Finding maximum for every window of size k in an array

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

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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] )