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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