// 786. K-th Smallest Prime Fraction #include "leetcode.h" /* * given a sorted integer array 'arr' containing '1' and prime numbers where * all the integers of 'arr' are unique. you are also given an integer 'k'. for * every 'i' and 'j' where '0 <= i < j < arr.len', we consider the fraction * 'arr[i] / arr[j]'. return the k'th smallest fraction considered. return your * answer as an array of integers of size '2'. */ int less_equal_position(int *nums, int n, int i, double val) { int left = i + 1, right = n - 1, mid; if (val < (double)nums[i] / nums[n - 1]) return n; while (left < right) { mid = left + (right - left) / 2; double tmp = (double)nums[i] / nums[mid]; if (tmp == val) return mid; if (tmp < val) right = mid; else left = mid + 1; } return left; } int *kthSmallestPrimeFraction(int *arr, int arrSize, int k, int *returnSize) { double left = 0, right = 1, mid; while (left < right) { mid = (left + right) / 2; int cnt = 0; for (int i = 0; i < arrSize - 1; i++) { int pos = less_equal_position(arr, arrSize, i, mid); cnt += (arrSize - pos); } if (cnt == k) break; if (cnt > k) right = mid; else left = mid; } double val = 0; int x = 0, y = 0; for (int i = 0; i < arrSize; i++) for (int j = i + 1; j < arrSize; j++) { double tmp = (double)arr[i] / arr[j]; if (tmp <= mid) { if (val < tmp) { val = tmp; x = arr[i]; y = arr[j]; } break; } } int *ans = (int *)malloc(2 * sizeof(int)); ans[0] = x; ans[1] = y; *returnSize = 2; return ans; } int main() { int a1[] = {1, 2, 3, 5}, a2[] = {1, 7}, rs1, rs2; int *ksmpf1 = kthSmallestPrimeFraction(a1, ARRAY_SIZE(a1), 3, &rs1); int *ksmpf2 = kthSmallestPrimeFraction(a2, ARRAY_SIZE(a2), 1, &rs2); for (int i = 0; i < rs1; i++) printf("%d ", ksmpf1[i]); // expect: 2 5 printf("\n"); for (int i = 0; i < rs2; i++) printf("%d ", ksmpf2[i]); // expect: 1 7 printf("\n"); }