# K’th Smallest/Largest Element using STL

• Difficulty Level : Easy
• Last Updated : 30 Jul, 2017

Given an array and a number k where k is smaller than size of array, we need to find the k’th smallest element in the given array.

Examples:

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```Input : arr[] = {7, 10, 4, 3, 20, 15}
k = 2
Output : 4
Smallest element is 3. Second smallest
is 4.

Input : arr[] = {7, 10, 4, 3, 3, 15}
k = 2
Output : 4
Even if there are more than one occurrences
of 3, answer should be 4.

Input :arr[] = {7, 10, 4, 3, 20, 15}
k = 4
Output : 10
```

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

We use set in C++ STL.
1) Insert all elements into a set.
2) Traverse the set and print k-th element.

 `// STL based C++ program to find k-th smallest``// element.``#include ``using` `namespace` `std;`` ` `int` `kthSmallest(``int` `arr[], ``int` `n, ``int` `k)``{``    ``// Insert all elements into the set``    ``set<``int``> s;``    ``for` `(``int` `i = 0; i < n; i++)``        ``s.insert(arr[i]);`` ` `    ``// Traverse set and print k-th element``    ``auto` `it = s.begin();``    ``for` `(``int` `i = 0; i < k - 1; i++)``        ``it++;``    ``return` `*it;``}`` ` `int` `main()``{``    ``int` `arr[] = { 12, 3, 5, 7, 3, 19 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr), k = 2;``    ``cout << ``"K'th smallest element is "``         ``<< kthSmallest(arr, n, k);``    ``return` `0;``}`

Output:

```K'th smallest element is 5
```

Time complexity of above solution is O(n Log n). Note that set in STL uses a self-balancing BST internally and therefore time complexity of search and insert operations is O(log n).

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