Sponsored
Sponsored
This approach uses a min-heap to efficiently find the kth smallest fraction. By iterating through each pair of elements (i, j) in the array where i < j, we can calculate the fraction arr[i]/arr[j] and push it into the heap. The heap is then used to retrieve the k-th smallest element efficiently.
Time Complexity: O(n^2 log(n^2))
Space Complexity: O(n^2)
1import java.util.PriorityQueue;
2
3public class KthSmallestPrimeFraction {
4 public int[] kthSmallestPrimeFraction(int[] arr, int k) {
5 int n = arr.length;
6 PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> arr[a[0]] * arr[b[1]] - arr[b[0]] * arr[a[1]]);
7 for (int i = 0; i < n - 1; i++) {
8 pq.offer(new int[]{i, n - 1});
9 }
10 int[] result = new int[2];
11 while (k-- > 0) {
12 result = pq.poll();
13 if (result[1] - 1 > result[0]) {
14 pq.offer(new int[]{result[0], result[1] - 1});
15 }
16 }
17 return new int[]{arr[result[0]], arr[result[1]]};
18 }
19}In Java, we use a priority queue to store potential fractions using the comparator to sort by fraction value. We repeatedly extract the smallest fraction and use the k-th extraction to obtain our result.
This approach uses binary search on the value of fractions. By implementing a search over the potential range of fraction values, the algorithm counts how many fractions are less than a given mid-point, narrowing down to the k-th one by manipulating the bounds of the potential fraction values. A two-pointer technique helps count the fractions efficiently.
Time Complexity: O(n log(max_fraction))
Space Complexity: O(1)
1
This Java approach implements binary searching over fraction values with a two-pointer count. The mid-point and count help narrow the range and identify the target k-th smallest fraction using refined boundaries.