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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// 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 



















                        chevron_right







                        filter_none









Python3














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















# 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 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// 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 



















                        chevron_right







                        filter_none











Output:


9 7 6 8 8 8 5






          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
monk789456123
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.



      
      
      
      

      

      
					
		

    Article Tags : 
Arrays
Stack

           	
Practice Tags : 
Arrays
Stack

        

         

         
 
                9

                

        



        
        
        
        
        

    

            

            3.6
            

            

             
                 Based on 8 vote(s)
                    

            
            
            
            
            
        

        
        

        

        

        

        

        

        


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.