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Finding Median of unsorted Array in linear time using C++ STL

Last Updated : 15 Jul, 2025
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Given an unsorted array arr[] having N elements, the task is to find out the median of the array in linear time complexity. Examples:
Input: N = 5, arr[] = {4, 1, 2, 6, 5} Output: 4 Explanation: Since N = 5, which is odd, therefore the median is the 3rd element in the sorted array. The 3rd element in the sorted arr[] is 4. Hence the median is 4. Input: N = 8, arr[] = {1, 3, 4, 2, 6, 5, 8, 7} Output: 4.5 Explanation: Since N = 8, which is even, therefore median is the average of 4th and 5th element in the sorted array. The 4th and 5th element in the sorted array is 4 and 5 respectively. Hence the median is (4+5)/2 = 4.5.
Approach: The idea is to use nth_element() function in C++ STL.
  1. If the number of element in the array is odd, then find the (N/2)th element using nth_element() function as illustrated below and then the value at index (N/2) is the median of the given array.
    nth_element(arr.begin(), arr.begin() + N/2, arr.end())
  2. Else find the (N/2)th and ((N - 1)/2)th element using nth_element() function as illustrated below and find the average of the values at index (N/2) and ((N - 1)/2) is the median of the given array.
    nth_element(arr.begin(), arr.begin() + N/2, arr.end()) nth_element(arr.begin(), arr.begin() + (N - 1)/2, arr.end())
Below is the implementation of the above approach: CPP 
// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function for calculating
// the median
double findMedian(vector<int> a,
                  int n)
{

    // If size of the arr[] is even
    if (n % 2 == 0) {

        // Applying nth_element
        // on n/2th index
        nth_element(a.begin(),
                    a.begin() + n / 2,
                    a.end());

        // Applying nth_element
        // on (n-1)/2 th index
        nth_element(a.begin(),
                    a.begin() + (n - 1) / 2,
                    a.end());

        // Find the average of value at
        // index N/2 and (N-1)/2
        return (double)(a[(n - 1) / 2]
                        + a[n / 2])
               / 2.0;
    }

    // If size of the arr[] is odd
    else {

        // Applying nth_element
        // on n/2
        nth_element(a.begin(),
                    a.begin() + n / 2,
                    a.end());

        // Value at index (N/2)th
        // is the median
        return (double)a[n / 2];
    }
}

// Driver Code
int main()
{
    // Given array arr[]
    vector<int> arr = { 1, 3, 4, 2,
                        7, 5, 8, 6 };

    // Function Call
    cout << "Median = "
         << findMedian(arr, arr.size())
         << endl;
    return 0;
}
Output: 
Median = 4.5
Time Complexity: O(N) Auxiliary Space Complexity: O(1)

C

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