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.

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// 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;
}

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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).

Related Posts :

K’th Smallest/Largest Element in Unsorted Array | Set 1
K’th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time
K’th Smallest/Largest Element in Unsorted Array | Set 3 (Worst Case Linear Time)

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