# Maximum of all Subarrays of size k using set in C++ STL

Given an array of size **N** and an integer **K**, the task is to find the maximum for each and every contiguous sub-array of size **K** and print the sum of all these values in the end.

**Examples:**

Input:arr[] = {4, 10, 54, 11, 8, 7, 9}, K = 3

Output:182

Input:arr[] = {1, 2, 3, 4, 1, 6, 7, 8, 2, 1}, K = 4

Output:45

**Prerequisite**:

**Approach:**

Set performs insertion and removal operation in **O(logK)** time and always stores the keys in the sorted order.

The idea is to use a set of pairs where the first item in the pair is the element itself and the second item in the pair contains the array index of the element.

- Pick first k elements and create a set of pair with these element and their index as described above.
- Now, set
**sum = 0**and use window sliding technique and Loop from**j = 0**to**n – k**:- Get the maximum element from the set (the last element) in the current window and update
**sum = sum + currMax**. - Search for the leftmost element of current window in the set and remove it.
- Insert the next element of the current window in the set to move to next window.

- Get the maximum element from the set (the last element) in the current window and update

Below is the implementation of the above approach:

`// C++ implementation of the approach ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to return the sum of maximum of ` `// all k size sub-arrays using set in C++ STL ` `int` `maxOfSubarrays(` `int` `arr[], ` `int` `n, ` `int` `k) ` `{ ` ` ` `// Create a set of pairs ` ` ` `set<pair<` `int` `, ` `int` `> > q; ` ` ` ` ` `// Create a reverse iterator to the set ` ` ` `set<pair<` `int` `, ` `int` `> >::reverse_iterator it; ` ` ` ` ` `// Insert the first k elements along ` ` ` `// with their indices into the set ` ` ` `for` `(` `int` `i = 0; i < k; i++) { ` ` ` `q.insert(pair<` `int` `, ` `int` `>(arr[i], i)); ` ` ` `} ` ` ` ` ` `// To store the sum ` ` ` `int` `sum = 0; ` ` ` `for` `(` `int` `j = 0; j < n - k + 1; j++) { ` ` ` ` ` `// Iterator to the end of the ` ` ` `// set since it has the maximum value ` ` ` `it = q.rbegin(); ` ` ` ` ` `// Add the maximum element ` ` ` `// of the current window ` ` ` `sum += it->first; ` ` ` ` ` `// Delete arr[j] (Leftmost element of ` ` ` `// current window) from the set ` ` ` `q.erase(pair<` `int` `, ` `int` `>(arr[j], j)); ` ` ` ` ` `// Insert next element ` ` ` `q.insert(pair<` `int` `, ` `int` `>(arr[j + k], j + k)); ` ` ` `} ` ` ` ` ` `// Return the required sum ` ` ` `return` `sum; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { 4, 10, 54, 11, 8, 7, 9 }; ` ` ` ` ` `int` `K = 3; ` ` ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `cout << maxOfSubarrays(arr, n, K); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

182

Time Complexity : O(n Log n)

Auxiliary Space : O(k)

The above problem can be solved in O(n) time. Please see below Dequeue based solution for the same.

Sliding Window Maximum (Maximum of all subarrays of size k)

## Recommended Posts:

- Maximum subarray size, such that all subarrays of that size have sum less than k
- Sliding Window Maximum (Maximum of all subarrays of size k) using stack in O(n) time
- Sliding Window Maximum (Maximum of all subarrays of size k)
- Maximum sum two non-overlapping subarrays of given size
- Sum of minimum and maximum elements of all subarrays of size k.
- Max sum of M non-overlapping subarrays of size K
- Counting inversions in all subarrays of given size
- Sum of maximum of all subarrays | Divide and Conquer
- Number of subarrays whose minimum and maximum are same
- Count of subarrays whose maximum element is greater than k
- Maximum sum of lengths of non-overlapping subarrays with k as the max element.
- Number of subarrays with maximum values in given range
- Maximum sum of non-overlapping subarrays of length atmost K
- Partition into two subarrays of lengths k and (N - k) such that the difference of sums is maximum
- Split array to three subarrays such that sum of first and third subarray is equal and maximum

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.