Find the Smallest Divisor Given a Threshold
Time complexity: $O(N\log M)$
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
|
#include <iostream>
#include <vector>
class Solution {
public:
int smallestDivisor(std::vector<int>& nums, int threshold) {
long long sum = 0;
int right = 0;
for (int i : nums) {
sum += i;
right = std::max(right, i);
}
int left = sum / threshold;
if (!left) {
return 1;
}
for (int mid = (left + right) / 2; right > left; mid = (left + right) / 2) {
int k = getSum(nums, mid);
if (k > threshold) {
left = mid + 1;
}
else {
right = mid;
}
}
return left;
}
private:
int getSum(std::vector<int>& nums, int divisor) {
int sum = nums.size();
for (int num : nums) {
sum += (num - 1) / divisor;
}
return sum;
}
};
|
