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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");
}
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