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

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:

 `// C++ program for the above approach ` ` `  `#include ` `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)
Auxillary Space Complexity: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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