LeetCode-in-Java

786. K-th Smallest Prime Fraction

Hard

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

public class Solution {
    public int[] kthSmallestPrimeFraction(int[] arr, int k) {
        int n = arr.length;
        double low = 0;
        double high = 1.0;
        while (low < high) {
            double mid = (low + high) / 2;
            int[] res = getFractionsLessThanMid(arr, n, mid);
            if (res[0] == k) {
                return new int[] {arr[res[1]], arr[res[2]]};
            } else if (res[0] > k) {
                high = mid;
            } else {
                low = mid;
            }
        }
        return new int[] {};
    }

    private int[] getFractionsLessThanMid(int[] arr, int n, double mid) {
        double maxLessThanMid = 0.0;
        // stores indices of max fraction less than mid;
        int x = 0;
        int y = 0;
        // for storing fractions less than mid
        int total = 0;
        int j = 1;
        for (int i = 0; i < n - 1; i++) {
            // while fraction is greater than mid increment j
            while (j < n && arr[i] > arr[j] * mid) {
                j++;
            }
            if (j == n) {
                break;
            }
            // j fractions greater than mid, n-j fractions smaller than mid, add fractions smaller
            // than mid to total
            total += (n - j);
            double fraction = (double) arr[i] / arr[j];
            if (fraction > maxLessThanMid) {
                maxLessThanMid = fraction;
                x = i;
                y = j;
            }
        }
        return new int[] {total, x, y};
    }
}

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