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// 1283. Find the Smallest Divisor Given a Threshold
#include "leetcode.h"
/*
* given an array of integers 'nums' and an integer 'threshold', we will choose
* a positive integer 'divisor', divide all the array by it and sum the
* divisions result. find the smallest divisor such that the result mentioned
* above is less than or equal to 'threshold'. each result of the division is
* rounded to the nearest integer greater than or equal to that element. the
* test cases are generated so that there will be an answer
*/
int divide(int *nums, int numsSize, double k) {
int res = 0;
for (int i = 0; i < numsSize; i++)
res += ceil(nums[i] / k);
return res;
}
int cmp(const void *a, const void *b) { return *(int *)a - *(int *)b; }
int smallestDivisor(int *nums, int numsSize, int threshold) {
qsort(nums, numsSize, sizeof(int), cmp);
int left = 1, right = 1000000, mid;
while (left < right) {
mid = (left + right) / 2;
if (divide(nums, numsSize, mid) <= threshold)
right = mid;
else
left = mid + 1;
}
mid = (left + right) / 2;
return mid;
}
int main() {
int n1[] = {1, 2, 5, 9}, n2[] = {44, 22, 33, 11, 1};
printf("%d\n", smallestDivisor(n1, ARRAY_SIZE(n1), 6)); // expect: 5
printf("%d\n", smallestDivisor(n2, ARRAY_SIZE(n2), 5)); // expect: 44
}
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