# 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 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++

 `// C++ implementation of the approach ` `#include ` `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 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
```

My Personal Notes arrow_drop_up 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 :
Practice Tags :

12

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