentLeftIndex);
} else
break;
}
if (leftSum == (totalSum / 2)) {
ArrayList<Integer> splitleft = new ArrayList<Integer>();
ArrayList<Integer> splitright = new ArrayList<Integer>();
for (int i = 0; i <= currentLeftIndex; i++) {
splitleft.add(inputArray.get(i));
}
for (int i = currentLeftIndex + 1; i < inputArray.size(); i++) {
splitright.add(inputArray.get(i));
}
System.out.println("splitleft is :" + splitleft);
System.out.println("splitright is :" + splitright);
}
else
System.out.println("Not possible");
}
public static ArrayList<Integer> getinputArray() {
Scanner scanner = new Scanner(System.in);
array = new ArrayList<Integer>();
int size;
System.out.println("Enter the Initial array size : ");
size = scanner.nextInt();
System.out.println("Enter elements in the array");
for (int j = 0; j < size; j++) {
int element;
element = scanner.nextInt();
array.add(element);
}
return array;
}
}
public boolean splitBetween(int[] x){
int sum=0;
int sum1=0;
if (x.length==1){
System.out.println("Not a valid value");
}
for (int i=0;i<x.length;i++){
sum=sum+x[i];
System.out.println(sum);
for (int j=i+1;j<x.length;j++){
sum1=sum1+x[j];
System.out.println("SUm1:"+sum1);
}
if(sum==sum1){
System.out.println("split possible");
System.out.println("Sum: " +sum +" Sum1:" + sum1);
return true;
}else{
System.out.println("Split not possible");
}
sum1=0;
}
return false;
}
package PACKAGE1;
import java.io.*;
import java.util.Arrays;
public class programToSplitAnArray {
public static void main(String args[]) throws NumberFormatException,
IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("enter the no. of elements to enter");
int n = Integer.parseInt(br.readLine());
int x[] = new int[n];
int half;
for (int i = 0; i < n; i++) {
x[i] = Integer.parseInt(br.readLine());
}
int sum = 0;
for (int i = 0; i < n; i++) {
sum = sum + x[i];
}
if (sum % 2 != 0) {
System.out.println("the sum is odd and cannot be divided");
System.out.println("The sum is " + sum);
}
else {
boolean div = false;
half = sum / 2;
int sum1 = 0;
for (int i = 0; i < n; i++) {
sum1 = sum1 + x[i];
if (sum1 == half) {
System.out.println("array can be divided");
div = true;
break;
}
}
if (div == true) {
int t = 0;
int[] array1 = new int[n];
int count = 0;
for (int i = 0; i < n; i++) {
t = t + x[i];
if (t <= half) {
array1[i] = x[i];
count++;
}
}
array1 = Arrays.copyOf(array1, count);
int array2[] = new int[n - count];
int k = 0;
for (int i = count; i < n; i++) {
array2[k] = x[i];
k++;
}
System.out.println("The first array is ");
for (int m : array1) {
System.out.println(m);
}
System.out.println("The second array is ");
for (int m : array2) {
System.out.println(m);
}
} else {
System.out.println("array cannot be divided");
}
}
}
}
A BAD greedy heuristic to solve this problem: try sorting the list from least to greatest, and split that list into two by having list1 = the odd elements, and list2 = the even elements.
very simple solution with recursion
public boolean splitArray(int[] nums){
return arrCheck(0, nums, 0);
}
public boolean arrCheck(int start, int[] nums, int tot){
if(start >= nums.length) return tot == 0;
if(arrCheck(start+1, nums, tot+nums[start])) return true;
if(arrCheck(start+1, nums, tot-nums[start])) return true;
return false;
}
https://github.com/ShubhamAgrahari/DRjj/blob/master/Subarray_Sum.java
package solution;
import java.util.Scanner;
public class Solution {
static int SplitPoint(int arr[], int n)
{
int leftSum = 0;
for (int i = 0 ; i < n ; i++)
leftSum += arr[i];
int rightSum = 0;
for (int i = n-1; i >= 0; i--)
{
rightSum += arr[i];
leftSum -= arr[i] ;
if (rightSum == leftSum)
return i ;
}
return -1;
}
static void output(int arr[], int n)
{
int s = SplitPoint(arr, n);
if (s == -1 || s == n )
{
System.out.println("Not Possible" );
return;
}
for (int i = 0; i < n; i++)
{
if(s == i)
System.out.println();
System.out.print(arr[i] + " ");
}
}
public static void main (String[] args) {
Scanner sc= new Scanner(System.in);
System.out.println("Enter Array Size");
int n = sc.nextInt();
int arr[]= new int[n];
for(int i=0;i<n;i++)
{
arr[i]=sc.nextInt();
}
output(arr, n);
} }

Identify the index corresponding to the smallest data in a set of arrays

This is a trivial algorithmic question, I believe, but I don't seem to be able to find an efficient and elegant solution.
We have 3 arrays of int (Aa, Ab, Ac) and 3 cursors (Ca, Cb, Cc) that indicate an index in the corresponding array. I want to identify and increment the cursor pointing to the smallest value. If this cursor is already at the end of the array, I will exclude it and increment the cursor pointing to the second smallest value. If there is only 1 cursor that is not at the end of the array, we increment this one.
The only solutions I can come up are complicated and/or not optimal. For example, I always end up with a huge if...else...
Does anyone see a neat solution to this problem ?
I am programming in C++ but feel free to discuss it in pseudo-code or any language you like.
Thank you
Pseudo-java code:
int[] values = new int[3];
values[0] = aa[ca];
values[1] = ab[cb];
values[2] = ac[cc];
Arrays.sort(values);
boolean done = false;
for (int i = 0; i < 3 && !done; i++) {
if (values[i] == aa[ca] && ca + 1 < aa.length) {
ca++;
done = true;
}
else if (values[i] == ab[cb] && cb + 1 < ab.length) {
cb++;
done = true;
}
else if (cc + 1 < ac.length) {
cc++;
done = true;
}
}
if (!done) {
System.out.println("cannot increment any index");
stop = true;
}
Essentially, it does the following:
initialize an array values with aa[ca], ab[cb] and ac[cc]
sort values
scan values and increment if possible (i.e. not already at the end of the array) the index of the corresponding value
I know, sorting is at best O(n lg n), but I'm only sorting an array of 3 elements.
what about this solution:
if (Ca != arraySize - 1) AND
((Aa[Ca] == min(Aa[Ca], Ab[Cb], Ac[Cc]) OR
(Aa[Ca] == min(Aa[Ca], Ab[Cb]) And Cc == arraySize - 1) OR
(Aa[Ca] == min(Aa[Ca], Ac[Cc]) And Cb == arraySize - 1) OR
(Cc == arraySize - 1 And Cb == arraySize - 1))
{
Ca++;
}
else if (Cb != arraySize - 1) AND
((Ab[Cb] == min(Ab[Cb], Ac[Cc]) OR (Cc == arraySize - 1))
{
Cb++;
}
else if (Cc != arraySize - 1)
{
Cc++;
}
Pseudo code: EDIT : tidied it up a bit
class CursoredArray
{
int index;
std::vector<int> array;
int val()
{
return array[index];
}
bool moveNext()
{
bool ret = true;
if( array.size() > index )
++index;
else
ret = false;
return ret;
}
}
std::vector<CursoredArray> arrays;
std::vector<int> order = { 0, 1, 2 };//have a default order to start with
if( arrays[0].val() > arrays[1].val() )
std::swap( order[0], order [1] );
if( arrays[2].val() < arrays[order[1]].val() )//if the third is less than the largest of the others
{
std::swap( order[1], order [2] );
if( arrays[2].val() < arrays[order[0]].val() )//if the third is less than the smallest of the others
std::swap( order[0], order [1] );
}
//else third pos of order is already correct
bool end = true;
for( i = 0; i < 3; ++i )
{
if( arrays[order[i]].MoveNext() )
{
end = false;
break;
}
}
if( end )//have gone through all the arrays

Find the 2nd largest element in an array with minimum number of comparisons

For an array of size N, what is the number of comparisons required?
The optimal algorithm uses n+log n-2 comparisons. Think of elements as competitors, and a tournament is going to rank them.
First, compare the elements, as in the tree
|
/ \
| |
/ \ / \
x x x x
this takes n-1 comparisons and each element is involved in comparison at most log n times. You will find the largest element as the winner.
The second largest element must have lost a match to the winner (he can't lose a match to a different element), so he's one of the log n elements the winner has played against. You can find which of them using log n - 1 comparisons.
The optimality is proved via adversary argument. See https://math.stackexchange.com/questions/1601 or http://compgeom.cs.uiuc.edu/~jeffe/teaching/497/02-selection.pdf or http://www.imada.sdu.dk/~jbj/DM19/lb06.pdf or https://www.utdallas.edu/~chandra/documents/6363/lbd.pdf
You can find the second largest value with at most 2·(N-1) comparisons and two variables that hold the largest and second largest value:
largest := numbers[0];
secondLargest := null
for i=1 to numbers.length-1 do
number := numbers[i];
if number > largest then
secondLargest := largest;
largest := number;
else
if number > secondLargest then
secondLargest := number;
end;
end;
end;
Use Bubble sort or Selection sort algorithm which sorts the array in descending order. Don't sort the array completely. Just two passes. First pass gives the largest element and second pass will give you the second largest element.
No. of comparisons for first pass: n-1
No. of comparisons for second pass: n-2
Total no. of comparison for finding second largest: 2n-3
May be you can generalize this algorithm. If you need the 3rd largest then you make 3 passes.
By above strategy you don't need any temporary variables as Bubble sort and Selection sort are in place sorting algorithms.
Here is some code that might not be optimal but at least actually finds the 2nd largest element:
if( val[ 0 ] > val[ 1 ] )
{
largest = val[ 0 ]
secondLargest = val[ 1 ];
}
else
{
largest = val[ 1 ]
secondLargest = val[ 0 ];
}
for( i = 2; i < N; ++i )
{
if( val[ i ] > secondLargest )
{
if( val[ i ] > largest )
{
secondLargest = largest;
largest = val[ i ];
}
else
{
secondLargest = val[ i ];
}
}
}
It needs at least N-1 comparisons if the largest 2 elements are at the beginning of the array and at most 2N-3 in the worst case (one of the first 2 elements is the smallest in the array).
case 1-->9 8 7 6 5 4 3 2 1
case 2--> 50 10 8 25 ........
case 3--> 50 50 10 8 25.........
case 4--> 50 50 10 8 50 25.......
public void second element()
{
int a[10],i,max1,max2;
max1=a[0],max2=a[1];
for(i=1;i<a.length();i++)
{
if(a[i]>max1)
{
max2=max1;
max1=a[i];
}
else if(a[i]>max2 &&a[i]!=max1)
max2=a[i];
else if(max1==max2)
max2=a[i];
}
}
Sorry, JS code...
Tested with the two inputs:
a = [55,11,66,77,72];
a = [ 0, 12, 13, 4, 5, 32, 8 ];
var first = Number.MIN_VALUE;
var second = Number.MIN_VALUE;
for (var i = -1, len = a.length; ++i < len;) {
var dist = a[i];
// get the largest 2
if (dist > first) {
second = first;
first = dist;
} else if (dist > second) { // && dist < first) { // this is actually not needed, I believe
second = dist;
}
}
console.log('largest, second largest',first,second);
largest, second largest 32 13
This should have a maximum of a.length*2 comparisons and only goes through the list once.
I know this is an old question, but here is my attempt at solving it, making use of the Tournament Algorithm. It is similar to the solution used by #sdcvvc , but I am using two-dimensional array to store elements.
To make things work, there are two assumptions:
1) number of elements in the array is the power of 2
2) there are no duplicates in the array
The whole process consists of two steps:
1. building a 2D array by comparing two by two elements. First row in the 2D array is gonna be the entire input array. Next row contains results of the comparisons of the previous row. We continue comparisons on the newly built array and keep building the 2D array until an array of only one element (the largest one) is reached.
2. we have a 2D-array where last row contains only one element: the largest one. We continue going from the bottom to the top, in each array finding the element that was "beaten" by the largest and comparing it to the current "second largest" value. To find the element beaten by the largest, and to avoid O(n) comparisons, we must store the index of the largest element in the previous row. That way we can easily check the adjacent elements. At any level (above root level),the adjacent elements are obtained as:
leftAdjacent = rootIndex*2
rightAdjacent = rootIndex*2+1,
where rootIndex is index of the largest(root) element at the previous level.
I know the question asks for C++, but here is my attempt at solving it in Java. (I've used lists instead of arrays, to avoid messy changing of the array size and/or unnecessary array size calculations)
public static Integer findSecondLargest(List<Integer> list) {
if (list == null) {
return null;
}
if (list.size() == 1) {
return list.get(0);
}
List<List<Integer>> structure = buildUpStructure(list);
System.out.println(structure);
return secondLargest(structure);
}
public static List<List<Integer>> buildUpStructure(List<Integer> list) {
List<List<Integer>> newList = new ArrayList<List<Integer>>();
List<Integer> tmpList = new ArrayList<Integer>(list);
newList.add(tmpList);
int n = list.size();
while (n>1) {
tmpList = new ArrayList<Integer>();
for (int i = 0; i<n; i=i+2) {
Integer i1 = list.get(i);
Integer i2 = list.get(i+1);
tmpList.add(Math.max(i1, i2));
}
n/= 2;
newList.add(tmpList);
list = tmpList;
}
return newList;
}
public static Integer secondLargest(List<List<Integer>> structure) {
int n = structure.size();
int rootIndex = 0;
Integer largest = structure.get(n-1).get(rootIndex);
List<Integer> tmpList = structure.get(n-2);
Integer secondLargest = Integer.MIN_VALUE;
Integer leftAdjacent = -1;
Integer rightAdjacent = -1;
for (int i = n-2; i>=0; i--) {
rootIndex*=2;
tmpList = structure.get(i);
leftAdjacent = tmpList.get(rootIndex);
rightAdjacent = tmpList.get(rootIndex+1);
if (leftAdjacent.equals(largest)) {
if (rightAdjacent > secondLargest) {
secondLargest = rightAdjacent;
}
}
if (rightAdjacent.equals(largest)) {
if (leftAdjacent > secondLargest) {
secondLargest = leftAdjacent;
}
rootIndex=rootIndex+1;
}
}
return secondLargest;
}
Suppose provided array is inPutArray = [1,2,5,8,7,3] expected O/P -> 7 (second largest)
take temp array
temp = [0,0], int dummmy=0;
for (no in inPutArray) {
if(temp[1]<no)
temp[1] = no
if(temp[0]<temp[1]){
dummmy = temp[0]
temp[0] = temp[1]
temp[1] = temp
}
}
print("Second largest no is %d",temp[1])
PHP version of the Gumbo algorithm: http://sandbox.onlinephpfunctions.com/code/51e1b05dac2e648fd13e0b60f44a2abe1e4a8689
$numbers = [10, 9, 2, 3, 4, 5, 6, 7];
$largest = $numbers[0];
$secondLargest = null;
for ($i=1; $i < count($numbers); $i++) {
$number = $numbers[$i];
if ($number > $largest) {
$secondLargest = $largest;
$largest = $number;
} else if ($number > $secondLargest) {
$secondLargest = $number;
}
}
echo "largest=$largest, secondLargest=$secondLargest";
Assuming space is irrelevant, this is the smallest I could get it. It requires 2*n comparisons in worst case, and n comparisons in best case:
arr = [ 0, 12, 13, 4, 5, 32, 8 ]
max = [ -1, -1 ]
for i in range(len(arr)):
if( arr[i] > max[0] ):
max.insert(0,arr[i])
elif( arr[i] > max[1] ):
max.insert(1,arr[i])
print max[1]
try this.
max1 = a[0].
max2.
for i = 0, until length:
if a[i] > max:
max2 = max1.
max1 = a[i].
#end IF
#end FOR
return min2.
it should work like a charm. low in complexity.
here is a java code.
int secondlLargestValue(int[] secondMax){
int max1 = secondMax[0]; // assign the first element of the array, no matter what, sorted or not.
int max2 = 0; // anything really work, but zero is just fundamental.
for(int n = 0; n < secondMax.length; n++){ // start at zero, end when larger than length, grow by 1.
if(secondMax[n] > max1){ // nth element of the array is larger than max1, if so.
max2 = max1; // largest in now second largest,
max1 = secondMax[n]; // and this nth element is now max.
}//end IF
}//end FOR
return max2;
}//end secondLargestValue()
Use counting sort and then find the second largest element, starting from index 0 towards the end. There should be at least 1 comparison, at most n-1 (when there's only one element!).
#include<stdio.h>
main()
{
int a[5] = {55,11,66,77,72};
int max,min,i;
int smax,smin;
max = min = a[0];
smax = smin = a[0];
for(i=0;i<=4;i++)
{
if(a[i]>max)
{
smax = max;
max = a[i];
}
if(max>a[i]&&smax<a[i])
{
smax = a[i];
}
}
printf("the first max element z %d\n",max);
printf("the second max element z %d\n",smax);
}
The accepted solution by sdcvvc in C++11.
#include <algorithm>
#include <iostream>
#include <vector>
#include <cassert>
#include <climits>
using std::vector;
using std::cout;
using std::endl;
using std::random_shuffle;
using std::min;
using std::max;
vector<int> create_tournament(const vector<int>& input) {
// make sure we have at least two elements, so the problem is interesting
if (input.size() <= 1) {
return input;
}
vector<int> result(2 * input.size() - 1, -1);
int i = 0;
for (const auto& el : input) {
result[input.size() - 1 + i] = el;
++i;
}
for (uint j = input.size() / 2; j > 0; j >>= 1) {
for (uint k = 0; k < 2 * j; k += 2) {
result[j - 1 + k / 2] = min(result[2 * j - 1 + k], result[2 * j + k]);
}
}
return result;
}
int second_smaller(const vector<int>& tournament) {
const auto& minimum = tournament[0];
int second = INT_MAX;
for (uint j = 0; j < tournament.size() / 2; ) {
if (tournament[2 * j + 1] == minimum) {
second = min(second, tournament[2 * j + 2]);
j = 2 * j + 1;
}
else {
second = min(second, tournament[2 * j + 1]);
j = 2 * j + 2;
}
}
return second;
}
void print_vector(const vector<int>& v) {
for (const auto& el : v) {
cout << el << " ";
}
cout << endl;
}
int main() {
vector<int> a;
for (int i = 1; i <= 2048; ++i)
a.push_back(i);
for (int i = 0; i < 1000; i++) {
random_shuffle(a.begin(), a.end());
const auto& v = create_tournament(a);
assert (second_smaller(v) == 2);
}
return 0;
}
I have gone through all the posts above but I am convinced that the implementation of the Tournament algorithm is the best approach. Let us consider the following algorithm posted by #Gumbo
largest := numbers[0];
secondLargest := null
for i=1 to numbers.length-1 do
number := numbers[i];
if number > largest then
secondLargest := largest;
largest := number;
else
if number > secondLargest then
secondLargest := number;
end;
end;
end;
It is very good in case we are going to find the second largest number in an array. It has (2n-1) number of comparisons. But what if you want to calculate the third largest number or some kth largest number. The above algorithm doesn't work. You got to another procedure.
So, I believe tournament algorithm approach is the best and here is the link for that.
The following solution would take 2(N-1) comparisons:
arr #array with 'n' elements
first=arr[0]
second=-999999 #large negative no
i=1
while i is less than length(arr):
if arr[i] greater than first:
second=first
first=arr[i]
else:
if arr[i] is greater than second and arr[i] less than first:
second=arr[i]
i=i+1
print second
It can be done in n + ceil(log n) - 2 comparison.
Solution:
it takes n-1 comparisons to get minimum.
But to get minimum we will build a tournament in which each element will be grouped in pairs. like a tennis tournament and winner of any round will go forward.
Height of this tree will be log n since we half at each round.
Idea to get second minimum is that it will be beaten by minimum candidate in one of previous round. So, we need to find minimum in potential candidates (beaten by minimum).
Potential candidates will be log n = height of tree
So, no. of comparison to find minimum using tournament tree is n-1
and for second minimum is log n -1
sums up = n + ceil(log n) - 2
Here is C++ code
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
using namespace std;
typedef pair<int,int> ii;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is for the case when x is 0 */
return x && (!(x&(x-1)));
}
// modified
int log_2(unsigned int n) {
int bits = 0;
if (!isPowerOfTwo(n))
bits++;
if (n > 32767) {
n >>= 16;
bits += 16;
}
if (n > 127) {
n >>= 8;
bits += 8;
}
if (n > 7) {
n >>= 4;
bits += 4;
}
if (n > 1) {
n >>= 2;
bits += 2;
}
if (n > 0) {
bits++;
}
return bits;
}
int second_minima(int a[], unsigned int n) {
// build a tree of size of log2n in the form of 2d array
// 1st row represents all elements which fights for min
// candidate pairwise. winner of each pair moves to 2nd
// row and so on
int log_2n = log_2(n);
long comparison_count = 0;
// pair of ints : first element stores value and second
// stores index of its first row
ii **p = new ii*[log_2n];
int i, j, k;
for (i = 0, j = n; i < log_2n; i++) {
p[i] = new ii[j];
j = j&1 ? j/2+1 : j/2;
}
for (i = 0; i < n; i++)
p[0][i] = make_pair(a[i], i);
// find minima using pair wise fighting
for (i = 1, j = n; i < log_2n; i++) {
// for each pair
for (k = 0; k+1 < j; k += 2) {
// find its winner
if (++comparison_count && p[i-1][k].first < p[i-1][k+1].first) {
p[i][k/2].first = p[i-1][k].first;
p[i][k/2].second = p[i-1][k].second;
}
else {
p[i][k/2].first = p[i-1][k+1].first;
p[i][k/2].second = p[i-1][k+1].second;
}
}
// if no. of elements in row is odd the last element
// directly moves to next round (row)
if (j&1) {
p[i][j/2].first = p[i-1][j-1].first;
p[i][j/2].second = p[i-1][j-1].second;
}
j = j&1 ? j/2+1 : j/2;
}
int minima, second_minima;
int index;
minima = p[log_2n-1][0].first;
// initialize second minima by its final (last 2nd row)
// potential candidate with which its final took place
second_minima = minima == p[log_2n-2][0].first ? p[log_2n-2][1].first : p[log_2n-2][0].first;
// minima original index
index = p[log_2n-1][0].second;
for (i = 0, j = n; i <= log_2n - 3; i++) {
// if its last candidate in any round then there is
// no potential candidate
if (j&1 && index == j-1) {
index /= 2;
j = j/2+1;
continue;
}
// if minima index is odd, then it fighted with its index - 1
// else its index + 1
// this is a potential candidate for second minima, so check it
if (index&1) {
if (++comparison_count && second_minima > p[i][index-1].first)
second_minima = p[i][index-1].first;
}
else {
if (++comparison_count && second_minima > p[i][index+1].first)
second_minima = p[i][index+1].first;
}
index/=2;
j = j&1 ? j/2+1 : j/2;
}
printf("-------------------------------------------------------------------------------\n");
printf("Minimum : %d\n", minima);
printf("Second Minimum : %d\n", second_minima);
printf("comparison count : %ld\n", comparison_count);
printf("Least No. Of Comparisons (");
printf("n+ceil(log2_n)-2) : %d\n", (int)(n+ceil(log(n)/log(2))-2));
return 0;
}
int main()
{
unsigned int n;
scanf("%u", &n);
int a[n];
int i;
for (i = 0; i < n; i++)
scanf("%d", &a[i]);
second_minima(a,n);
return 0;
}
function findSecondLargeNumber(arr){
var fLargeNum = 0;
var sLargeNum = 0;
for(var i=0; i<arr.length; i++){
if(fLargeNum < arr[i]){
sLargeNum = fLargeNum;
fLargeNum = arr[i];
}else if(sLargeNum < arr[i]){
sLargeNum = arr[i];
}
}
return sLargeNum;
}
var myArray = [799, -85, 8, -1, 6, 4, 3, -2, -15, 0, 207, 75, 785, 122, 17];
Ref: http://www.ajaybadgujar.com/finding-second-largest-number-from-array-in-javascript/
A good way with O(1) time complexity would be to use a max-heap. Call the heapify twice and you have the answer.
int[] int_array = {4, 6, 2, 9, 1, 7, 4, 2, 9, 0, 3, 6, 1, 6, 8};
int largst=int_array[0];
int second=int_array[0];
for (int i=0; i<int_array.length; i++){
if(int_array[i]>largst) {
second=largst;
largst=int_array[i];
}
else if(int_array[i]>second && int_array[i]<largst) {
second=int_array[i];
}
}
I suppose, follow the "optimal algorithm uses n+log n-2 comparisons" from above, the code that I came up with that doesn't use binary tree to store the value would be the following:
During each recursive call, the array size is cut in half.
So the number of comparison is:
1st iteration: n/2 comparisons
2nd iteration: n/4 comparisons
3rd iteration: n/8 comparisons
...
Up to log n iterations?
Hence, total => n - 1 comparisons?
function findSecondLargestInArray(array) {
let winner = [];
if (array.length === 2) {
if (array[0] < array[1]) {
return array[0];
} else {
return array[1];
}
}
for (let i = 1; i <= Math.floor(array.length / 2); i++) {
if (array[2 * i - 1] > array[2 * i - 2]) {
winner.push(array[2 * i - 1]);
} else {
winner.push(array[2 * i - 2]);
}
}
return findSecondLargestInArray(winner);
}
Assuming array contain 2^n number of numbers.
If there are 6 numbers, then 3 numbers will move to the next level, which is not right.
Need like 8 numbers => 4 number => 2 number => 1 number => 2^n number of number
package com.array.orderstatistics;
import java.util.Arrays;
import java.util.Collections;
public class SecondLargestElement {
/**
* Total Time Complexity will be n log n + O(1)
* #param str
*/
public static void main(String str[]) {
Integer[] integerArr = new Integer[] { 5, 1, 2, 6, 4 };
// Step1 : Time Complexity will be n log(n)
Arrays.sort(integerArr, Collections.reverseOrder());
// Step2 : Array.get Second largestElement
int secondLargestElement = integerArr[1];
System.out.println(secondLargestElement);
}
}
Sort the array into ascending order then assign a variable to the (n-1)th term.

skyline algorithm

How do I find the vertices of the broken line that surrounds the silhouette in this image?
A possible input for the example above is:
WIDTH HEIGHT POSITION
3 9 17
5 9 9
12 4 8
3 11 3
10 7 1
2 3 19
So for this example the solution would be
[(1, 0), (1, 7), (3, 7), (3, 11), (6, 11), (6, 7),
(9, 7), (9, 9), (14, 9), (14, 4), (17, 4), (17, 9),
(20, 9), (20, 3), (21, 3), (21, 0)]
This is pretty simple. Make an array that is the length of the X axis, initialize to 0. As you read in the inputs, write the heights into this array if the height is >= the current value at that location in the array.
Then just loop over the array, and every time the value changes it is a vertex.
Basically:
int heights[SIZE] = {0};
int i, width, pos, height, prev = -1;
while (scanf("%d %d %d", &width, &height, &pos) == 3) {
for (i = 0; i < width; ++i) {
if (heights[pos+i] < height)
heights[pos+i] = height;
}
}
for (i = 0; i < SIZE; ++i) {
if (heights[i] != prev) {
printf("(%d,%d) ", i+1, heights[i]);
prev = heights[i];
}
}
printf("\n");
In the naive case, this doesn't seem like a very difficult algorithm. Do you know if the input size will get large/how large?
My initial attempt: Try to move from left to right. First pick the block with the leftmost edge that exists on the origin line. Climb to its top. Find all blocks with a left edge between the current point and the upper right point of the current block. Of that set, pick the closest (but check for edge cases, pun not intended). If the set is empty, start working your way down the right side of the block, looking for other blocks you may intercept.
Basically this is just how you'd trace it with your eye.
You can do some simple optimization by keeping sorted lists and then searching the lists rather than finding sets and digging around. For example, you might keep 4 sorted lists of the blocks, each sorted by the x or y coordinate of one of the sides.
If you have many, many blocks, you could consider using a multi-dimensional data structure to further organize the information.
I solved this problem using the sweep-line algorithm. This is a python class solution.
there two keys:
1) using the variable "points" to save all the left and right points and their heights and the sign of the height to indicate whether the points are left or right.
2) the variable "active" is used to save all the active lines that has been scanned.
class Solution:
# #param {integer[][]} buildings
# #return {integer[][]}
def getSkyline(self, buildings):
if len(buildings)==0: return []
if len(buildings)==1: return [[buildings[0][0], buildings[0][2]], [buildings[0][1], 0]]
points=[]
for building in buildings:
points+=[[building[0],building[2]]]
points+=[[building[1],-building[2]]] # the negative sign means this point is a right point
points=sorted(points, key=lambda x: x[0])
moving, active, res, current=0, [0], [],-1
while moving<len(points):
i=moving
while i<=len(points):
if i<len(points) and points[i][0]==points[moving][0]:
if points[i][1]>0:
active+=[points[i][1]]
if points[i][1]>current:
current=points[i][1]
if len(res)>0 and res[-1][0]==points[i][0]:
res[-1][1]=current
else:
res+=[[points[moving][0], current]]
else:
active.remove(-points[i][1]) #remove height of the lines than have been finished with scanning
i+=1
else:
break
if max(active)<current:
current=max(active)
res+=[[points[moving][0], current]]
moving=i
return res
I made a Java class to try and solve this. The class includes methods for generating, solving and printing data-sets. I haven't tested extensively, there may be a few bugs remaining. Also, my solution may be needlessly complicated, but it's designed to work (in theory) for non-discrete height and coordinate values.
import java.util.Random;
public class Skyline {
private int[][] buildings;
private int[][] skyline;
private int maxLength;
private int maxHeight;
public Skyline(int buildings, int maxLength, int maxHeight) {
this.maxLength = maxLength;
this.maxHeight = maxHeight;
makeRandom(buildings);
}
public Skyline(int[][] buildings, int dimensions) {
this.maxLength = maxLength;
this.maxHeight = maxHeight;
this.buildings = buildings;
}
public void makeRandom(int buildings) {
this.buildings = new int[buildings][3];
Random rand = new Random();
for(int i = 0; i < buildings; i++) {
int start = rand.nextInt(maxLength-3);
int end = rand.nextInt(maxLength - start - 1) + start + 1;
int height = rand.nextInt(maxHeight-1) + 1;
this.buildings[i][0] = start;
this.buildings[i][1] = height;
this.buildings[i][2] = end;
}
boolean swapped = true;
while(swapped) {
swapped = false;
for(int i = 0; i < this.buildings.length-1; i++) {
if(this.buildings[i][0] > this.buildings[i+1][0]) {
swapped = true;
int[] temp = this.buildings[i];
this.buildings[i] = this.buildings[i+1];
this.buildings[i+1] = temp;
}
}
}
// this.buildings[0][0] = 2;
// this.buildings[0][1] = 3;
// this.buildings[0][2] = 8;
}
public void printBuildings() {
print(this.buildings, false);
}
public void printSkyline() {
print(this.buildings, true);
}
public void print(int[][] buildings, boolean outline) {
char[][] str = new char[this.maxLength][this.maxHeight];
for(int i = 0; i < this.maxLength; i++) {
for(int j = 0; j < this.maxHeight; j++) {
str[i][j] = '.';
}
}
for(int i = 0; i < buildings.length; i++) {
int start = buildings[i][0];
int height = buildings[i][1];
int end = buildings[i][2];
//print the starting vertical
for(int j = 0; j < height; j++) {
if(outline) str[start][j] = str[start][j] == '|' ? '.' : '|';
else str[start][j] = '|';
}
//print the ending vertical
for(int j = 0; j < height; j++) {
if(outline) str[end][j] = str[end][j] == '|' ? '.' : '|';
else str[end][j] = '|';
}
//print the horizontal
if(height > 0) {
for(int j = start; j <= end; j++) {
str[j][height] = str[j][height] == '|' ? '|' : '-';
}
}
}
for(int i = maxHeight-1; i >= 0; i--) {
for(int j = 0; j < maxLength; j++) {
System.out.print(str[j][i]);
}
System.out.println();
}
System.out.println();
}
public void solveSkyline() {
for(int i = 0; i < buildings.length; i++) {
boolean reduced = true;
while(reduced) {
reduced = false;
for(int j = i+1; j < buildings.length; j++) {
if(buildings[j][0] < buildings[i][2] && buildings[j][1] > buildings[i][1] && buildings[j][2] >= buildings[i][2]) { //if intersecting building is taller, and longer
buildings[i][2] = buildings[j][0];
reduced = true;
break;
} else if(buildings[j][0] < buildings[i][2] && buildings[j][1] <= buildings[i][1] && buildings[j][2] >= buildings[i][2]) { //intersecting building is shorter, but longer
buildings[j][0] = buildings[i][2];
reduced = true;
break;
} else if(buildings[j][0] < buildings[i][2] && buildings[j][1] > 0 && buildings[j][1] < buildings[i][1] && buildings[j][2] <= buildings[i][2]) { //building is invisible, so ignore it
buildings[j][1] = 0;
reduced = true;
break;
} else if(buildings[j][0] < buildings[i][2] && buildings[j][2] <= buildings[i][2] && buildings[j][1] > buildings[i][1]) {
int[] newBuilding = new int[]{buildings[j][2], buildings[i][1], buildings[i][2]};
int[][] newBuildings = new int[buildings.length+1][3];
boolean inserted = false;
buildings[i][2] = buildings[j][0];
for(int k = 0; k < buildings.length; k++) {
if(inserted == false) {
if(newBuilding[0] < buildings[k][0]) {
newBuildings[k] = newBuilding;
newBuildings[k+1] = buildings[k];
inserted = true;
} else {
newBuildings[k] = buildings[k];
}
}
if(inserted == false && k == buildings.length - 1) {
newBuildings[k+1] = newBuilding;
} else {
newBuildings[k+1] = buildings[k];
}
}
buildings = newBuildings;
reduced = true;
break;
}
}
}
}
}
public static void main(String args[]) {
Skyline s = new Skyline(5, 100, 10);
s.printBuildings();
s.solveSkyline();
s.printBuildings();
s.printSkyline();
}
}
My solution to the problem as described here https://leetcode.com/problems/the-skyline-problem/ it iterates the list of buildings twice, however this could be combined into a single iteration. However, there are more optimal approaches if you consider the pure algorithm solution explained here http://www.algorithmist.com/index.php/UVa_105
class Solution {
public:
vector<pair<int, int>> getSkyline(vector<vector<int>>& buildings) {
// The final result.
vector<pair<int, int>> result;
// To hold information about the buildings
std::set<BuildingInformation> buildingInformation;
// Go through each building, and store information about the start and end heights.
for ( vector<vector<int>>::iterator buildingIt = buildings.begin( ); buildingIt != buildings.end( ); ++buildingIt ) {
BuildingInformation buildingStart;
buildingStart.x = (*buildingIt)[0];
buildingStart.h = (*buildingIt)[2];
buildingStart.StartOrEnd = Start;
buildingInformation.insert(buildingStart);
buildingStart.x = (*buildingIt)[1];
buildingStart.StartOrEnd = End;
buildingInformation.insert(buildingStart);
}
// Keep track of the current height.
int currentHeight = 0;
// A map of active building heights against number of buildings (to handle multiple buildings overlapping with same height).
// As it is a map, it'll be sorted by key, which is the height.
std::map<int, int> heights;
// Go through each building information that we generated earlier.
for ( std::set<BuildingInformation>::iterator it = buildingInformation.begin( ); it != buildingInformation.end( ); ++it ) {
if ( it->StartOrEnd == Start ) {
// This is a start point, do we have this height already in our map?
if ( heights.find( it->h ) != heights.end( ) ) {
// Yes, increment count of active buildings with this height/
heights[ it->h ] += 1;
} else {
// Nope, add this building to our map.
heights[ it->h ] = 1;
}
// Check if building height is taller than current height.
if ( it->h > currentHeight ) {
// Update current height and add marker to results.
currentHeight = it->h;
result.push_back( pair<int, int>( it->x, currentHeight ) );
}
} else {
// This is an end point, get iterator into our heights map.
std::map<int, int>::iterator heightIt = heights.find( it->h );
// Reduce by one.
heightIt->second -= 1;
// If this was the last building of the current height in the map...
if ( heightIt->second == 0 ) {
// Remove from heights map.
heights.erase( heightIt );
// If our height was the current height...
if ( it->h == currentHeight ) {
// If we have no more active buildings...
if ( heights.size( ) == 0 ) {
// Current height is zero.
currentHeight = 0;
} else {
// Otherwise, get iterator to one past last.
heightIt = heights.end( );
// Go back to get last valid iterator.
--heightIt;
// Store current height.
currentHeight = heightIt->first;
}
// Add marker to results.
result.push_back( pair<int, int>( it->x, currentHeight ) );
}
}
}
}
return result;
}
private:
// Is this a building start or end?
enum BuildingStartOrEnd
{
Start = 0,
End
};
// Information about building, there are two of these for each building, one for start, one for end.
struct BuildingInformation
{
int x;
int h;
BuildingStartOrEnd StartOrEnd;
// The ordering algorithm for the key, the rules we want to implement is keys are put in X order, and
// in the case of a tie (x values the same), we want Start pieces to come before End pieces (this is
// to handle cases where an old building ends and a new building begins on same X index, in which case
// we want to process the new start before processing the old end), however if we have two Start pieces
// at the same index, we wish to favour taller pieces (in this scenario we want to add a marker for the
// tallest building), finally if we have two End pieces at the same index, we wish to prefer lower
// pieces, as when multiple buildings end, we only want to add one result for the ultimate lowest point.
bool operator < ( const BuildingInformation & rhs ) const
{
if ( x == rhs.x )
{
if ( StartOrEnd == rhs.StartOrEnd ) {
if ( StartOrEnd == Start )
return h > rhs.h;
else
return h < rhs.h;
} else {
return StartOrEnd < rhs.StartOrEnd;
}
}
return x < rhs.x;
}
};
};

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