Given an array of size N and an integer K, find the minimum for each and every contiguous subarray of size K.

**Examples**:

Input : arr[] = {5, 3, 4, 1, 1}, K = 3 Output : 3 1 1 Input : arr[] = {1, 2, 3, 4, 1, 6, 7, 8, 2, 1}, K = 4 Output : 1 1 1 1 1 2 1

**Prerequisite**:

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.

The following approach is being used in the program:

- Pick first k elements and create a set of pair with these element and their index as described above.
- Now, use window sliding technique and Loop from j=0 to n-k:
- Get the minimum element from the set in the current window and print it.(The first element)
- 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.

**Why do we use a set of pairs instead of a set?**

A set doesn’t allow insertion of duplicate elements and a window of size k may have any number of duplicate elements. So, We insert a pair of the element and its index into the set.

Below is the implementation of the above approach:

`// CPP program to pint Minimum of` `// all Subarrays using set in C++ STL` ` ` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Function to pint Minimum of` `// all Subarrays using set in C++ STL` `void` `minOfSubarrays(` `int` `arr[], ` `int` `n, ` `int` `k)` `{` ` ` `// Create a set of pairs` ` ` `set<pair<` `int` `, ` `int` `> > q;` ` ` ` ` `// Create an iterator to the set` ` ` `set<pair<` `int` `, ` `int` `> >::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));` ` ` `}` ` ` ` ` `for` `(` `int` `j = 0; j < n - k + 1; j++) {` ` ` ` ` `// Iterator to the beginning of the` ` ` `// set since it has the minimum value` ` ` `it = q.begin();` ` ` ` ` `// Print the minimum element` ` ` `// of current window` ` ` `cout << 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));` ` ` `}` `}` ` ` `// Driver Code` `int` `main()` `{` ` ` `int` `arr[] = { 5, 3, 4, 1, 1 };` ` ` ` ` `int` `K = 3;` ` ` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` ` ` `minOfSubarrays(arr, n, K);` ` ` ` ` `return` `0;` `}` |

**Output:**

3 1 1

**Time Complexity**: O(N * LogK)

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