Finding maximum for every window of size k in an array

Question

Given an array of size n and k, how do you find the maximum for every contiguous subarray of size k?

For example

arr = 1 5 2 6 3 1 24 7
k = 3
ans = 5 6 6 6 24 24

I was thinking of having an array of size k and each step evict the last element out and add the new element and find maximum among that. It leads to a running time of O(nk). Is there a better way to do this?

Answer 1

You have heard about doing it in O(n) using dequeue.

Well that is a well known algorithm for this question to do in O(n).

The method i am telling is quite simple and has time complexity O(n).

Your Sample Input:
n=10 , W = 3

10 3
1 -2 5 6 0 9 8 -1 2 0

Answer = 5 6 6 9 9 9 8 2

Concept: Dynamic Programming

Algorithm:

1. N is number of elements in an array and W is window size. So, Window number = N-W+1

2. Now divide array into blocks of W starting from index 1.

   Here divide into blocks of size 'W'=3. For your sample input:

3. We have divided into blocks because we will calculate maximum in 2 ways A.) by traversing from left to right B.) by traversing from right to left. but how ??

4. Firstly, Traversing from Left to Right. For each element ai in block we will find maximum till that element ai starting from START of Block to END of that block. So here,

5. Secondly, Traversing from Right to Left. For each element 'ai' in block we will find maximum till that element 'ai' starting from END of Block to START of that block. So Here,

6. Now we have to find maximum for each subarray or window of size 'W'. So, starting from index = 1 to index = N-W+1 .

   max_val[index] = max(RL[index], LR[index+w-1]);

    for index=1: max_val[1] = max(RL[1],LR[3]) = max(5,5)= 5
   

Simliarly, for all index i, (i<=(n-k+1)), value at RL[i] and LR[i+w-1] are compared and maximum among those two is answer for that subarray.

So Final Answer : 5 6 6 9 9 9 8 2

Time Complexity: O(n)

Implementation code:

// Shashank Jain
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

#define LIM 100001 

using namespace std;

int arr[LIM]; // Input Array
int LR[LIM]; // maximum from Left to Right
int RL[LIM]; // maximum from Right to left
int max_val[LIM]; // number of subarrays(windows) will be n-k+1

int main(){
    int n, w, i, k; // 'n' is number of elements in array
                    // 'w' is Window's Size 
    cin >> n >> w;

    k = n - w + 1; // 'K' is number of Windows

    for(i = 1; i <= n; i++)
        cin >> arr[i];

    for(i = 1; i <= n; i++){ // for maximum Left to Right
        if(i % w == 1) // that means START of a block
            LR[i] = arr[i];
        else
            LR[i] = max(LR[i - 1], arr[i]);        
    }

    for(i = n; i >= 1; i--){ // for maximum Right to Left
        if(i == n) // Maybe the last block is not of size 'W'. 
            RL[i] = arr[i]; 
        else if(i % w == 0) // that means END of a block
            RL[i] = arr[i];
        else
            RL[i] = max(RL[i+1], arr[i]);
    }

    for(i = 1; i <= k; i++)    // maximum
        max_val[i] = max(RL[i], LR[i + w - 1]);

    for(i = 1; i <= k ; i++)
        cout << max_val[i] << " ";

    cout << endl;

    return 0;
}  

Running Code Link

---

I'll try to proof: (by @johnchen902)

If k % w != 1 (k is not the begin of a block)

Let k* = The begin of block containing k
ans[k] = max( arr[k], arr[k + 1], arr[k + 2], ..., arr[k + w - 1])
       = max( max( arr[k],  arr[k + 1],  arr[k + 2],  ..., arr[k*]), 
              max( arr[k*], arr[k* + 1], arr[k* + 2], ..., arr[k + w - 1]) )
       = max( RL[k], LR[k+w-1] )

Otherwise (k is the begin of a block)

ans[k] = max( arr[k], arr[k + 1], arr[k + 2], ..., arr[k + w - 1])
       = RL[k] = LR[k+w-1]
       = max( RL[k], LR[k+w-1] )

Answer 2

Dynamic programming approach is very neatly explained by Shashank Jain. I would like to explain how to do the same using dequeue. The key is to maintain the max element at the top of the queue(for a window ) and discarding the useless elements and we also need to discard the elements that are out of index of current window.
useless elements = If Current element is greater than the last element of queue than the last element of queue is useless .
Note : We are storing the index in queue not the element itself. It will be more clear from the code itself.
1. If Current element is greater than the last element of queue than the last element of queue is useless . We need to delete that last element. (and keep deleting until the last element of queue is smaller than current element).
2. If if current_index - k >= q.front() that means we are going out of window so we need to delete the element from front of queue.

vector<int> max_sub_deque(vector<int> &A,int k)
{
    deque<int> q;
    for(int i=0;i<k;i++)
    {
        while(!q.empty() && A[i] >= A[q.back()])
            q.pop_back();
        q.push_back(i);
    }
    vector<int> res;
    for(int i=k;i<A.size();i++)
    {
        res.push_back(A[q.front()]);
        while(!q.empty() && A[i] >= A[q.back()] )
            q.pop_back();
        while(!q.empty() && q.front() <= i-k)
            q.pop_front();
        q.push_back(i); 
    }
    res.push_back(A[q.front()]);
    return res;
}

Since each element is enqueued and dequeued atmost 1 time to time complexity is
O(n+n) = O(2n) = O(n).
And the size of queue can not exceed the limit k . so space complexity = O(k).

Answer 3

An O(n) time solution is possible by combining the two classic interview questions:

• Make a stack data-structure (called MaxStack) which supports push, pop and max in O(1) time.
  

  This can be done using two stacks, the second one contains the minimum seen so far.

• Model a queue with a stack.
  

  This can done using two stacks. Enqueues go into one stack, and dequeues come from the other.

For this problem, we basically need a queue, which supports enqueue, dequeue and max in O(1) (amortized) time.

We combine the above two, by modelling a queue with two MaxStacks.

To solve the question, we queue k elements, query the max, dequeue, enqueue k+1 th element, query the max etc. This will give you the max for every k sized sub-array.

I believe there are other solutions too.

1)

I believe the queue idea can be simplified. We maintain a queue and a max for every k. We enqueue a new element, and dequeu all elements which are not greater than the new element.

2) Maintain two new arrays which maintain the running max for each block of k, one array for one direction (left to right/right to left).

3) Use a hammer: Preprocess in O(n) time for range maximum queries.

The 1) solution above might be the most optimal.

Answer 4

You need a fast data structure that can add, remove and query for the max element in less than O(n) time (you can just use an array if O(n) or O(nlogn) is acceptable). You can use a heap, a balanced binary search tree, a skip list, or any other sorted data structure that performs these operations in O(log(n)).

The good news is that most popular languages have a sorted data structure implemented that supports these operations for you. C++ has std::set and std::multiset (you probably need the latter) and Java has PriorityQueue and TreeSet.

Answer 5

Here is the java implementation

public static Integer[] maxsInEveryWindows(int[] arr, int k) {
    Deque<Integer> deque = new ArrayDeque<Integer>();
    /* Process first k (or first window) elements of array */
    for (int i = 0; i < k; i++) {
        // For very element, the previous smaller elements are useless so
        // remove them from deque
        while (!deque.isEmpty() && arr[i] >= arr[deque.peekLast()]) {
            deque.removeLast(); // Remove from rear
        }
        // Add new element at rear of queue
        deque.addLast(i);
    }
    List<Integer> result = new ArrayList<Integer>();
    // Process rest of the elements, i.e., from arr[k] to arr[n-1]
    for (int i = k; i < arr.length; i++) {
        // The element at the front of the queue is the largest element of
        // previous window, so add to result.
        result.add(arr[deque.getFirst()]);
        // Remove all elements smaller than the currently
        // being added element (remove useless elements)
        while (!deque.isEmpty() && arr[i] >= arr[deque.peekLast()]) {
            deque.removeLast();
        }
        // Remove the elements which are out of this window
        while (!deque.isEmpty() && deque.getFirst() <= i - k) {
            deque.removeFirst();
        }
        // Add current element at the rear of deque
        deque.addLast(i);
    }
    // Print the maximum element of last window
    result.add(arr[deque.getFirst()]);

    return result.toArray(new Integer[0]);
}

Here is the corresponding test case

@Test
public void maxsInWindowsOfSizeKTest() {
    Integer[] result = ArrayUtils.maxsInEveryWindows(new int[]{1, 2, 3, 1, 4, 5, 2, 3, 6}, 3);
    assertThat(result, equalTo(new Integer[]{3, 3, 4, 5, 5, 5, 6}));

    result = ArrayUtils.maxsInEveryWindows(new int[]{8, 5, 10, 7, 9, 4, 15, 12, 90, 13}, 4);
    assertThat(result, equalTo(new Integer[]{10, 10, 10, 15, 15, 90, 90}));
}

Answer 6

Using a heap (or tree), you should be able to do it in O(n * log(k)). I'm not sure if this would be indeed better.

Answer 7

here is the Python implementation in O(1)...Thanks to @Shahshank Jain in advance..

from sys import stdin,stdout
from operator import *
n,w=map(int , stdin.readline().strip().split())
Arr=list(map(int , stdin.readline().strip().split()))
k=n-w+1           # window size = k
leftA=[0]*n
rightA=[0]*n
result=[0]*k
for i in range(n):
    if i%w==0:
        leftA[i]=Arr[i]
    else:
        leftA[i]=max(Arr[i],leftA[i-1])
for i in range(n-1,-1,-1):
    if i%w==(w-1) or i==n-1:
        rightA[i]=Arr[i]
    else:
        rightA[i]=max(Arr[i],rightA[i+1])

for i in range(k):
    result[i]=max(rightA[i],leftA[i+w-1])
print(*result,sep=' ')

Answer 8

Method 1: O(n) time, O(k) space
We use a deque (it is like a list but with constant-time insertion and deletion from both ends) to store the index of useful elements.
The index of the current max is kept at the leftmost element of deque. The rightmost element of deque is the smallest. In the following, for easier explanation we say an element from the array is in the deque, while in fact the index of that element is in the deque.

Let's say {5, 3, 2} are already in the deque (again, if fact their indexes are).

• If the next element we read from the array is bigger than 5 (remember, the leftmost element of deque holds the max), say 7: We delete the deque and create a new one with only 7 in it (we do this because the current elements are useless, we have found a new max).

• If the next element is less than 2 (which is the smallest element of deque), say 1: We add it to the right ({5, 3, 2, 1})

• If the next element is bigger than 2 but less than 5, say 4: We remove elements from right that are smaller than the element and then add the element from right ({5, 4}).

• Also we keep elements of the current window only (we can do this in constant time because we are storing the indexes instead of elements).

from collections import deque

def max_subarray(array, k):
  deq = deque()

  for index, item in enumerate(array):

    if len(deq) == 0:
      deq.append(index)

    elif index - deq[0] >= k:  # the max element is out of the window
      deq.popleft()

    elif item > array[deq[0]]:  # found a new max
      deq = deque()
      deq.append(index)

    elif item < array[deq[-1]]:  # the array item is smaller than all the deque elements
      deq.append(index)

    elif item > array[deq[-1]] and item < array[deq[0]]:
      while item > array[deq[-1]]:
        deq.pop()
      deq.append(index)

    if index >= k - 1:  # start printing when the first window is filled
      print(array[deq[0]])

Proof of O(n) time: The only part we need to check is the while loop. In the whole runtime of the code, the while loop can perform at most O(n) operations in total. The reason is that the while loop pops elements from the deque, and since in other parts of the code, we do at most O(n) insertions into the deque, the while loop cannot exceed O(n) operations in total. So the total runtime is O(n) + O(n) = O(n)

---

Method 2: O(n) time, O(n) space
This is the explanation of the method suggested by S Jain (as mentioned in the comments of his post, this method doesn't work with data streams, which most sliding window questions are designed for).

The reason that method works is explained using the following example:

array = [5, 6, 2, 3, 1, 4, 2, 3]
k = 4

    [5, 6, 2, 3   1, 4, 2, 3 ]
LR:  5  6  6  6   1  4  4  4 
RL:  6  6  3  3   4  4  3  3  
     6  6  4  4   4  

To get the max for the window [2, 3, 1, 4], we can get the max of [2, 3] and max of [1, 4], and return the bigger of the two. Max of [2, 3] is calculated in the RL pass and max of [1, 4] is calculated in LR pass.

Answer 9

Using Fibonacci heap, you can do it in O(n + (n-k) log k), which is equal to O(n log k) for small k, for k close to n this becomes O(n).

The algorithm: in fact, you need:

• n inserts to the heap
• n-k deletions
• n-k findmax's

How much these operations cost in Fibonacci heaps? Insert and findmax is O(1) amortized, deletion is O(log n) amortized. So, we have

O(n + (n-k) log k + (n-k)) = O(n + (n-k) log k)

Answer 10

Sorry, this should have been a comment but I am not allowed to comment for now. @leo and @Clay Goddard You can save yourselves from re-computing the maximum by storing both maximum and 2nd maximum of the window in the beginning (2nd maximum will be the maximum only if there are two maximums in the initial window). If the maximum slides out of the window you still have the next best candidate to compare with the new entry. So you get O(n) , otherwise if you allowed the whole re-computation again the worst case order would be O(nk), k is the window size.

Answer 11

class MaxFinder
{
    // finds the max and its index
    static int[] findMaxByIteration(int arr[], int start, int end)
    {
        int max, max_ndx; 

        max = arr[start];
        max_ndx = start;
        for (int i=start; i<end; i++)
        {
            if (arr[i] > max)
            {
                max = arr[i];
                max_ndx = i;
            }    
        }

        int result[] = {max, max_ndx};

        return result;
    }

    // optimized to skip iteration, when previous windows max element 
    // is present in current window
    static void optimizedPrintKMax(int arr[], int n, int k)
    {
        int i, j, max, max_ndx;

        // for first window - find by iteration.    
        int result[] = findMaxByIteration(arr, 0, k);

        System.out.printf("%d ", result[0]);

        max = result[0];
        max_ndx = result[1];   

         for (j=1; j <= (n-k); j++)
         {
            // if previous max has fallen out of current window, iterate and find
            if (max_ndx < j)  
            {
                result = findMaxByIteration(arr, j, j+k);
                max = result[0];
                max_ndx = result[1];   
            } 
            // optimized path, just compare max with new_elem that has come into the window 
            else 
            {
                int new_elem_ndx = j + (k-1);
                if (arr[new_elem_ndx] > max)
                {
                    max = arr[new_elem_ndx];
                    max_ndx = new_elem_ndx;
                }      
            }

            System.out.printf("%d ", max);
         }   
    }     

    public static void main(String[] args)
    {
        int arr[] = {10, 9, 8, 7, 6, 5, 4, 3, 2, 1};
        //int arr[] = {1,5,2,6,3,1,24,7};
        int n = arr.length;
        int k = 3;
        optimizedPrintKMax(arr, n, k);
    }
}    

Answer 12

package com;
public class SlidingWindow {

    public static void main(String[] args) {

        int[] array = { 1, 5, 2, 6, 3, 1, 24, 7 };
        int slide = 3;//say
        List<Integer> result = new ArrayList<Integer>();

        for (int i = 0; i < array.length - (slide-1); i++) {
            result.add(getMax(array, i, slide));

        }
        System.out.println("MaxList->>>>" + result.toString());
    }

    private static Integer getMax(int[] array, int i, int slide) {

        List<Integer> intermediate = new ArrayList<Integer>();
        System.out.println("Initial::" + intermediate.size());
        while (intermediate.size() < slide) {
            intermediate.add(array[i]);
            i++;
        }
        Collections.sort(intermediate);
        return intermediate.get(slide - 1);
    }
}

Answer 13

Here is the solution in O(n) time complexity with auxiliary deque

public class TestSlidingWindow {

    public static void main(String[] args) {
        int[] arr = { 1, 5, 7, 2, 1, 3, 4 };
        int k = 3;

        printMaxInSlidingWindow(arr, k);

    }

    public static void printMaxInSlidingWindow(int[] arr, int k) {

        Deque<Integer> queue = new ArrayDeque<Integer>();
        Deque<Integer> auxQueue = new ArrayDeque<Integer>();

        int[] resultArr = new int[(arr.length - k) + 1];

        int maxElement = 0;
        int j = 0;
        for (int i = 0; i < arr.length; i++) {

            queue.add(arr[i]);

            if (arr[i] > maxElement) {
                maxElement = arr[i];
            }

    /** we need to maintain the auxiliary deque to maintain max element  in case max element is removed. 
        We add the element to deque straight away if subsequent element is less than the last element
        (as there is a probability if last element is removed this element can be max element) otherwise 
        remove all lesser element then insert current element **/

            if (auxQueue.size() > 0) {
                if (arr[i] <  auxQueue.peek()) {
                    auxQueue.push(arr[i]);
                } else {
                    while (auxQueue.size() > 0 && (arr[i] >  auxQueue.peek())) {
                        auxQueue.pollLast();
                    }

                    auxQueue.push(arr[i]);
                }
            }else {
                auxQueue.push(arr[i]);
            }

            if (queue.size() > 3) {
                int removedEl = queue.removeFirst();
                if (maxElement == removedEl) {
                    maxElement = auxQueue.pollFirst();
                }
            }

            if (queue.size() == 3) {
                resultArr[j++] = maxElement;
            }

        }

        for (int i = 0; i < resultArr.length; i++) {
            System.out.println(resultArr[i]);
        }
    }

}

Answer 14

    static void countDistinct(int arr[], int n, int k) 
    { 
        System.out.print("\nMaximum integer in the window : ");
        // Traverse through every window
        for (int i = 0; i <= n - k; i++) {
               System.out.print(findMaximuminAllWindow(Arrays.copyOfRange(arr, i, arr.length), k)+ " ");
           }
    } 

    private static int findMaximuminAllWindow(int[] win, int k) {
        // TODO Auto-generated method stub
        int max= Integer.MIN_VALUE;

        for(int i=0; i<k;i++) {
            if(win[i]>max)
                max=win[i];
        }

        return max;
    }

Answer 15

arr = 1 5 2 6 3 1 24 7

We have to find the maximum of subarray, Right? So, What is meant by subarray? SubArray = Partial set and it should be in order and contiguous.

From the above array {1,5,2} {6,3,1} {1,24,7} all are the subarray examples

n = 8 // Array length
k = 3 // window size

For finding the maximum, we have to iterate through the array, and find the maximum. From the window size k,

{1,5,2} = 5 is the maximum
{5,2,6} = 6 is the maximum
{2,6,3} = 6 is the maximum
and so on..
ans = 5 6 6 6 24 24

It can be evaluated as the n-k+1 Hence, 8-3+1 = 6 And the length of an answer is 6 as we seen.

How can we solve this now? When the data is moving from the pipe, the first thought for the data structure came in mind is the Queue

But, rather we are not discussing much here, we directly jump on the deque

Thinking Would be: 
Window is fixed and data is in and out 
Data is fixed and window is sliding 
EX: Time series database 

While (Queue is not empty and arr[Queue.back() < arr[i]] { 
Queue.pop_back(); 
Queue.push_back(); 

For the rest:

Print the front of queue

// purged expired element 
While (queue not empty and queue.front() <= I-k) { 
  Queue.pop_front(); 
  While (Queue is not empty and arr[Queue.back() < arr[i]] { 
  Queue.pop_back(); 
  Queue.push_back(); 
  }
}

Answer 16

arr = [1, 2, 3, 1, 4, 5, 2, 3, 6]
k = 3
for i in range(len(arr)-k):
  k=k+1
  print (max(arr[i:k]),end=' ') #3 3 4 5 5 5 6

Answer 17

Two approaches.

1. Segment Tree O(nlog(n-k))
   • Build a maximum segment-tree.
   • Query between [i, i+k)

Something like..

public static void printMaximums(int[] a, int k) {
    int n = a.length;
    SegmentTree tree = new SegmentTree(a);
    for (int i=0; i<=n-k; i++) System.out.print(tree.query(i, i+k));
}

2. Deque O(n)
   • If the next element is greater than the rear element, remove the rear element.
   • If the element in the front of the deque is out of the window, remove the front element.

public static void printMaximums(int[] a, int k) {
     int n = a.length;
     Deque<int[]> deck = new ArrayDeque<>();
     List<Integer> result = new ArrayList<>();
     for (int i=0; i<n; i++) {
         while (!deck.isEmpty() && a[i] >= deck.peekLast()[0]) deck.pollLast();
         deck.offer(new int[] {a[i], i});
         while (!deck.isEmpty() && deck.peekFirst()[1] <= i - k) deck.pollFirst();
         if (i >= k - 1) result.add(deck.peekFirst()[0]);
     }
     System.out.println(result);
}

Answer 18

Here is an optimized version of the naive (conditional) nested loop approach I came up with which is much faster and doesn't require any auxiliary storage or data structure. As the program moves from window to window, the start index and end index moves forward by 1. In other words, two consecutive windows have adjacent start and end indices.

For the first window of size W , the inner loop finds the maximum of elements with index (0 to W-1). (Hence i == 0 in the if in 4th line of the code). Now instead of computing for the second window which only has one new element, since we have already computed the maximum for elements of indices 0 to W-1, we only need to compare this maximum to the only new element in the new window with the index W.

But if the element at 0 was the maximum which is the only element not part of the new window, we need to compute the maximum using the inner loop from 1 to W again using the inner loop (hence the second condition maxm == arr[i-1] in the if in line 4), otherwise just compare the maximum of the previous window and the only new element in the new window.

void print_max_for_each_subarray(int arr[], int n, int k) 
{
    int maxm;
    for(int i = 0; i < n - k + 1 ; i++) 
{
        if(i == 0 || maxm == arr[i-1]) {
            maxm = arr[i];
            for(int j = i+1; j < i+k; j++) 
                if(maxm < arr[j]) maxm = arr[j];

            }
        else  {
            maxm = maxm < arr[i+k-1] ? arr[i+k-1] : maxm;
        }
        cout << maxm << ' ';
    }
    cout << '\n';
}

Answer 19

You can use Deque data structure to implement this. Deque has an unique facility that you can insert and remove elements from both the ends of the queue unlike the traditional queue where you can only insert from one end and remove from other.

Following is the code for the above problem.

    public int[] maxSlidingWindow(int[] nums, int k) {
    int n = nums.length;
    int[] maxInWindow = new int[n - k + 1];
    Deque<Integer> dq = new LinkedList<Integer>();
    int i = 0;
    for(; i<k; i++){
        while(!dq.isEmpty() && nums[dq.peekLast()] <= nums[i]){
            dq.removeLast();
        }
        dq.addLast(i);
    }
    for(; i <n; i++){
        maxInWindow[i - k] = nums[dq.peekFirst()];
        while(!dq.isEmpty() && dq.peekFirst() <= i - k){
            dq.removeFirst();
        }
        
        while(!dq.isEmpty() && nums[dq.peekLast()] <= nums[i]){
            dq.removeLast();
        }
        dq.addLast(i);
    }
    maxInWindow[i - k] = nums[dq.peekFirst()];
    return maxInWindow;
}

the resultant array will have n - k + 1 elements where n is length of the given array, k is the given window size.

Answer 20

We can solve it using the Python , applying the slicing.

def sliding_window(a,k,n):
  max_val =[]
  val =[]
  val1=[]

for i in range(n-k-1):
    if i==0:
        val = a[0:k+1]
        print("The value in val variable",val)
        val1 = max(val)
        max_val.append(val1)
    else:
        val = a[i:i*k+1]
        val1 =max(val)
        max_val.append(val1)
return max_val

Driver Code

a = [15,2,3,4,5,6,2,4,9,1,5]
n = len(a)
k = 3
sl=s liding_window(a,k,n)


print(sl)

Answer 21

Create a TreeMap of size k. Put first k elements as keys in it and assign any value like 1(doesn't matter). TreeMap has the property to sort the elements based on key so now, first element in map will be min and last element will be max element. Then remove 1 element from the map whose index in the arr is i-k. Here, I have considered that Input elements are taken in array arr and from that array we are filling the map of size k. Since, we can't do anything with sorting happening inside TreeMap, therefore this approach will also take O(n) time.

Answer 22

100% working Tested (Swift)

 func maxOfSubArray(arr:[Int],n:Int,k:Int)->[Int]{
        var lenght = arr.count
        var resultArray = [Int]()
        for i in 0..<arr.count{
            if lenght+1 > k{
           let tempArray =  Array(arr[i..<k+i])
                resultArray.append(tempArray.max()!)
            }
            lenght = lenght - 1
        }
        print(resultArray)
    return resultArray
    }

This way we can use:

maxOfSubArray(arr: [1,2,3,1,4,5,2,3,6], n: 9, k: 3)

Result:

[3, 3, 4, 5, 5, 5, 6]

Answer 23

for how big k? for reasonable-sized k. you can create k k-sized buffers and just iterate over the array keeping track of max element pointers in the buffers - needs no data structures and is O(n) k^2 pre-allocation.

Answer 24

Just notice that you only have to find in the new window if: * The new element in the window is smaller than the previous one (if it's bigger, it's for sure this one). OR * The element that just popped out of the window was the current bigger.

In this case, re-scan the window.

Answer 25

A complete working solution in Amortised Constant O(1) Complexity. https://github.com/varoonverma/code-challenge.git

Answer 26

Compare the first k elements and find the max, this is your first number

then compare the next element to the previous max. If the next element is bigger, that is your max of the next subarray, if its equal or smaller, the max for that sub array is the same

then move on to the next number

max(1 5 2) = 5
max(5 6) = 6
max(6 6) = 6
... and so on
max(3 24) = 24
max(24 7) = 24

It's only slightly better than your answer