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: 182Input: 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.
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; } |
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)
