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authorjack <[email protected]>2024-06-14 13:13:15 -0600
committerjack <[email protected]>2024-06-14 13:13:15 -0600
commitd343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch)
tree96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /kth_smallest_prime_fraction.c
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+// 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");
+}