Matrix Median

Given a matrix of integers A of size N x M in which each row is sorted.

Find an return the overall median of the matrix A.

Note: No extra memory is allowed.

Note: Rows are numbered from top to bottom and columns are numbered from left to right.

Input Format

The first and only argument given is the integer matrix A.

Output Format

Return the overall median of the matrix A.

Constraints

1 <= N, M <= 10^5
1 <= N*M  <= 10^6
1 <= A[i] <= 10^9
N*M is odd

For Example

Input 1:
    A = [   [1, 3, 5],
            [2, 6, 9],
            [3, 6, 9]   ]
Output 1:
    5
Explanation 1:
    A = [1, 2, 3, 3, 5, 6, 6, 9, 9]
    Median is 5. So, we return 5.

Input 2:
    A = [   [5, 17, 100]    ]
Output 2:
    17 

public class Solution {
    public int findMedian(int[][] A) {
        int r = A.length, c = A[0].length;

        int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;
        for (int i = 0; i < r; i++) {
            min = Math.min(min, A[i][0]);
            max = Math.max(max, A[i][c - 1]);
        }
        // A median has (r*c/2) + 1 elements <= median (Including median)
        int desired = 1 + (r * c) / 2;
        while (min < max) {
            int mid = (min + max) / 2;
            // Find number of elements <= mid
            // We can reduce this step to RlogC, but it is not required here
            int count = smallerNumbers(mid, A);
            if (count < desired)
                min = mid + 1;
            else
                max = mid;
        }
        return min;
    }

    public int smallerNumbers(int k, int[][] A) {
        int r = A.length, c = A[0].length;
        int counter = 0;

        for (int i = 0; i < r; i++)
            for (int j = 0; j < c; j++)
                if (A[i][j] <= k)
                    counter = counter + 1;

        return counter;
    }
}