K’th Smallest/Largest Element using STL

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:

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

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

Implementation:

C++

// STL based C++ program to find k-th smallest
// element.
#include <bits/stdc++.h>
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[0]), k = 2;
cout << "K'th smallest element is "
<< kthSmallest(arr, n, k);
return 0;
}

Java

import java.util.Set;
import java.util.TreeSet;

public class KthSmallest {
    public static int kthSmallest(int[] arr, int n, int k) {
        // Insert all elements into the set
        Set<Integer> s = new TreeSet<Integer>();
        for (int i = 0; i < n; i++) {
            s.add(arr[i]);
        }

        // Traverse set and print k-th element
        int i = 0;
        for (Integer num : s) {
            if (i == k - 1) {
                return num;
            }
            i++;
        }
        return -1;
    }

    public static void main(String[] args) {
        int[] arr = {12, 3, 5, 7, 3, 19};
        int n = arr.length, k = 2;
        System.out.println("K'th smallest element is " + kthSmallest(arr, n, k));
    }
}

Python3

# STL based python program to find k-th smallest
# element.
from sortedcontainers import SortedList, SortedSet, SortedDict

def kthSmallest(arr, n, k) :
    # Insert all elements into the set
    s=SortedSet();
    for i in range(0,n):
        s.add(arr[i]);

    # Traverse set and print k-th element
    i=1;
    ans=-1;
    for val in s:
        if(i==k):
            ans=val;
            break;
        i+=1;

    return ans;

arr = [ 12, 3, 5, 7, 3, 19 ];
n = len(arr);
k = 2;
print("K'th smallest element is ",kthSmallest(arr, n, k));

C#

using System;
using System.Linq;
using System.Collections.Generic;

class KthSmallest
{
    public static int kthSmallest(int[] arr, int k)
    {
        // Insert all elements into the set
        var s = new SortedSet<int>(arr);

        // Traverse set and print k-th element
        return s.ElementAt(k - 1);
    }

    public static void Main(string[] args)
    {
        int[] arr = { 12, 3, 5, 7, 3, 19 };
        int k = 2;
        Console.WriteLine("K'th smallest element is " + kthSmallest(arr, k));
    }
}

Javascript

// Javascript program to find k-th smallest element.

function kthSmallest(arr, n, k)
{
// Insert all elements into the set
    let s= new Set();
    for (let i = 0; i < n; i++)
    s.add(arr[i]);

    // Traverse set and print k-th element
    let i = 0;
    for (const entry of s.values())
    {
        if(i==k)
        {
            i=entry;
            break;
        }
        i++;
    }

    return i;
}

let arr = [ 12, 3, 5, 7, 3, 19 ];
let n = arr.length, k = 2;
document.write("K'th smallest element is "+ kthSmallest(arr, n, k));

Output

K'th smallest element is 5

Time complexity: 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). 
Auxiliary Space: O(n) where n is size of array, since n extra space has been taken.

This article is contributed by Roshan Halwai. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.