# Print Minimum of all Subarrays using set in C++ STL

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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

3 1 1

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

## Recommended Posts:

- Sum of minimum and maximum elements of all subarrays of size k.
- Differences between number of increasing subarrays and decreasing subarrays in k sized windows
- Print the pyramid pattern with given height and minimum number of stars
- Number of subarrays having sum less than K
- Sum of Bitwise-OR of all subarrays of a given Array | Set 2
- Maximum of all Subarrays of size k using set in C++ STL
- Number of subarrays having product less than K
- Number of subarrays having sum in a given range
- Number of subarrays with given product
- Number of subarrays with maximum values in given range
- Count subarrays having total distinct elements same as original array
- Sliding Window Maximum (Maximum of all subarrays of size k)
- Count subarrays where second highest lie before highest
- Print numbers with digits 0 and 1 only such that their sum is N
- Program to print the given Z Pattern

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.