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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
+}