# 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
Explanation:
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
Explanation:
Window 1: {6, 7}, max = 7
Window 2: {7, 5}, max = 7
Window 3: {5, 2}, max = 5
Window 4: {2, 1}, max = 2
Window 5: {1, 7}, max = 7
Window 6: {7, 2}, max = 7
Window 7: {2, 1}, max = 2
Window 8: {1, 10}, max = 10
```

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

Prerequisite: Next greater element

Approach:
For every index calculate the index upto which the current element is maximum when the subarray starts from this index, i.e 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 the maximum element in the sub-array of length k starting from ith index, can be found by checking every index starting from i to i + k – 1 for which max_upto[i] ≥ i + k – 1 (last index of that window).
Stack data-structure can be used to store the values in an window, i.e. the last visited or the previous inserted element will be at the top (element with closest index to current element).

Algorithm:

1. Create an array max_upto and a stack to store indices. Push 0 in the stack.
2. Run a loop from index 1 to index n-1.
3. Pop all the indices from the stack, which elements (array[s.top()]) is less than the current element and update max_upto[s.top()] = i – 1 and then insert i in the stack.
4. Pop all the indices from the stack and assign max_upto[s.top()] = n – 1.
5. Create a variable j = 0
6. Run a loop from 0 to n – k, loop counter is i
7. Run a nested loop until j < i or max_upto[j] < i + k – 1, increment j in every iteration.
8. Print the jth array element.

Implementation:

## 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
```

Complexity Analysis:

• Time Complexity: O(n).
Only two traversal of the array is needed. So Time Complexity is O(n).
• Space Complexity: O(n).
Two extra space of size n is required.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

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 :

25

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