diff options
author | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
---|---|---|
committer | jack <[email protected]> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /kth_smallest_prime_fraction.c | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
initial
Diffstat (limited to 'kth_smallest_prime_fraction.c')
-rw-r--r-- | kth_smallest_prime_fraction.c | 76 |
1 files changed, 76 insertions, 0 deletions
diff --git a/kth_smallest_prime_fraction.c b/kth_smallest_prime_fraction.c new file mode 100644 index 0000000..c3a8126 --- /dev/null +++ b/kth_smallest_prime_fraction.c @@ -0,0 +1,76 @@ +// 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"); +} |