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668. Kth Smallest Number in Multiplication Table

Description

Nearly everyone has used the Multiplication Table. The multiplication table of size m x n is an integer matrix mat where mat[i][j] == i * j (1-indexed).

Given three integers m, n, and k, return the kth smallest element in the m x n multiplication table.

 

Example 1:

Input: m = 3, n = 3, k = 5
Output: 3
Explanation: The 5th smallest number is 3.

Example 2:

Input: m = 2, n = 3, k = 6
Output: 6
Explanation: The 6th smallest number is 6.

 

Constraints:

  • 1 <= m, n <= 3 * 104
  • 1 <= k <= m * n

Solutions

Solution 1

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class Solution:
    def findKthNumber(self, m: int, n: int, k: int) -> int:
        left, right = 1, m * n
        while left < right:
            mid = (left + right) >> 1
            cnt = 0
            for i in range(1, m + 1):
                cnt += min(mid // i, n)
            if cnt >= k:
                right = mid
            else:
                left = mid + 1
        return left

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class Solution {
    public int findKthNumber(int m, int n, int k) {
        int left = 1, right = m * n;
        while (left < right) {
            int mid = (left + right) >>> 1;
            int cnt = 0;
            for (int i = 1; i <= m; ++i) {
                cnt += Math.min(mid / i, n);
            }
            if (cnt >= k) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

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class Solution {
public:
    int findKthNumber(int m, int n, int k) {
        int left = 1, right = m * n;
        while (left < right) {
            int mid = (left + right) >> 1;
            int cnt = 0;
            for (int i = 1; i <= m; ++i) cnt += min(mid / i, n);
            if (cnt >= k)
                right = mid;
            else
                left = mid + 1;
        }
        return left;
    }
};

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func findKthNumber(m int, n int, k int) int {
    left, right := 1, m*n
    for left < right {
        mid := (left + right) >> 1
        cnt := 0
        for i := 1; i <= m; i++ {
            cnt += min(mid/i, n)
        }
        if cnt >= k {
            right = mid
        } else {
            left = mid + 1
        }
    }
    return left
}

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