Related
I saw this interview question on a LinkedIn group here
To summarise, if I have an array
[1,2,3,4,5]
and input
3
I require the output
[1,2,3], [3,2,1], [2,3,1], [2,1,3], [1,3,2], [3,1,2], [2,3,4], [4,3,2],...
In no particular order.
I have been thinking about this one for a while now. I have come up with various different ways of solving but all methods use for-loops.
I think it's clear that in order to eliminate loops it must be recursive.
I thought I got close to doing this recursively partitioning the array and joining elements, but with great frustration I ended up requiring another for loop.
Im beginning to think this is impossible (which it can't be, otherwise why the interview question?).
Any ideas or links? The amount of possible outputs should be 5PN, where N is the input.
The following recursive algorithm will attempt to print every subset of {1,.., n}. These subsets are in one to one with numbers between 0 and 2^n-1 via the following bijection: to an integer x between 0 and 2^n-1, associate the set that contains 1 if the first bit of x is set to one, 2 if the second bit of x is set to one, ..
void print_all_subsets (int n, int m, int x) {
if (x==pow(2,n)) {
return;
}
else if (x has m bits set to one) {
print the set corresponding to x;
}
print_all_subsets(n,m,x+1);
}
You need to call it with n = 5 (in your case), m=3 (in your case), and x = 0.
Then you need to implement the two functions "print the set corresponding to x" and "x has m bits set to one" without for loops... but this is easily done using again recursion.
However, I think this is more of a challenge -- there is no point in completely eliminating for-loops, what makes sense is just to use them in a smart way.
Your first thought is right. Every loop can be replaced with recursion. In some languages (for example Scheme), loops are actually implemented with recursion. So just start with any solution, and keep on turning loops into recursion. Eventually you will be done.
Here is a working solution in Python.
def subsets_of_size (array, size, start=0, prepend=None):
if prepend is None:
prepend = [] # Standard Python precaution with modifiable defaults.
if 0 == size:
return [[] + prepend] # Array with one thing. The + forces a copy.
elif len(array) < start + size:
return [] # Array with no things.
else:
answer = subsets_of_size(array, size, start=start + 1, prepend=prepend)
prepend.append(array[start])
answer = answer + subsets_of_size(array, size-1, start=start + 1, prepend=prepend)
prepend.pop()
return answer
print subsets_of_size([1,2,3,4,5], 3)
I don't think the solution is not to use for-loop but there is an optimum way to use for-loop.
And so, there is the Heap's Algorithm. Below from wiki http://en.wikipedia.org/wiki/Heap%27s_algorithm
procedure generate(n : integer, A : array of any):
if n = 1 then
output(A)
else
for i := 0; i < n; i += 1 do
generate(n - 1, A)
if n is even then
swap(A[i], A[n - 1])
else
swap(A[0], A[n-1])
end if
end for
end if
define listPermutations:
input: int p_l , int[] prevP , int atElement , int[] val , int nextElement
output: list
if nextElement > length(val) OR atElement == p_l OR contains(prevP , val[nextElement]
return EMPTY
list result
int[] tmp = copy(prevP)
tmp[atElement] = val[nextElement]
add(result , tmp)
//create the next permutation stub with the last sign different to this sign
//(node with the same parent)
addAll(result , listPermutations(p_l , tmp , atElement , val , nextElement + 1))
//create the next permutation stub with an additional sign
//(child node of the current permutation
addAll(result , listPermutations(p_l , tmp , atElement + 1 , val , 0))
return result
//this will return the permutations for your example input:
listPermutations(3 , new int[3] , 0 , int[]{1 , 2 , 3 , 4 , 5} , 0)
Basic idea: all permutations of a given number of elements form a tree, where the node is the empty permutation and all childnodes of a node have one additional element. Now all the algorithm has to do is to traverse this tree level by level, until the level is equal to the required length of the permutation and list all nodes on that level
You could use recursion here, and every time you call an inner level, you give it the location it is in the array and when it returns it return an increased location. You'd be using one while loop for this.
Pseudo code:
int[] input = [1,2,3,4,5];
int level = 3;
int PrintArrayPermutation(int level, int location, string base)
{
if (level == 0)
{
print base + input[location];
return location + 1;
}
while (location <= input.Length)
{
location =
PrintArrayPermutation(level - 1, location, base + input[location]);
}
}
This is a very basic outline of my idea.
Here are two recursive functions in JavaScript. The first is the combinatorial choose function to which we apply the second function, permuting each result (permutator is adapted from the SO user, delimited's, answer here: Permutations in JavaScript?)
function c(n,list){
var result = [];
function _c(p,r){
if (p > list.length)
return
if (r.length == n){
result = result.concat(permutator(r));
} else {
var next = list[p],
_r = r.slice();
_r.push(next)
_c(p+1,_r);
_c(p+1,r);
}
}
_c(0,[])
return result;
}
function permutator(inputArr) {
var results = [];
function permute(arr, memo) {
var cur, memo = memo || [];
function _permute (i,arr,l){
if (i == l)
return
cur = arr.splice(i,1);
if (arr.length === 0){
results.push(memo.concat(cur));
}
permute(arr.slice(), memo.concat(cur));
arr.splice(i, 0, cur[0]);
_permute(i + 1,arr,l)
}
_permute(0,arr,arr.length);
return results;
}
return permute(inputArr);
}
Output:
console.log(c(3,[1,2,3,4,5]))
[[1,2,3],[1,3,2],[2,1,3]...[4,5,3],[5,3,4],[5,4,3]]
I have to implement a divide-and-conquer algorithm in C++ for a max function, which returns the maximum value in an array. I understand the algorithm and have designed the function already, but I am running into issues with the array indices.
In pseudocode, here is my function:
def max(array, startIndex, endIndex)
// if there is only one element, return it
if startIdx = endIdx
return array[startIdx];
leftHigh = max(array, startIdx, endIdx/2);
rightHigh = max(array, endIdx/2 + 1, endIdx);
return maximum of leftHigh and rightHigh;
However, I run into an issue with these values for the recursive call parameters. The following paragraph demonstrates what I found when I mentally stepped through the algorithm:
The simplest case is an array of 4 elements. The first call to max will take index parameters 0, 3, and will make the calls with parameters 0, 1 and 2, 3. The first recursive call will result in calls with 0, 0 and 1, 1 which will terminate correctly. However, the second recursive call will result in calls with 2, 1 and 2, 3. The first eventually results in overstepping the array bounds and the second results in an infinite loop since those parameters have already been used.
I have tried messing with it, for example using (startIdx, endIdx/2 -1) for the first bounds and (endIdx/2, endIdx) for the second bounds, and this fixes the second branch of recursive calls but messes up the first.
Is there a way to find these indices resulting in the correct behavior? I appreciate the help.
It should be
leftHigh = max(array, startIdx, (startIdx + endIdx)/2);
rightHigh = max(array, (startIdx + endIdx)/2 + 1, endIdx);
If I give you two numbers: a < c, then how would you characterise any numbers in between the two. I.E, what can we say about b when a < b < c ?
a < b < c
0 < b - a < c - a
You are picking b = c/2. Which we can see only meets the above criteria in some cases:
b = c / 2
and b > a
then c / 2 > a
So we can see that your method works as long as a > c/2. In your case a = startIdx and c = endIdx, so your algorithm works only while startIdx < endIdx / 2.
Consider this finding carefully:
a < b < c
0 < b - a < c - a (subtract a from all parts)
If b is half-way between a and c, then what is its value? In this case, how does (b - a) relate to (c - a)?
Suggested solution in Python:
from math import floor, ceil
def maximum(arr, left, right):
if left >= right:
if left < len(arr):
return arr[left]
else:
return arr[right]
else:
left = maximum(arr, left, int((left+right)/2)) # pay attention to the midpoint!
right = maximum(arr, int((left+right)/2), right)
return max(left, right)
print maximum([1,8,2,9,3,15,5,3,2], 0, 8)
OUTPUT
15
I'm having a trouble with what should be a very simple algorithm, but for some reason my head is not working properly (too much work?)
I have an Array of numeric values: [ 10, 20, 30, 40, 100, 1000, 5000, 100000] I want to check which is the next "item" in the array.
For example,
given the number 10, my algorithm should return 10.
given the number 1, my algorithm should return 10
given the number 50, my algorithm should return 100.
given the number 99999999, my algorithm should return 100000
In pseudo-code I was thinking:
for previousValue, nextValue in values:
if ( previousValue < value && nextValue >= value ):
return nextValue
return values[max]
If anyone can point out to my exhausted brain what I'm missing it would be great. thanks!
Assuming the array is already sorted, your solution works fine. But to make it more clear I would rewrite it as follows:
prev = 0
for(v in values){
if(target > prev && target <= v)
return v
prev = v
}
return values[values.length-1]
In ruby :
find 0,values.length,val
def find left,right,val
if left == right || left > right then return values[left]
if val > values[(right + left)/2]
find (right + left)/2, right, val
else
find left, (right + left)/2, val
end
end
the array need not be sorted. sorting an array has O(nlogn) time complexity. You can check each element in the array in theta(n) time.
Assuming the array is arr and the comparison item is item and the array arr contains at least one element greater than or equal to item:
In java
public static int find_greater(int[] arr,int item)
{
int j=0;
int previous =0;
for(int i=0;i<arr.length;i++){
if(arr[i]>=item){
if(j==0){
previous = arr[i];
}else{
if(arr[i]<previous)
{
previous = arr[i];
}
}
j++;
}
}
return previous;
}
Here I go through the array and keep a variable previous which is the smallest number greater or equal item.
The running time for this procedure is theta(N) where N is arr.length.
I would probably represent this array as a tree.
To find out the closest biggest value, I will just look in the tree the same way I will do to find a key, and always keep track of my closest value. When I reach the end of the node, I return the closest value.
This is another alternative, compared to the other answers. You can check it in this thread How to find the closest element to a given key value in a binary search tree?
Saw this question recently:
Given 2 arrays, the 2nd array containing some of the elements of the 1st array, return the minimum window in the 1st array which contains all the elements of the 2nd array.
Eg :
Given A={1,3,5,2,3,1} and B={1,3,2}
Output : 3 , 5 (where 3 and 5 are indices in the array A)
Even though the range 1 to 4 also contains the elements of A, the range 3 to 5 is returned Since it contains since its length is lesser than the previous range ( ( 5 - 3 ) < ( 4 - 1 ) )
I had devised a solution but I am not sure if it works correctly and also not efficient.
Give an Efficient Solution for the problem. Thanks in Advance
A simple solution of iterating through the list.
Have a left and right pointer, initially both at zero
Move the right pointer forwards until [L..R] contains all the elements (or quit if right reaches the end).
Move the left pointer forwards until [L..R] doesn't contain all the elements. See if [L-1..R] is shorter than the current best.
This is obviously linear time. You'll simply need to keep track of how many of each element of B is in the subarray for checking whether the subarray is a potential solution.
Pseudocode of this algorithm.
size = bestL = A.length;
needed = B.length-1;
found = 0; left=0; right=0;
counts = {}; //counts is a map of (number, count)
for(i in B) counts.put(i, 0);
//Increase right bound
while(right < size) {
if(!counts.contains(right)) continue;
amt = count.get(right);
count.set(right, amt+1);
if(amt == 0) found++;
if(found == needed) {
while(found == needed) {
//Increase left bound
if(counts.contains(left)) {
amt = count.get(left);
count.set(left, amt-1);
if(amt == 1) found--;
}
left++;
}
if(right - left + 2 >= bestL) continue;
bestL = right - left + 2;
bestRange = [left-1, right] //inclusive
}
}
An interesting interview question that a colleague of mine uses:
Suppose that you are given a very long, unsorted list of unsigned 64-bit integers. How would you find the smallest non-negative integer that does not occur in the list?
FOLLOW-UP: Now that the obvious solution by sorting has been proposed, can you do it faster than O(n log n)?
FOLLOW-UP: Your algorithm has to run on a computer with, say, 1GB of memory
CLARIFICATION: The list is in RAM, though it might consume a large amount of it. You are given the size of the list, say N, in advance.
If the datastructure can be mutated in place and supports random access then you can do it in O(N) time and O(1) additional space. Just go through the array sequentially and for every index write the value at the index to the index specified by value, recursively placing any value at that location to its place and throwing away values > N. Then go again through the array looking for the spot where value doesn't match the index - that's the smallest value not in the array. This results in at most 3N comparisons and only uses a few values worth of temporary space.
# Pass 1, move every value to the position of its value
for cursor in range(N):
target = array[cursor]
while target < N and target != array[target]:
new_target = array[target]
array[target] = target
target = new_target
# Pass 2, find first location where the index doesn't match the value
for cursor in range(N):
if array[cursor] != cursor:
return cursor
return N
Here's a simple O(N) solution that uses O(N) space. I'm assuming that we are restricting the input list to non-negative numbers and that we want to find the first non-negative number that is not in the list.
Find the length of the list; lets say it is N.
Allocate an array of N booleans, initialized to all false.
For each number X in the list, if X is less than N, set the X'th element of the array to true.
Scan the array starting from index 0, looking for the first element that is false. If you find the first false at index I, then I is the answer. Otherwise (i.e. when all elements are true) the answer is N.
In practice, the "array of N booleans" would probably be encoded as a "bitmap" or "bitset" represented as a byte or int array. This typically uses less space (depending on the programming language) and allows the scan for the first false to be done more quickly.
This is how / why the algorithm works.
Suppose that the N numbers in the list are not distinct, or that one or more of them is greater than N. This means that there must be at least one number in the range 0 .. N - 1 that is not in the list. So the problem of find the smallest missing number must therefore reduce to the problem of finding the smallest missing number less than N. This means that we don't need to keep track of numbers that are greater or equal to N ... because they won't be the answer.
The alternative to the previous paragraph is that the list is a permutation of the numbers from 0 .. N - 1. In this case, step 3 sets all elements of the array to true, and step 4 tells us that the first "missing" number is N.
The computational complexity of the algorithm is O(N) with a relatively small constant of proportionality. It makes two linear passes through the list, or just one pass if the list length is known to start with. There is no need to represent the hold the entire list in memory, so the algorithm's asymptotic memory usage is just what is needed to represent the array of booleans; i.e. O(N) bits.
(By contrast, algorithms that rely on in-memory sorting or partitioning assume that you can represent the entire list in memory. In the form the question was asked, this would require O(N) 64-bit words.)
#Jorn comments that steps 1 through 3 are a variation on counting sort. In a sense he is right, but the differences are significant:
A counting sort requires an array of (at least) Xmax - Xmin counters where Xmax is the largest number in the list and Xmin is the smallest number in the list. Each counter has to be able to represent N states; i.e. assuming a binary representation it has to have an integer type (at least) ceiling(log2(N)) bits.
To determine the array size, a counting sort needs to make an initial pass through the list to determine Xmax and Xmin.
The minimum worst-case space requirement is therefore ceiling(log2(N)) * (Xmax - Xmin) bits.
By contrast, the algorithm presented above simply requires N bits in the worst and best cases.
However, this analysis leads to the intuition that if the algorithm made an initial pass through the list looking for a zero (and counting the list elements if required), it would give a quicker answer using no space at all if it found the zero. It is definitely worth doing this if there is a high probability of finding at least one zero in the list. And this extra pass doesn't change the overall complexity.
EDIT: I've changed the description of the algorithm to use "array of booleans" since people apparently found my original description using bits and bitmaps to be confusing.
Since the OP has now specified that the original list is held in RAM and that the computer has only, say, 1GB of memory, I'm going to go out on a limb and predict that the answer is zero.
1GB of RAM means the list can have at most 134,217,728 numbers in it. But there are 264 = 18,446,744,073,709,551,616 possible numbers. So the probability that zero is in the list is 1 in 137,438,953,472.
In contrast, my odds of being struck by lightning this year are 1 in 700,000. And my odds of getting hit by a meteorite are about 1 in 10 trillion. So I'm about ten times more likely to be written up in a scientific journal due to my untimely death by a celestial object than the answer not being zero.
As pointed out in other answers you can do a sort, and then simply scan up until you find a gap.
You can improve the algorithmic complexity to O(N) and keep O(N) space by using a modified QuickSort where you eliminate partitions which are not potential candidates for containing the gap.
On the first partition phase, remove duplicates.
Once the partitioning is complete look at the number of items in the lower partition
Is this value equal to the value used for creating the partition?
If so then it implies that the gap is in the higher partition.
Continue with the quicksort, ignoring the lower partition
Otherwise the gap is in the lower partition
Continue with the quicksort, ignoring the higher partition
This saves a large number of computations.
To illustrate one of the pitfalls of O(N) thinking, here is an O(N) algorithm that uses O(1) space.
for i in [0..2^64):
if i not in list: return i
print "no 64-bit integers are missing"
Since the numbers are all 64 bits long, we can use radix sort on them, which is O(n). Sort 'em, then scan 'em until you find what you're looking for.
if the smallest number is zero, scan forward until you find a gap. If the smallest number is not zero, the answer is zero.
For a space efficient method and all values are distinct you can do it in space O( k ) and time O( k*log(N)*N ). It's space efficient and there's no data moving and all operations are elementary (adding subtracting).
set U = N; L=0
First partition the number space in k regions. Like this:
0->(1/k)*(U-L) + L, 0->(2/k)*(U-L) + L, 0->(3/k)*(U-L) + L ... 0->(U-L) + L
Find how many numbers (count{i}) are in each region. (N*k steps)
Find the first region (h) that isn't full. That means count{h} < upper_limit{h}. (k steps)
if h - count{h-1} = 1 you've got your answer
set U = count{h}; L = count{h-1}
goto 2
this can be improved using hashing (thanks for Nic this idea).
same
First partition the number space in k regions. Like this:
L + (i/k)->L + (i+1/k)*(U-L)
inc count{j} using j = (number - L)/k (if L < number < U)
find first region (h) that doesn't have k elements in it
if count{h} = 1 h is your answer
set U = maximum value in region h L = minimum value in region h
This will run in O(log(N)*N).
I'd just sort them then run through the sequence until I find a gap (including the gap at the start between zero and the first number).
In terms of an algorithm, something like this would do it:
def smallest_not_in_list(list):
sort(list)
if list[0] != 0:
return 0
for i = 1 to list.last:
if list[i] != list[i-1] + 1:
return list[i-1] + 1
if list[list.last] == 2^64 - 1:
assert ("No gaps")
return list[list.last] + 1
Of course, if you have a lot more memory than CPU grunt, you could create a bitmask of all possible 64-bit values and just set the bits for every number in the list. Then look for the first 0-bit in that bitmask. That turns it into an O(n) operation in terms of time but pretty damned expensive in terms of memory requirements :-)
I doubt you could improve on O(n) since I can't see a way of doing it that doesn't involve looking at each number at least once.
The algorithm for that one would be along the lines of:
def smallest_not_in_list(list):
bitmask = mask_make(2^64) // might take a while :-)
mask_clear_all (bitmask)
for i = 1 to list.last:
mask_set (bitmask, list[i])
for i = 0 to 2^64 - 1:
if mask_is_clear (bitmask, i):
return i
assert ("No gaps")
Sort the list, look at the first and second elements, and start going up until there is a gap.
We could use a hash table to hold the numbers. Once all numbers are done, run a counter from 0 till we find the lowest. A reasonably good hash will hash and store in constant time, and retrieves in constant time.
for every i in X // One scan Θ(1)
hashtable.put(i, i); // O(1)
low = 0;
while (hashtable.get(i) <> null) // at most n+1 times
low++;
print low;
The worst case if there are n elements in the array, and are {0, 1, ... n-1}, in which case, the answer will be obtained at n, still keeping it O(n).
You can do it in O(n) time and O(1) additional space, although the hidden factor is quite large. This isn't a practical way to solve the problem, but it might be interesting nonetheless.
For every unsigned 64-bit integer (in ascending order) iterate over the list until you find the target integer or you reach the end of the list. If you reach the end of the list, the target integer is the smallest integer not in the list. If you reach the end of the 64-bit integers, every 64-bit integer is in the list.
Here it is as a Python function:
def smallest_missing_uint64(source_list):
the_answer = None
target = 0L
while target < 2L**64:
target_found = False
for item in source_list:
if item == target:
target_found = True
if not target_found and the_answer is None:
the_answer = target
target += 1L
return the_answer
This function is deliberately inefficient to keep it O(n). Note especially that the function keeps checking target integers even after the answer has been found. If the function returned as soon as the answer was found, the number of times the outer loop ran would be bound by the size of the answer, which is bound by n. That change would make the run time O(n^2), even though it would be a lot faster.
Thanks to egon, swilden, and Stephen C for my inspiration. First, we know the bounds of the goal value because it cannot be greater than the size of the list. Also, a 1GB list could contain at most 134217728 (128 * 2^20) 64-bit integers.
Hashing part
I propose using hashing to dramatically reduce our search space. First, square root the size of the list. For a 1GB list, that's N=11,586. Set up an integer array of size N. Iterate through the list, and take the square root* of each number you find as your hash. In your hash table, increment the counter for that hash. Next, iterate through your hash table. The first bucket you find that is not equal to it's max size defines your new search space.
Bitmap part
Now set up a regular bit map equal to the size of your new search space, and again iterate through the source list, filling out the bitmap as you find each number in your search space. When you're done, the first unset bit in your bitmap will give you your answer.
This will be completed in O(n) time and O(sqrt(n)) space.
(*You could use use something like bit shifting to do this a lot more efficiently, and just vary the number and size of buckets accordingly.)
Well if there is only one missing number in a list of numbers, the easiest way to find the missing number is to sum the series and subtract each value in the list. The final value is the missing number.
int i = 0;
while ( i < Array.Length)
{
if (Array[i] == i + 1)
{
i++;
}
if (i < Array.Length)
{
if (Array[i] <= Array.Length)
{//SWap
int temp = Array[i];
int AnoTemp = Array[temp - 1];
Array[temp - 1] = temp;
Array[i] = AnoTemp;
}
else
i++;
}
}
for (int j = 0; j < Array.Length; j++)
{
if (Array[j] > Array.Length)
{
Console.WriteLine(j + 1);
j = Array.Length;
}
else
if (j == Array.Length - 1)
Console.WriteLine("Not Found !!");
}
}
Here's my answer written in Java:
Basic Idea:
1- Loop through the array throwing away duplicate positive, zeros, and negative numbers while summing up the rest, getting the maximum positive number as well, and keep the unique positive numbers in a Map.
2- Compute the sum as max * (max+1)/2.
3- Find the difference between the sums calculated at steps 1 & 2
4- Loop again from 1 to the minimum of [sums difference, max] and return the first number that is not in the map populated in step 1.
public static int solution(int[] A) {
if (A == null || A.length == 0) {
throw new IllegalArgumentException();
}
int sum = 0;
Map<Integer, Boolean> uniqueNumbers = new HashMap<Integer, Boolean>();
int max = A[0];
for (int i = 0; i < A.length; i++) {
if(A[i] < 0) {
continue;
}
if(uniqueNumbers.get(A[i]) != null) {
continue;
}
if (A[i] > max) {
max = A[i];
}
uniqueNumbers.put(A[i], true);
sum += A[i];
}
int completeSum = (max * (max + 1)) / 2;
for(int j = 1; j <= Math.min((completeSum - sum), max); j++) {
if(uniqueNumbers.get(j) == null) { //O(1)
return j;
}
}
//All negative case
if(uniqueNumbers.isEmpty()) {
return 1;
}
return 0;
}
As Stephen C smartly pointed out, the answer must be a number smaller than the length of the array. I would then find the answer by binary search. This optimizes the worst case (so the interviewer can't catch you in a 'what if' pathological scenario). In an interview, do point out you are doing this to optimize for the worst case.
The way to use binary search is to subtract the number you are looking for from each element of the array, and check for negative results.
I like the "guess zero" apprach. If the numbers were random, zero is highly probable. If the "examiner" set a non-random list, then add one and guess again:
LowNum=0
i=0
do forever {
if i == N then leave /* Processed entire array */
if array[i] == LowNum {
LowNum++
i=0
}
else {
i++
}
}
display LowNum
The worst case is n*N with n=N, but in practice n is highly likely to be a small number (eg. 1)
I am not sure if I got the question. But if for list 1,2,3,5,6 and the missing number is 4, then the missing number can be found in O(n) by:
(n+2)(n+1)/2-(n+1)n/2
EDIT: sorry, I guess I was thinking too fast last night. Anyway, The second part should actually be replaced by sum(list), which is where O(n) comes. The formula reveals the idea behind it: for n sequential integers, the sum should be (n+1)*n/2. If there is a missing number, the sum would be equal to the sum of (n+1) sequential integers minus the missing number.
Thanks for pointing out the fact that I was putting some middle pieces in my mind.
Well done Ants Aasma! I thought about the answer for about 15 minutes and independently came up with an answer in a similar vein of thinking to yours:
#define SWAP(x,y) { numerictype_t tmp = x; x = y; y = tmp; }
int minNonNegativeNotInArr (numerictype_t * a, size_t n) {
int m = n;
for (int i = 0; i < m;) {
if (a[i] >= m || a[i] < i || a[i] == a[a[i]]) {
m--;
SWAP (a[i], a[m]);
continue;
}
if (a[i] > i) {
SWAP (a[i], a[a[i]]);
continue;
}
i++;
}
return m;
}
m represents "the current maximum possible output given what I know about the first i inputs and assuming nothing else about the values until the entry at m-1".
This value of m will be returned only if (a[i], ..., a[m-1]) is a permutation of the values (i, ..., m-1). Thus if a[i] >= m or if a[i] < i or if a[i] == a[a[i]] we know that m is the wrong output and must be at least one element lower. So decrementing m and swapping a[i] with the a[m] we can recurse.
If this is not true but a[i] > i then knowing that a[i] != a[a[i]] we know that swapping a[i] with a[a[i]] will increase the number of elements in their own place.
Otherwise a[i] must be equal to i in which case we can increment i knowing that all the values of up to and including this index are equal to their index.
The proof that this cannot enter an infinite loop is left as an exercise to the reader. :)
The Dafny fragment from Ants' answer shows why the in-place algorithm may fail. The requires pre-condition describes that the values of each item must not go beyond the bounds of the array.
method AntsAasma(A: array<int>) returns (M: int)
requires A != null && forall N :: 0 <= N < A.Length ==> 0 <= A[N] < A.Length;
modifies A;
{
// Pass 1, move every value to the position of its value
var N := A.Length;
var cursor := 0;
while (cursor < N)
{
var target := A[cursor];
while (0 <= target < N && target != A[target])
{
var new_target := A[target];
A[target] := target;
target := new_target;
}
cursor := cursor + 1;
}
// Pass 2, find first location where the index doesn't match the value
cursor := 0;
while (cursor < N)
{
if (A[cursor] != cursor)
{
return cursor;
}
cursor := cursor + 1;
}
return N;
}
Paste the code into the validator with and without the forall ... clause to see the verification error. The second error is a result of the verifier not being able to establish a termination condition for the Pass 1 loop. Proving this is left to someone who understands the tool better.
Here's an answer in Java that does not modify the input and uses O(N) time and N bits plus a small constant overhead of memory (where N is the size of the list):
int smallestMissingValue(List<Integer> values) {
BitSet bitset = new BitSet(values.size() + 1);
for (int i : values) {
if (i >= 0 && i <= values.size()) {
bitset.set(i);
}
}
return bitset.nextClearBit(0);
}
def solution(A):
index = 0
target = []
A = [x for x in A if x >=0]
if len(A) ==0:
return 1
maxi = max(A)
if maxi <= len(A):
maxi = len(A)
target = ['X' for x in range(maxi+1)]
for number in A:
target[number]= number
count = 1
while count < maxi+1:
if target[count] == 'X':
return count
count +=1
return target[count-1] + 1
Got 100% for the above solution.
1)Filter negative and Zero
2)Sort/distinct
3)Visit array
Complexity: O(N) or O(N * log(N))
using Java8
public int solution(int[] A) {
int result = 1;
boolean found = false;
A = Arrays.stream(A).filter(x -> x > 0).sorted().distinct().toArray();
//System.out.println(Arrays.toString(A));
for (int i = 0; i < A.length; i++) {
result = i + 1;
if (result != A[i]) {
found = true;
break;
}
}
if (!found && result == A.length) {
//result is larger than max element in array
result++;
}
return result;
}
An unordered_set can be used to store all the positive numbers, and then we can iterate from 1 to length of unordered_set, and see the first number that does not occur.
int firstMissingPositive(vector<int>& nums) {
unordered_set<int> fre;
// storing each positive number in a hash.
for(int i = 0; i < nums.size(); i +=1)
{
if(nums[i] > 0)
fre.insert(nums[i]);
}
int i = 1;
// Iterating from 1 to size of the set and checking
// for the occurrence of 'i'
for(auto it = fre.begin(); it != fre.end(); ++it)
{
if(fre.find(i) == fre.end())
return i;
i +=1;
}
return i;
}
Solution through basic javascript
var a = [1, 3, 6, 4, 1, 2];
function findSmallest(a) {
var m = 0;
for(i=1;i<=a.length;i++) {
j=0;m=1;
while(j < a.length) {
if(i === a[j]) {
m++;
}
j++;
}
if(m === 1) {
return i;
}
}
}
console.log(findSmallest(a))
Hope this helps for someone.
With python it is not the most efficient, but correct
#!/usr/bin/env python3
# -*- coding: UTF-8 -*-
import datetime
# write your code in Python 3.6
def solution(A):
MIN = 0
MAX = 1000000
possible_results = range(MIN, MAX)
for i in possible_results:
next_value = (i + 1)
if next_value not in A:
return next_value
return 1
test_case_0 = [2, 2, 2]
test_case_1 = [1, 3, 44, 55, 6, 0, 3, 8]
test_case_2 = [-1, -22]
test_case_3 = [x for x in range(-10000, 10000)]
test_case_4 = [x for x in range(0, 100)] + [x for x in range(102, 200)]
test_case_5 = [4, 5, 6]
print("---")
a = datetime.datetime.now()
print(solution(test_case_0))
print(solution(test_case_1))
print(solution(test_case_2))
print(solution(test_case_3))
print(solution(test_case_4))
print(solution(test_case_5))
def solution(A):
A.sort()
j = 1
for i, elem in enumerate(A):
if j < elem:
break
elif j == elem:
j += 1
continue
else:
continue
return j
this can help:
0- A is [5, 3, 2, 7];
1- Define B With Length = A.Length; (O(1))
2- initialize B Cells With 1; (O(n))
3- For Each Item In A:
if (B.Length <= item) then B[Item] = -1 (O(n))
4- The answer is smallest index in B such that B[index] != -1 (O(n))