Kth Smallest Element in a Sorted Matrix

Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.

Note that it is the kth smallest element in the sorted order, not the kth distinct element.

Example:

matrix = [
   [ 1,  5,  9],
   [10, 11, 13],
   [12, 13, 15]
],
k = 8,

return 13.

Note: 
You may assume k is always valid, 1 ≤ k ≤ n2.

The best solution is to use binary search. Based on the nature of the matrix, we know the minimum of the matrix is its upper left element and the max is the lower right. Now we try to get the mid of the element and find the position of the mid in the matrix. If the position is less than k, we let left = mid + 1, otherwise right = mid.

Now how to find the position of the mid? We start from lower left element. If the element is smaller than target, we know all elements in current column is smaller than target, so we add pos by #row + 1. And we increment column by 1. Otherwise we decrease row by 1.

public int kthSmallest(int[][] matrix, int k) {
        if (matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        int r = matrix.length;
        int c = matrix[0].length;
        int left = matrix[0][0], right = matrix[r - 1][c - 1];
        while (left < right) {
            int mid = (left + right) / 2;
            int pos = getPos(matrix, mid);
            if (pos < k) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }
    
    private int getPos(int[][] matrix, int target) {
        int r = matrix.length;
        int c = matrix[0].length;
        int pos = 0;
        for (int i = r - 1, j = 0; i >= 0 && j < c;) {
            if (matrix[i][j] <= target) {
                pos += i + 1;
                j++;
            } else {
                i--;
            }
        }
        return pos;
    }