I am new to Algorithms and Competitive Programming. I am learning about Dynamic programming and I have a problem as below:
Given an array with n numbers. Define a sub-array is a[i, j] = {a[i], a[i + 1], ..., a[j]}, in other words, elements must be contiguous.
The problem is the find the maximum weight of a sub-array such that
that weight is an even number.
The input is 2 <= n <= 1000000; -100 <= a[i] <= 100
Sample test:
5
-2 1 -4 4 9
Output: 10
For this problem, I can do brute force but with a large value of n, I can not do it with the time limit is 1 second. Therefore, I want to change it to Dynamic programming.
I have an idea but I do not know if it works. I think I can divide this problem into two sub-problems. For each element/number, I consider if it is odd/even and then find the largest sum with its corresponding property (odd + odd or even + even to get a even sum). However, that is just what I think and I really need your help.
Here is C++ algorithm with O(n) time complexity:
const int Inf = 1e9;
int main() {
int n = 5;
vector<int> inputArray = {-2, 1, -4, 4, 9};
int minEvenPrefixSum = 0, minOddPrefixSum = Inf;
bool isOddPrefixSumFound = false;
int prefixSum = 0, answer = -Inf;
for(int i = 0; i < n; ++i) {
prefixSum += inputArray[i];
if(abs(prefixSum) % 2 == 0) {
answer = max(answer, prefixSum - minEvenPrefixSum);
minEvenPrefixSum = min(minEvenPrefixSum, prefixSum);
} else {
if(isOddPrefixSumFound) {
answer = max(answer, prefixSum - minOddPrefixSum);
}
isOddPrefixSumFound = true;
minOddPrefixSum = min(minOddPrefixSum, prefixSum);
}
}
if(answer == -Inf) {
cout << "There is no subarray with even sum";
} else {
cout << answer;
}
}
Explanation:
As #nico-schertler mentioned in commentary this task is very similar with more basic problem of the maximum-sum contiguous sub array. How to solve basic task with O(n) time complexity you can read here.
Now let's store not just one value of the minimum prefix sum, but two. One is for minimum even prefix sum, and the other is for minimum odd prefix sum. As a result, when we process the next number, we look at what the value of the prefix sum becomes. If it is even, we try to update the answer using the minimum even value of the prefix sum, in the other case using the minimum odd value of the prefix sum.
Problem: Given a square matrix, A, whose dimension, N, is a divisor of 64, find a fast way to smoothe A by letting a 3 x 3 grid slide over the A, s.t. for each entry A_{ij}, we replace this with the average of the 8 nearby neighbors.
My solution: Let's assume A is a 4x4 matrix, we initialize as:
int A[NxN] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
Which is equivalent to:
In the first iteration, we see A_ij is equivalent to 0, which is in the center of the 3x3 matrix. We realize that indices (0,0), (0,1), (0,2), (1,0), (1,3) of the 3x3 matrix is trying to find numbers not existing. Therefore, we see that the result must be that if $A_{ij}$ = 0, then we must replace it by $0+1+4+5 = \frac{10}{4}$.
Next iteration, we see A_ij is equivalent to 1, which is now the new center of the 3x3 matrix, and we see that the numbers to be included in the averaging is: 0,1,2,4,5,6.
So when we are in a corner of our matrix, A, we need 4 numbers for averaging, in the top of the matrix, we need 6 numbers, and if A_{ij} = 6, meaning in the middle of the matrix, we will need 9 numbers.
What I am trying to find is a general pattern for these sequences of needed numbers in order to do this in hopefully a single \texttt{for loop}.
I found that when i % dim == 0, we are in the left side of the matrix, when i % dim == 3 we are in the left side, but I have tried to find a general index pattern to define my averaging functions to pair with my if conditions.
#include <stdio.h>
int smooth(int dim);
int main() {
smooth(4);
}
int smooth(int dim) {
int A[16] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15};
int K[dim*dim]; // Replacement array, otherwise we average averages
for (int i = 0; i < dim; i++) {
for (int j = 0; j < dim; j++)
{
// Corner cases
if ((i % dim == 0 && j % dim == 0) || (i % dim == dim - 1 && j % dim == 0) || (i % dim == 0 && j % dim == dim -1) || (i % dim == dim -1 && j % dim == dim-1))
{
// Average function on corners
}
// else if not corner and in "edge row", top/bottom {
// Average 6 numbers
}
}
}
return 0;
}
Your mre is a bit theoretical, I will leave all parts which are not practically plausible out of this. They obviously only serve to provide a MRE quick.
Interpret your input as 2D.
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
More abstractly as
0 ... ( 0*dim + (dim-1))
(1 * dim) ... ( 1*dim + (dim-1))
(2 * dim) ... ( 2*dim + (dim-1))
.
.
.
((dim-1) * dim) ... ((dim-1)*dim + (dim-1))
Here is the structure I propose for your function
int smooth(int dim)
{
int A[16] = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}; //obvious MRE weirdness, ignored
int K[dim*dim]; // Replacement array, otherwise we average averages
// Corner cases
K[0] = A[0]; /* dummy, to be calculated */
K[dim-1] = A[dim-1]; /* dummy, to be calculated */
K[dim*(dim-1)] = A[dim*(dim-1)]; /* dummy, to be calculated */
K[dim*dim-1] = A[dim*dim-1]; /* dummy, to be calculated */
for (int i = 0; i < dim-2; i++)
{
// edge cases
{
K[1+i] = A[1+i]; /* dummy, average to be calculated */
K[dim*(1+i)] = A[dim*(1+i)]; /* dummy, average to be calculated */
K[dim*(1+i)+dim-1] = A[dim*(1+i)+dim-1]; /* dummy, average to be calculated */
K[dim*(dim-1)+i+1] = A[dim*(dim-1)+i+1]; /* dummy, average to be calculated */
}
// center part
for (int j = 0; j < dim-2; j++)
{
K[(i+1)*dim+1+j] = A[(i+1)*dim+1+j]; /* dummy, average to be calculated */
}
}
return 0;
}
For all cases of A[...]; /* dummy, average to be calculated */ you can use the same index for reading from A as for writing into K and
subtract 1 to get to the pixel to the left
add one to get to the pixel to the right
add dim to get to the pixel below
subtract dim to get to the pixel above
do that for all neigbouring pixels you decide to use
sum the values up to determine the average
In all cases where you need a neighbour pixel you will always arrive at a valid pixel, because either there is one or you are not going to need it.
(Note: I strictly practice starting for loops from 0 and count up with <needed_length. There is an alternative to start from 1 and count up to needed length +1. It will in this case here make the index calculations read easier. This is a matter of habits. I come from an environment where you have to prepare strong reasoning for loops not starting at 0. Habits are hard to shed. Going through the alternative concept will probably provide you with some helpful learning opportunity.)
I want to know a algorithm to find out the maximum xor value of three elements of an array.
I have read about the maximum xor for two elements from an array but cannot understand how to apply it on finding the maximum value of XOR taking 3 elements of an array . Can someone point out a hint ?
Required complexity : less than O(N^3) where N is the number of elements in the array.
Example:
A = [1,2,3,4]
All Possible Triplets :-
1^2^3 = 0
1^2^4 = 7
1^3^4 = 6
2^3^4 = 5
Thus, the maximum XOR value is 7.
Edit :
I have thought of a solution having complexity O(N^2 * log(MAX)) and it has solved my purpose :D .
MAX = Maximum Value in the Array
Well, I have found a solution with complexity O(N^2 * log(MAX)) where MAX is the largest value in the array .
Let there be 3 elements X,Y,Z fron the array A.
where X = A[i] , Y = A[j] , Z = A[k] and i != j != k
We want the maximum value of (X^Y^Z) .
Let us assume W = X*Y.
Then we would like to find such a Z which give maximum value for W^Z and Z != X and Z != Y
Now this has been reduced to the problem of finding "Two elements whose XOR is maximum" which can be done for a given W in O(log(MAX)) using a Trie .
Explanation for Trie :
Let us assume W = 10001 here W is in binary .
Now we know 1^0 = 1 , 0^0 = 0 , 1^1 = 0 , so the maximum value we
can get for W^Z is when Z is 01110 because
W^Z will give = 11111.
But it is not necessary to have 15 or Base2(11111) in our array so
we would take the best possible option available.
So we will create a Trie of all the elements of the array
according to their binary representation.
If A = [1,2,7] , then 1 = 001 , 2 = 010 , 7 = 111 in
binary .
Then the Trie will look like :-
Top
/ \
0 1
/ \ \
0 1 1
\ / \
1 0 1
Now to lets assume W = 7 , and we want to find Z such that
W^Z is maximum (when Z = 000 ) then we will start at the Top and look if we have branch leading to 0 since the first bit of 7 is 1
, then we will down through that branch and then again look if we have
branch leading to 0 at 2nd bit , again we find it , then for the last
time we search for branch leading to a 0 at 3rd bit but we do not find
it , so we go down through the other branch which gives us Z =
001. Thus, the maximum W^Z will be 7^1 = 6 . Now , the
complexity of finding Z will be maximum height of the Trie which
will be log(MAX).
Thus , we have N*(N-1)/2 number of W's and for each W we can find the Maximum value of W^Z and if we take the Maximum from all the values of W^Z we will have our answer.
With three nested loop:
int max2=0,max3=0;
for (int i=0;i<arr.size();i++)
for (int j=0;j<arr.size();j++)
for (int k=0;k<arr.size();k++)
{
if (arr[i]^arr[j]>max2) // for 2 elements
max2 = arr[i]^arr[j];
if (arr[i]^arr[j]^arr[k]>max3) // for 3 elements
max3 = arr[i]^arr[j]^arr[k];
}
int max = max2; // for both
if (max3>max2)
max = max3;
following will do the O(N^3), but in an more optimized approach - not testing same combination more than once, not testing element against itself,
and somewhat optimized evaluation (xoring the first two elements once for all possible third elements)
Number of Xor operations performed will be:
n(n-1)(n-2)/6 + n(n-1)/2
Complexity is still n(n-1)(n-2)/6 ===> O(N^3) though.
unsigned int maxXor3(unsigned int* element, int len)
{
unsigned int max = 0;
unsigned int xor2 = 0;
unsigned int xor3 = 0;
int j = k = 0;
for (int i = 0 ; i < len ; i++)
{
for (j = i + 1 ; j < len ; j++)
{
xor2 = element[i] ^ element[j];
for(k = j + 1; k < len; k++)
{
xor3 = xor2 ^ element[k];
if (xor3 > max)
max = xor3;
}
}
}
return max;
}
I did this in c :
#include<stdio.h>
int main (void)
{
int n,i;
scanf("%d", &n);
for(i=2;i<=n;i=i+2)
{
if((i*i)%2==0 && (i*i)<= n)
printf("%d \n",(i*i));
}
return 0;
}
What would be a better/faster approach to tackle this problem?
Let me illustrate not only a fast solution, but also how to derive it. Start with a fast way of listing all squares and work from there (pseudocode):
max = n*n
i = 1
d = 3
while i < max:
print i
i += d
d += 2
So, starting from 4 and listing only even squares:
max = n*n
i = 4
d = 5
while i < max:
print i
i += d
d += 2
i += d
d += 2
Now we can shorten that mess on the end of the while loop:
max = n*n
i = 4
d = 5
while i < max:
print i
i += 2 + 2*d
d += 4
Note that we are constantly using 2*d, so it's better to just keep calculating that:
max = n*n
i = 4
d = 10
while i < max:
print i
i += 2 + d
d += 8
Now note that we are constantly adding 2 + d, so we can do better by incorporating this into d:
max = n*n
i = 4
d = 12
while i < max:
print i
i += d
d += 8
Blazing fast. It only takes two additions to calculate each square.
I like your solution. The only suggestions I would make would be:
Put the (i*i)<=n as the middle clause of your for loop, then it's checked earlier and you break out of the loop sooner.
You don't need to check and see if (i*i)%2==0, since 'i' is always positive and a positive squared is always positive.
With those two changes in mind you can get rid of the if statement in your for loop and just print.
Square of even is even. So, you really do not need to check it again. Following is the code, I would suggest:
for (i = 2; i*i <= n; i+=2)
printf ("%d\t", i*i);
The largest value for i in your loop should be the floor of the square root of n.
The reason is that the square of any i (integer) larger than this will be greater than n. So, if you make this change, you don't need to check that i*i <= n.
Also, as others have pointed out, there is no point in checking that i*i is even since the square of all even numbers is even.
And you are right in ignoring odd i since for any odd i, i*i is odd.
Your code with the aforementioned changes follows:
#include "stdio.h"
#include "math.h"
int main ()
{
int n,i;
scanf("%d", &n);
for( i = 2; i <= (int)floor(sqrt(n)); i = i+2 ) {
printf("%d \n",(i*i));
}
return 0;
}
given an array of 0s and 1s, find maximum subarray such that number of zeros and 1s are equal.
This needs to be done in O(n) time and O(1) space.
I have an algo which does it in O(n) time and O(n) space. It uses a prefix sum array and exploits the fact that if the number of 0s and 1s are same then
sumOfSubarray = lengthOfSubarray/2
#include<iostream>
#define M 15
using namespace std;
void getSum(int arr[],int prefixsum[],int size) {
int i;
prefixsum[0]=arr[0]=0;
prefixsum[1]=arr[1];
for (i=2;i<=size;i++) {
prefixsum[i]=prefixsum[i-1]+arr[i];
}
}
void find(int a[],int &start,int &end) {
while(start < end) {
int mid = (start +end )/2;
if((end-start+1) == 2 * (a[end] - a[start-1]))
break;
if((end-start+1) > 2 * (a[end] - a[start-1])) {
if(a[start]==0 && a[end]==1)
start++; else
end--;
} else {
if(a[start]==1 && a[end]==0)
start++; else
end--;
}
}
}
int main() {
int size,arr[M],ps[M],start=1,end,width;
;
cin>>size;
arr[0]=0;
end=size;
for (int i=1;i<=size;i++)
cin>>arr[i];
getSum(arr,ps,size);
find(ps,start,end);
if(start!=end)
cout<<(start-1)<<" "<<(end-1)<<endl; else cout<<"No soln\n";
return 0;
}
Now my algorithm is O(n) time and O(Dn) space where Dn is the total imblance in the list.
This solution doesn't modify the list.
let D be the difference of 1s and 0s found in the list.
First, let's step linearily through the list and calculate D, just to see how it works:
I'm gonna use this list as an example : l=1100111100001110
Element D
null 0
1 1
1 2 <-
0 1
0 0
1 1
1 2
1 3
1 4
0 3
0 2
0 1
0 0
1 1
1 2
1 3
0 2 <-
Finding the longest balanced subarray is equivalent to finding 2 equal elements in D that are the more far appart. (in this example the 2 2s marked with arrows.)
The longest balanced subarray is between first occurence of element +1 and last occurence of element. (first arrow +1 and last arrow : 00111100001110)
Remark:
The longest subarray will always be between 2 elements of D that are
between [0,Dn] where Dn is the last element of D. (Dn = 2 in the
previous example) Dn is the total imbalance between 1s and 0s in the
list. (or [Dn,0] if Dn is negative)
In this example it means that I don't need to "look" at 3s or 4s
Proof:
Let Dn > 0 .
If there is a subarray delimited by P (P > Dn). Since 0 < Dn < P,
before reaching the first element of D which is equal to P we reach one
element equal to Dn. Thus, since the last element of the list is equal to Dn, there is a longest subarray delimited by Dns than the one delimited by Ps.And therefore we don't need to look at Ps
P cannot be less than 0 for the same reasons
the proof is the same for Dn <0
Now let's work on D, D isn't random, the difference between 2 consecutive element is always 1 or -1. Ans there is an easy bijection between D and the initial list. Therefore I have 2 solutions for this problem:
the first one is to keep track of first and last appearance of each
element in D that are between 0 and Dn (cf remark).
second is to transform the list into D, and then work on D.
FIRST SOLUTION
For the time being I cannot find a better approach than the first one:
First calculate Dn (in O(n)) . Dn=2
Second instead of creating D, create a dictionnary where the keys are the value of D (between [0 and Dn]) and the value of each keys is a couple (a,b) where a is the first occurence of the key and b the last.
Element D DICTIONNARY
null 0 {0:(0,0)}
1 1 {0:(0,0) 1:(1,1)}
1 2 {0:(0,0) 1:(1,1) 2:(2,2)}
0 1 {0:(0,0) 1:(1,3) 2:(2,2)}
0 0 {0:(0,4) 1:(1,3) 2:(2,2)}
1 1 {0:(0,4) 1:(1,5) 2:(2,2)}
1 2 {0:(0,4) 1:(1,5) 2:(2,6)}
1 3 { 0:(0,4) 1:(1,5) 2:(2,6)}
1 4 {0:(0,4) 1:(1,5) 2:(2,6)}
0 3{0:(0,4) 1:(1,5) 2:(2,6) }
0 2 {0:(0,4) 1:(1,5) 2:(2,9) }
0 1 {0:(0,4) 1:(1,10) 2:(2,9) }
0 0 {0:(0,11) 1:(1,10) 2:(2,9) }
1 1 {0:(0,11) 1:(1,12) 2:(2,9) }
1 2 {0:(0,11) 1:(1,12) 2:(2,13)}
1 3 {0:(0,11) 1:(1,12) 2:(2,13)}
0 2 {0:(0,11) 1:(1,12) 2:(2,15)}
and you chose the element with the largest difference : 2:(2,15) and is l[3:15]=00111100001110 (with l=1100111100001110).
Time complexity :
2 passes, the first one to caclulate Dn, the second one to build the
dictionnary.
find the max in the dictionnary.
Total is O(n)
Space complexity:
the current element in D : O(1) the dictionnary O(Dn)
I don't take 3 and 4 in the dictionnary because of the remark
The complexity is O(n) time and O(Dn) space (in average case Dn <<
n).
I guess there is may be a better way than a dictionnary for this approach.
Any suggestion is welcome.
Hope it helps
SECOND SOLUTION (JUST AN IDEA NOT THE REAL SOLUTION)
The second way to proceed would be to transform your list into D. (since it's easy to go back from D to the list it's ok). (O(n) time and O(1) space, since I transform the list in place, even though it might not be a "valid" O(1) )
Then from D you need to find the 2 equal element that are the more far appart.
it looks like finding the longest cycle in a linked list, A modification of Richard Brent algorithm might return the longest cycle but I don't know how to do it, and it would take O(n) time and O(1) space.
Once you find the longest cycle, go back to the first list and print it.
This algorithm would take O(n) time and O(1) space complexity.
Different approach but still O(n) time and memory. Start with Neil's suggestion, treat 0 as -1.
Notation: A[0, …, N-1] - your array of size N, f(0)=0, f(x)=A[x-1]+f(x-1) - a function
If you'd plot f, you'll see, that what you look for are points for which f(m)=f(n), m=n-2k where k-positive natural. More precisely, only for x such that A[x]!=A[x+1] (and the last element in an array) you must check whether f(x) already occurred. Unfortunately, now I see no improvement over having array B[-N+1…N-1] where such information would be stored.
To complete my thought: B[x]=-1 initially, B[x]=p when p = min k: f(k)=x . And the algorithm is (double-check it, as I'm very tired):
fx = 0
B = new array[-N+1, …, N-1]
maxlen = 0
B[0]=0
for i=1…N-1 :
fx = fx + A[i-1]
if B[fx]==-1 :
B[fx]=i
else if ((i==N-1) or (A[i-1]!=A[i])) and (maxlen < i-B[fx]):
We found that A[B[fx], …, i] is best than what we found so far
maxlen = i-B[fx]
Edit: Two bed-thoughts (= figured out while laying in bed :P ):
1) You could binary search the result by the length of subarray, which would give O(n log n) time and O(1) memory algorithm. Let's use function g(x)=x - x mod 2 (because subarrays which sum to 0 are always of even length). Start by checking, if the whole array sums to 0. If yes -- we're done, otherwise continue. We now assume 0 as starting point (we know there's subarray of such length and "summing-to-zero property") and g(N-1) as ending point (we know there's no such subarray). Let's do
a = 0
b = g(N-1)
while a<b :
c = g((a+b)/2)
check if there is such subarray in O(n) time
if yes:
a = c
if no:
b = c
return the result: a (length of maximum subarray)
Checking for subarray with "summing-to-zero property" of some given length L is simple:
a = 0
b = L
fa = fb = 0
for i=0…L-1:
fb = fb + A[i]
while (fa != fb) and (b<N) :
fa = fa + A[a]
fb = fb + A[b]
a = a + 1
b = b + 1
if b==N:
not found
found, starts at a and stops at b
2) …can you modify input array? If yes and if O(1) memory means exactly, that you use no additional space (except for constant number of elements), then just store your prefix table values in your input array. No more space used (except for some variables) :D
And again, double check my algorithms as I'm veeery tired and could've done off-by-one errors.
Like Neil, I find it useful to consider the alphabet {±1} instead of {0, 1}. Assume without loss of generality that there are at least as many +1s as -1s. The following algorithm, which uses O(sqrt(n log n)) bits and runs in time O(n), is due to "A.F."
Note: this solution does not cheat by assuming the input is modifiable and/or has wasted bits. As of this edit, this solution is the only one posted that is both O(n) time and o(n) space.
A easier version, which uses O(n) bits, streams the array of prefix sums and marks the first occurrence of each value. It then scans backward, considering for each height between 0 and sum(arr) the maximal subarray at that height. Some thought reveals that the optimum is among these (remember the assumption). In Python:
sum = 0
min_so_far = 0
max_so_far = 0
is_first = [True] * (1 + len(arr))
for i, x in enumerate(arr):
sum += x
if sum < min_so_far:
min_so_far = sum
elif sum > max_so_far:
max_so_far = sum
else:
is_first[1 + i] = False
sum_i = 0
i = 0
while sum_i != sum:
sum_i += arr[i]
i += 1
sum_j = sum
j = len(arr)
longest = j - i
for h in xrange(sum - 1, -1, -1):
while sum_i != h or not is_first[i]:
i -= 1
sum_i -= arr[i]
while sum_j != h:
j -= 1
sum_j -= arr[j]
longest = max(longest, j - i)
The trick to get the space down comes from noticing that we're scanning is_first sequentially, albeit in reverse order relative to its construction. Since the loop variables fit in O(log n) bits, we'll compute, instead of is_first, a checkpoint of the loop variables after each O(√(n log n)) steps. This is O(n/√(n log n)) = O(√(n/log n)) checkpoints, for a total of O(√(n log n)) bits. By restarting the loop from a checkpoint, we compute on demand each O(√(n log n))-bit section of is_first.
(P.S.: it may or may not be my fault that the problem statement asks for O(1) space. I sincerely apologize if it was I who pulled a Fermat and suggested that I had a solution to a problem much harder than I thought it was.)
If indeed your algorithm is valid in all cases (see my comment to your question noting some corrections to it), notice that the prefix array is the only obstruction to your constant memory goal.
Examining the find function reveals that this array can be replaced with two integers, thereby eliminating the dependence on the length of the input and solving your problem. Consider the following:
You only depend on two values in the prefix array in the find function. These are a[start - 1] and a[end]. Yes, start and end change, but does this merit the array?
Look at the progression of your loop. At the end, start is incremented or end is decremented only by one.
Considering the previous statement, if you were to replace the value of a[start - 1] by an integer, how would you update its value? Put another way, for each transition in the loop that changes the value of start, what could you do to update the integer accordingly to reflect the new value of a[start - 1]?
Can this process can be repeated with a[end]?
If, in fact, the values of a[start - 1] and a[end] can be reflected with two integers, doesn't the whole prefix array no longer serve a purpose? Can't it therefore be removed?
With no need for the prefix array and all storage dependencies on the length of the input removed, your algorithm will use a constant amount of memory to achieve its goal, thereby making it O(n) time and O(1) space.
I would prefer you solve this yourself based on the insights above, as this is homework. Nevertheless, I have included a solution below for reference:
#include <iostream>
using namespace std;
void find( int *data, int &start, int &end )
{
// reflects the prefix sum until start - 1
int sumStart = 0;
// reflects the prefix sum until end
int sumEnd = 0;
for( int i = start; i <= end; i++ )
sumEnd += data[i];
while( start < end )
{
int length = end - start + 1;
int sum = 2 * ( sumEnd - sumStart );
if( sum == length )
break;
else if( sum < length )
{
// sum needs to increase; get rid of the lower endpoint
if( data[ start ] == 0 && data[ end ] == 1 )
{
// sumStart must be updated to reflect the new prefix sum
sumStart += data[ start ];
start++;
}
else
{
// sumEnd must be updated to reflect the new prefix sum
sumEnd -= data[ end ];
end--;
}
}
else
{
// sum needs to decrease; get rid of the higher endpoint
if( data[ start ] == 1 && data[ end ] == 0 )
{
// sumStart must be updated to reflect the new prefix sum
sumStart += data[ start ];
start++;
}
else
{
// sumEnd must be updated to reflect the new prefix sum
sumEnd -= data[ end ];
end--;
}
}
}
}
int main() {
int length;
cin >> length;
// get the data
int data[length];
for( int i = 0; i < length; i++ )
cin >> data[i];
// solve and print the solution
int start = 0, end = length - 1;
find( data, start, end );
if( start == end )
puts( "No soln" );
else
printf( "%d %d\n", start, end );
return 0;
}
This algorithm is O(n) time and O(1) space. It may modify the source array, but it restores all the information back. So it is not working with const arrays. If this puzzle has several solutions, this algorithm picks the solution nearest to the array beginning. Or it might be modified to provide all solutions.
Algorithm
Variables:
p1 - subarray start
p2 - subarray end
d - difference of 1s and 0s in the subarray
Calculate d, if d==0, stop. If d<0, invert the array and after balanced subarray is found invert it back.
While d > 0 advance p2: if the array element is 1, just decrement both p2 and d. Otherwise p2 should pass subarray of the form 11*0, where * is some balanced subarray. To make backtracking possible, 11*0? is changed to 0?*00 (where ? is the value next to the subarray). Then d is decremented.
Store p1 and p2.
Backtrack p2: if the array element is 1, just increment p2. Otherwise we found element, changed on step 2. Revert the changes and pass subarray of the form 11*0.
Advance p1: if the array element is 1, just increment p1. Otherwise p1 should pass subarray of the form 0*11.
Store p1 and p2, if p2 - p1 improved.
If p2 is at the end of the array, stop. Otherwise continue with step 4.
How does it work
Algorithm iterates through all possible positions of the balanced subarray in the input array. For each subarray position p1 and p2 are kept as far from each other as possible, providing locally longest subarray. Subarray with maximum length is chosen between all these subarrays.
To determine the next best position for p1, it is advanced to the first position where the balance between 1s and 0s is changed by one. (Step 5).
To determine the next best position for p2, it is advanced to the last position where the balance between 1s and 0s is changed by one. To make it possible, step 2 detects all such positions (starting from the array's end) and modifies the array in such a way, that it is possible to iterate through these positions with linear search. (Step 4).
While performing step 2, two possible conditions may be met. Simple one: when value '1' is found; pointer p2 is just advanced to the next value, no special treatment needed. But when value '0' is found, balance is going in wrong direction, it is necessary to pass through several bits until correct balance is found. All these bits are of no interest to the algorithm, stopping p2 there will give either a balanced subarray, which is too short, or a disbalanced subarray. As a result, p2 should pass subarray of the form 11*0 (from right to left, * means any balanced subarray). There is no chance to go the same way in other direction. But it is possible to temporary use some bits from the pattern 11*0 to allow backtracking. If we change first '1' to '0', second '1' to the value next to the rightmost '0', and clear the value next to the rightmost '0': 11*0? -> 0?*00, then we get the possibility to (first) notice the pattern on the way back, since it starts with '0', and (second) find the next good position for p2.
C++ code:
#include <cstddef>
#include <bitset>
static const size_t N = 270;
void findLargestBalanced(std::bitset<N>& a, size_t& p1s, size_t& p2s)
{
// Step 1
size_t p1 = 0;
size_t p2 = N;
int d = 2 * a.count() - N;
bool flip = false;
if (d == 0) {
p1s = 0;
p2s = N;
return;
}
if (d < 0) {
flip = true;
d = -d;
a.flip();
}
// Step 2
bool next = true;
while (d > 0) {
if (p2 < N) {
next = a[p2];
}
--d;
--p2;
if (a[p2] == false) {
if (p2+1 < N) {
a[p2+1] = false;
}
int dd = 2;
while (dd > 0) {
dd += (a[--p2]? -1: 1);
}
a[p2+1] = next;
a[p2] = false;
}
}
// Step 3
p2s = p2;
p1s = p1;
do {
// Step 4
if (a[p2] == false) {
a[p2++] = true;
bool nextToRestore = a[p2];
a[p2++] = true;
int dd = 2;
while (dd > 0 && p2 < N) {
dd += (a[p2++]? 1: -1);
}
if (dd == 0) {
a[--p2] = nextToRestore;
}
}
else {
++p2;
}
// Step 5
if (a[p1++] == false) {
int dd = 2;
while (dd > 0) {
dd += (a[p1++]? -1: 1);
}
}
// Step 6
if (p2 - p1 > p2s - p1s) {
p2s = p2;
p1s = p1;
}
} while (p2 < N);
if (flip) {
a.flip();
}
}
Sum all elements in the array, then diff = (array.length - sum) will be the difference in number of 0s and 1s.
If diff is equal to array.length/2, then the maximum subarray = array.
If diff is less than array.length/2 then there are more 1s than 0s.
If diff is greater than array.length/2 then there are more 0s than 1s.
For cases 2 & 3, initialize two pointers, start & end pointing to beginning and end of array. If we have more 1s, then move the pointers inward (start++ or end--) based on whether array[start] = 1 or array[end] = 1, and update sum accordingly. At each step check if sum = (end - start) / 2. If this condition is true, then start and end represent the bounds of your maximum subarray.
Here we end up doing two passes of the array, once to calculate sum, and once which moving the pointers inward. And we are using constant space as we just need to store sum and two index values.
If anyone wants to knock up some pseudocode, you're more than welcome :)
Here's an actionscript solution that looked like it was scaling O(n). Though it might be more like O(n log n). It definitely uses only O(1) memory.
Warning I haven't checked how complete it is. I could be missing some cases.
protected function findLongest(array:Array, start:int = 0, end:int = -1):int {
if (end < start) {
end = array.length-1;
}
var startDiff:int = 0;
var endDiff:int = 0;
var diff:int = 0;
var length:int = end-start;
for (var i:int = 0; i <= length; i++) {
if (array[i+start] == '1') {
startDiff++;
} else {
startDiff--;
}
if (array[end-i] == '1') {
endDiff++;
} else {
endDiff--;
}
//We can stop when there's no chance of equalizing anymore.
if (Math.abs(startDiff) > length - i) {
diff = endDiff;
start = end - i;
break;
} else if (Math.abs(endDiff) > length - i) {
diff = startDiff;
end = i+start;
break;
}
}
var bit:String = diff > 0 ? '1': '0';
var diffAdjustment:int = diff > 0 ? -1: 1;
//Strip off the bad vars off the ends.
while (diff != 0 && array[start] == bit) {
start++;
diff += diffAdjustment;
}
while(diff != 0 && array[end] == bit) {
end--;
diff += diffAdjustment;
}
//If we have equalized end. Otherwise recurse within the sub-array.
if (diff == 0)
return end-start+1;
else
return findLongest(array, start, end);
}
I would argue that it is impossible, that an algorithm with O(1) exists, in the following way. Assume you iterate ONCE over every bit. This requires a counter which needs the space of O(log n). Possibly one could argue that n itself is part of the problem instance, then you have as input length for a binary string of the length k: k + 2-log k. Regardless how you look over them you need an additional variable, on case you need an index into that array, that already makes it non O(1).
Usually you dont have this problem, because you have for an problem of the size n, an input of n numbers of the size log k, which adds up to nlog k. Here a variable of length log k is just O(1). But here our log k is just 1. So we can only introduce a help variable that has constant length (and I mean really constant, it must be limited regardless how big the n is).
Here one problem is the description of the problem comes visible. In computer theory you have to be very careful about your encoding. E.g. you can make NP problems polynomial if you switch to unary encoding (because then input size is exponential bigger than in a n-ary (n>1) encoding.
As for n the input has just the size 2-log n, one must be careful. When you speak in this case of O(n) - this is really an algorithm that is O(2^n) (This is no point we need to discuss about - because one can argue whether the n itself is part of the description or not).
I have this algorithm running in O(n) time and O(1) space.
It makes use of simple "shrink-then-expand" trick. Comments in codes.
public static void longestSubArrayWithSameZerosAndOnes() {
// You are given an array of 1's and 0's only.
// Find the longest subarray which contains equal number of 1's and 0's
int[] A = new int[] {1, 0, 1, 1, 1, 0, 0,0,1};
int num0 = 0, num1 = 0;
// First, calculate how many 0s and 1s in the array
for(int i = 0; i < A.length; i++) {
if(A[i] == 0) {
num0++;
}
else {
num1++;
}
}
if(num0 == 0 || num1 == 0) {
System.out.println("The length of the sub-array is 0");
return;
}
// Second, check the array to find a continuous "block" that has
// the same number of 0s and 1s, starting from the HEAD and the
// TAIL of the array, and moving the 2 "pointer" (HEAD and TAIL)
// towards the CENTER of the array
int start = 0, end = A.length - 1;
while(num0 != num1 && start < end) {
if(num1 > num0) {
if(A[start] == 1) {
num1--; start++;
}
else if(A[end] == 1) {
num1--; end--;
}
else {
num0--; start++;
num0--; end--;
}
}
else if(num1 < num0) {
if(A[start] == 0) {
num0--; start++;
}
else if(A[end] == 0) {
num0--; end--;
}
else {
num1--; start++;
num1--; end--;
}
}
}
if(num0 == 0 || num1 == 0) {
start = end;
end++;
}
// Third, expand the continuous "block" just found at step #2 by
// moving "HEAD" to head of the array and "TAIL" to the end of
// the array, while still keeping the "block" balanced(containing
// the same number of 0s and 1s
while(0 < start && end < A.length - 1) {
if(A[start - 1] == 0 && A[end + 1] == 0 || A[start - 1] == 1 && A[end + 1] == 1) {
break;
}
start--;
end++;
}
System.out.println("The length of the sub-array is " + (end - start + 1) + ", starting from #" + start + " to #" + end);
}
linear time, constant space. Let me know if there is any bug I missed.
tested in python3.
def longestBalancedSubarray(A):
lo,hi = 0,len(A)-1
ones = sum(A);zeros = len(A) - ones
while lo < hi:
if ones == zeros: break
else:
if ones > zeros:
if A[lo] == 1: lo+=1; ones-=1
elif A[hi] == 1: hi+=1; ones-=1
else: lo+=1; zeros -=1
else:
if A[lo] == 0: lo+=1; zeros-=1
elif A[hi] == 0: hi+=1; zeros-=1
else: lo+=1; ones -=1
return(A[lo:hi+1])