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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 i^th 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++

// 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; 
} 

Java

// 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 

Python3

# 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 

C#

// 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 

Output:

9 7 6 8 8 8 5

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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