Sliding Window Maximum (Maximum of all subarrays of size k) using stack in O(n) time

Give an array arr[] of N integers and another integer k ≤ N. The task is to find the maximum element of every sub-array of size k.

Examples:

Input: arr[] = {9, 7, 2, 4, 6, 8, 2, 1, 5}, k = 3
Output: 9 7 6 8 8 8 5
Window 1: {9, 7, 2}, max = 9
Window 2: {7, 2, 4}, max = 7
Window 3: {2, 4, 6}, max = 6
Window 4: {4, 6, 8}, max = 8
Window 5: {6, 8, 2}, max = 8
Window 6: {8, 2, 1}, max = 8
Window 7: {2, 1, 5}, max = 5



Input: arr[] = {6, 7, 5, 2, 1, 7, 2, 1, 10}, k = 2
Output: 7 7 5 2 7 7 2 10

Prerequisite: Next greater element
Approach: If you know for every index i an index j ≥ i such that max(a[i], a[i + 1], … a[j]) = a[i]. Lets call it max_upto[i]. Then if you want to know the maximum element in the sub-array of length k starting from ith index you can get it by checking every index starting from i to i + k – 1 for which max_upto[i] ≥ i + k – 1 (last index of that window).

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to print the maximum for
// every k size sub-array
void print_max(int a[], int n, int k)
{
    // max_upto array stores the index
    // upto which the maximum element is a[i]
    // i.e. max(a[i], a[i + 1], ... a[max_upto[i]]) = a[i]
  
    int max_upto[n];
  
    // Update max_upto array similar to
    // finding next greater element
    stack<int> s;
    s.push(0);
    for (int i = 1; i < n; i++) {
        while (!s.empty() && a[s.top()] < a[i]) {
            max_upto[s.top()] = i - 1;
            s.pop();
        }
        s.push(i);
    }
    while (!s.empty()) {
        max_upto[s.top()] = n - 1;
        s.pop();
    }
    int j = 0;
    for (int i = 0; i <= n - k; i++) {
  
        // j < i is to check whether the
        // jth element is outside the window
        while (j < i || max_upto[j] < i + k - 1)
            j++;
        cout << a[j] << " ";
    }
    cout << endl;
}
  
// Driver code
int main()
{
    int a[] = { 9, 7, 2, 4, 6, 8, 2, 1, 5 };
    int n = sizeof(a) / sizeof(int);
    int k = 3;
    print_max(a, n, k);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
    // Function to print the maximum for
    // every k size sub-array
    static void print_max(int a[], int n, int k)
    {
        // max_upto array stores the index
        // upto which the maximum element is a[i]
        // i.e. max(a[i], a[i + 1], ... a[max_upto[i]]) = a[i]
  
        int[] max_upto = new int[n];
  
        // Update max_upto array similar to
        // finding next greater element
        Stack<Integer> s = new Stack<>();
        s.push(0);
        for (int i = 1; i < n; i++)
        {
            while (!s.empty() && a[s.peek()] < a[i])
            {
                max_upto[s.peek()] = i - 1;
                s.pop();
            }
            s.push(i);
        }
        while (!s.empty())
        {
            max_upto[s.peek()] = n - 1;
            s.pop();
        }
        int j = 0;
        for (int i = 0; i <= n - k; i++)
        {
  
            // j < i is to check whether the
            // jth element is outside the window
            while (j < i || max_upto[j] < i + k - 1)
            {
                j++;
            }
            System.out.print(a[j] + " ");
        }
        System.out.println();
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int a[] = {9, 7, 2, 4, 6, 8, 2, 1, 5};
        int n = a.length;
        int k = 3;
        print_max(a, n, k);
  
    }
}
  
// This code has been contributed by 29AjayKumar

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Python3

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# Python3 implementation of the approach
  
# Function to print the maximum for
# every k size sub-array
def print_max(a, n, k):
      
    # max_upto array stores the index
    # upto which the maximum element is a[i]
    # i.e. max(a[i], a[i + 1], ... a[max_upto[i]]) = a[i]
  
    max_upto=[0 for i in range(n)]
  
    # Update max_upto array similar to
    # finding next greater element
    s=[]
    s.append(0)
  
    for i in range(1,n):
        while (len(s) > 0 and a[s[-1]] < a[i]):
            max_upto[s[-1]] = i - 1
            del s[-1]
          
        s.append(i)
  
    while (len(s) > 0):
        max_upto[s[-1]] = n - 1
        del s[-1]
  
    j = 0
    for i in range(n - k + 1):
  
        # j < i is to check whether the
        # jth element is outside the window
        while (j < i or max_upto[j] < i + k - 1):
            j += 1
        print(a[j], end=" ")
    print() 
  
# Driver code
  
a = [9, 7, 2, 4, 6, 8, 2, 1, 5]
n = len(a)
k = 3
print_max(a, n, k)
  
# This code is contributed by mohit kumar

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG 
{
  
    // Function to print the maximum for
    // every k size sub-array
    static void print_max(int []a, int n, int k)
    {
        // max_upto array stores the index
        // upto which the maximum element is a[i]
        // i.e. max(a[i], a[i + 1], ... a[max_upto[i]]) = a[i]
  
        int[] max_upto = new int[n];
  
        // Update max_upto array similar to
        // finding next greater element
        Stack<int> s = new Stack<int>();
        s.Push(0);
        for (int i = 1; i < n; i++)
        {
            while (s.Count!=0 && a[s.Peek()] < a[i])
            {
                max_upto[s.Peek()] = i - 1;
                s.Pop();
            }
            s.Push(i);
        }
        while (s.Count!=0)
        {
            max_upto[s.Peek()] = n - 1;
            s.Pop();
        }
        int j = 0;
        for (int i = 0; i <= n - k; i++)
        {
  
            // j < i is to check whether the
            // jth element is outside the window
            while (j < i || max_upto[j] < i + k - 1)
            {
                j++;
            }
            Console.Write(a[j] + " ");
        }
        Console.WriteLine();
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        int []a = {9, 7, 2, 4, 6, 8, 2, 1, 5};
        int n = a.Length;
        int k = 3;
        print_max(a, n, k);
  
    }
}
  
// This code has been contributed by 29AjayKumar

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Output:

9 7 6 8 8 8 5


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