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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;
}
};


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
}