We are given a row wise sorted matrix of size r*c, we need to find the median of the matrix given. It is assumed that r*c is always odd.

Examples:

Input : 1 3 5 2 6 9 3 6 9 Output : Median is 5 If we put all the values in a sorted array A[] = 1 2 3 3 5 6 6 9 9) Input: 1 4 3 2 5 6 7 8 9 Output: Median is 5

**Simple Method**: The simplest method to solve this problem is to store all the elements of the given matrix in an array of size r*c. Then find the median element in this array.This seems to be the simplest method but it would take extra memory.

Time Complexity = O(r*c)

Auxiliary Space = O(r*c)

An **efficient approach** for this problem is to use binary search algorithm. The idea is that for a number to be median there should be exactly (n/2) numbers which are less than this number. So, we try to find the count of numbers less than all the numbers. Below is the step by step algorithm for this approach:

**Algorithm**:

- First we find the minimum and maximum elements in the matrix. Minimum element can be easily found by comparing the first element of each row, and similarly the maximum element can be found by comparing the last element of each row.
- Then we use binary search on our range of numbers from minimum to maximum, we find the mid of the min and max, and get count of numbers less than our mid. And accordingly change the min or max.
- For a number to be median, there should be (r*c)/2 numbers smaller than that number. So for every number, we get the count of numbers less than that by using upper_bound() in each row of matrix, if it is less than the required count, the median must be greater than the selected number, else the median must be less than or equal to the selected number.

Below is implementation of above approach:

## C++

// C++ program to find median of a matrix // sorted row wise #include<bits/stdc++.h> using namespace std; const int MAX = 100; // function to find median in the matrix int binaryMedian(int m[][MAX], int r ,int c) { int min = INT_MAX, max = INT_MIN; for (int i=0; i<r; i++) { // Finding the minimum element if (m[i][0] < min) min = m[i][0]; // Finding the maximum element if (m[i][c-1] > max) max = m[i][c-1]; } int desired = (r * c + 1) / 2; while (min < max) { int mid = min + (max - min) / 2; int place = 0; // Find count of elements smaller than mid for (int i = 0; i < r; ++i) place += upper_bound(m[i], m[i]+c, mid) - m[i]; if (place < desired) min = mid + 1; else max = mid; } return min; } // driver program to check above functions int main() { int r = 3, c = 3; int m[][MAX]= { {1,3,5}, {2,6,9}, {3,6,9} }; cout << "Median is " << binaryMedian(m, r, c) << endl; return 0; }

## Python

# Python program to find median of matrix # sorted row wise from bisect import bisect_right as upper_bound MAX = 100; # Function to find median in the matrix def binaryMedian(m, r, c): mi = m[0][0] mx = 0 for i in range(r): if m[i][0] < mi: mi = m[i][0] if m[i][c-1] > mx : mx = m[i][c-1] desired = (r * c + 1) / 2 while (mi < mx): mid = mi + (mx - mi) / 2 place = [0]; # Find count of elements smaller than mid for i in range(r): j = upper_bound(m[i], mid) place[0] = place[0] + j if place[0] < desired: mi = mid + 1 else: mx = mid print "Median is", mi return # Driver code r = 3 c = 3 m = [ [1, 3, 5], [2, 6, 9], [3, 6, 9]] binaryMedian(m, r, c) # This code is contributed by Sachin BIsht

Output:

Median is 5

**Time Complexity**: O(32 * r * log(c)). The upper bound function will take log(c) time and is performed for each row. And since the numbers will be max of 32 bit, so binary search of numbers from min to max will be performed in at most 32 ( log2(2^32) = 32 ) operations.

**Auxiliary Space **: O(1)

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

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