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Formatted question description: https://leetcode.ca/all/668.html

668. Kth Smallest Number in Multiplication Table (Hard)

Nearly every one have used the Multiplication Table. But could you find out the k-th smallest number quickly from the multiplication table?

Given the height m and the length n of a m * n Multiplication Table, and a positive integer k, you need to return the k-th smallest number in this table.

Example 1:

Input: m = 3, n = 3, k = 5
Output: 
Explanation: 
The Multiplication Table:
1	2	3
2	4	6
3	6	9

The 5-th smallest number is 3 (1, 2, 2, 3, 3).

Example 2:

Input: m = 2, n = 3, k = 6
Output: 
Explanation: 
The Multiplication Table:
1	2	3
2	4	6

The 6-th smallest number is 6 (1, 2, 2, 3, 4, 6).

Note:

  1. The m and n will be in the range [1, 30000].
  2. The k will be in the range [1, m * n]

Related Topics:
Binary Search

Similar Questions:

TLE version: Next Heap

// OJ: https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/
// Time: O(k * MlogM) = O(M^2 * NlogM)
// Space: O(M)
class Solution {
public:
    int findKthNumber(int m, int n, int k) {
        auto cmp = [](pair<int, int> &a, pair<int, int> &b) { return a.first * a.second > b.first * b.second; };
        priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(cmp)> q(cmp);
        for (int i = 1; i <= m; ++i) q.emplace(i, 1);
        int ans = 0;
        while (k--) {
            auto p = q.top();
            q.pop();
            ans = p.first * p.second;
            if (p.second < n) q.emplace(p.first, p.second + 1);
        }
        return ans;
    }
};

Solution 1. Binary Answer

The range of the answer is [1, m * n]. We can use binary answer to find it.

Let L = 1, R = m * n, M = (L + R) / 2.

Define cnt(M) as the count of numbers less than or equal to M.

For the answer ans, the corresponding cnt(ans) could be exactly k (when there is only one ans in the table) or greater than k (when there are multiple ans in the table).

The goal is to find the first element M whose cnt(M) is greater than or equal to k.

So let the left part of the array be those elements whose cnt < k, and the right part be cnt >= k.

In the end, L will point to the first element whose cnt >= k and it is the answer.

// OJ: https://leetcode.com/problems/kth-smallest-number-in-multiplication-table/
// Time: O(Mlog(MN))
// Space: O(1)
class Solution {
public:
    int findKthNumber(int m, int n, int k) {
        long long L = 1, R = m * n;
        while (L <= R) {
            long long M = (L + R) / 2, cnt = 0;
            for (int i = 1; i <= m; ++i) cnt += min(M / i, (long long) n);
            if (cnt < k) L = M + 1;
            else R = M - 1;
        }
        return L;
    }
};

  • class Solution {
    public int findKthNumber(int m, int n, int k) {
        int low = 1, high = m * n;
        while (low < high) {
            int mid = (high - low) / 2 + low;
            int rank = rank(mid, m, n);
            if (rank < k)
                low = mid + 1;
            else
                high = mid;
        }
        return low;
    }

    public int rank(int num, int m, int n) {
        int rank = 0;
        int maxRow = Math.min(num, m);
        for (int i = 1; i <= maxRow; i++)
            rank += Math.min(num / i, n);
        return rank;
    }
}

############

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;
    }
}

  • 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



  • 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
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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