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g0701_0800.s0786_k_th_smallest_prime_fraction.readme.md Maven / Gradle / Ivy

786\. K-th Smallest Prime Fraction

Medium

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`

**Follow up:** Can you solve the problem with better than O(n2) complexity?