Note:

Basic idea could be same as Kth smallest number in sorted matrix.

But this could be different since all the numbers fall in [1, m*n + 1] range. But not all the numbers in the range belongs to the multiplication table.

So we need a helper function to calculate how many numbers from [1, mid]. It is counts += Math.min(value / i, n), since max value no larger than m.

class Solution {    public int findKthNumber(int m, int n, int k) {        int start = 1;        int end = m * n + 1;        while (start < end) {            int mid = start + (end - start) / 2;            int count = getCount(mid, m, n);            if (count >= k) {                end = mid;            } else {                start = mid + 1;            }        }        return end;    }        private int getCount(int mid, int m, int n) {        int count = 0;        for (int i = 1; i <= m; i++) {            count += Math.min(mid / i, n);        }        return count;    }}