Formatted question description: https://leetcode.ca/all/1283.html

# 1283. Find the Smallest Divisor Given a Threshold (Medium)

Given an array of integers nums and an integer threshold, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the smallest divisor such that the result mentioned above is less than or equal to threshold.

Each result of division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

It is guaranteed that there will be an answer.

Example 1:

Input: nums = [1,2,5,9], threshold = 6
Output: 5
Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1.
If the divisor is 4 we can get a sum to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).


Example 2:

Input: nums = [2,3,5,7,11], threshold = 11
Output: 3


Example 3:

Input: nums = [19], threshold = 5
Output: 4


Constraints:

• 1 <= nums.length <= 5 * 10^4
• 1 <= nums[i] <= 10^6
• nums.length <= threshold <= 10^6

Related Topics:
Binary Search

// OJ: https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/

// Time: O(Nlog(max(A)))
// Space: O(1)
class Solution {
bool valid(vector<int> &A, int d, int threshold) {
for (int n : A) {
if ((threshold -= (n + d - 1) / d) < 0) return false;
}
return true;
}
public:
int smallestDivisor(vector<int>& A, int threshold) {
int L = 1, R = 1e6;
while (L <= R) {
int M = (L + R) / 2;
if (valid(A, M, threshold)) R = M - 1;
else L = M + 1;
}
return L;
}
};


Java

• class Solution {
public int smallestDivisor(int[] nums, int threshold) {
long arraySum = 0;
for (int num : nums)
arraySum += num;
long low = 1, high = arraySum;
while (low < high) {
long mid = (high - low) / 2 + low;
int sum = 0;
for (int num : nums) {
int quotient = (int) Math.ceil(num * 1.0 / mid);
sum += quotient;
}
if (sum > threshold)
low = mid + 1;
else
high = mid;
}
return (int) low;
}
}

• // OJ: https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/
// Time: O(Nlog(max(A)))
// Space: O(1)
class Solution {
bool valid(vector<int> &A, int d, int threshold) {
for (int n : A) {
if ((threshold -= (n + d - 1) / d) < 0) return false;
}
return true;
}
public:
int smallestDivisor(vector<int>& A, int threshold) {
int L = 1, R = 1e6;
while (L <= R) {
int M = (L + R) / 2;
if (valid(A, M, threshold)) R = M - 1;
else L = M + 1;
}
return L;
}
};

• # 1283. Find the Smallest Divisor Given a Threshold
# https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/

class Solution:
def smallestDivisor(self, nums: List[int], threshold: int) -> int:

def good(d):
return sum(ceil(num/mid) for num in nums) <= threshold

left, right = 1, 10**6

while left < right:
mid = (left+right)//2

if good(mid):
right = mid
else:
left = mid + 1

return left