K-th Smallest Prime Fraction

You are 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.length, we consider the fraction arr[i] / arr[j].

Return the kth smallest fraction considered. Return your answer as an array of integers of size 2, where answer[0] == arr[i] and answer[1] == arr[j].

 

Example 1:

Input: arr = [1,2,3,5], k = 3
Output: [2,5]
Explanation: The fractions to be considered in sorted order are:
1/5, 1/3, 2/5, 1/2, 3/5, and 2/3.
The third fraction is 2/5.

Example 2:

Input: arr = [1,7], k = 1
Output: [1,7]

 

Constraints:

  • 2 <= arr.length <= 1000
  • 1 <= arr[i] <= 3 * 104
  • arr[0] == 1
  • arr[i] is a prime number for i > 0.
  • All the numbers of arr are unique and sorted in strictly increasing order.
  • 1 <= k <= arr.length * (arr.length - 1) / 2
SOLUTION:
import heapq

class Solution:
    def kthSmallestPrimeFraction(self, arr: List[int], k: int) -> List[int]:
        n = len(arr)
        heap = []
        for i in range(n):
            for j in range(n - 1, i, -1):
                heapq.heappush(heap, (-arr[i] / arr[j], (arr[i], arr[j])))
                if len(heap) > k:
                    heapq.heappop(heap)
        return list(heap[0][1])

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