LeetCode #786 — MEDIUM

K-th Smallest Prime Fraction

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 k^th 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?

Step 01

Brute Force Baseline

Problem summary: 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].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search

Example 1

[1,2,3,5]
3

Example 2

[1,7]
1

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #786: K-th Smallest Prime Fraction
class Solution {
    public int[] kthSmallestPrimeFraction(int[] arr, int k) {
        int n = arr.length;
        Queue<Frac> pq = new PriorityQueue<>();
        for (int i = 1; i < n; i++) {
            pq.offer(new Frac(1, arr[i], 0, i));
        }
        for (int i = 1; i < k; i++) {
            Frac f = pq.poll();
            if (f.i + 1 < f.j) {
                pq.offer(new Frac(arr[f.i + 1], arr[f.j], f.i + 1, f.j));
            }
        }
        Frac f = pq.peek();
        return new int[] {f.x, f.y};
    }

    static class Frac implements Comparable {
        int x, y, i, j;

        public Frac(int x, int y, int i, int j) {
            this.x = x;
            this.y = y;
            this.i = i;
            this.j = j;
        }

        @Override
        public int compareTo(Object o) {
            return x * ((Frac) o).y - ((Frac) o).x * y;
        }
    }
}

// Accepted solution for LeetCode #786: K-th Smallest Prime Fraction
type frac struct{ x, y, i, j int }
type hp []frac

func (a hp) Len() int           { return len(a) }
func (a hp) Swap(i, j int)      { a[i], a[j] = a[j], a[i] }
func (a hp) Less(i, j int) bool { return a[i].x*a[j].y < a[j].x*a[i].y }
func (a *hp) Push(x any)        { *a = append(*a, x.(frac)) }
func (a *hp) Pop() any          { l := len(*a); tmp := (*a)[l-1]; *a = (*a)[:l-1]; return tmp }

func kthSmallestPrimeFraction(arr []int, k int) []int {
	n := len(arr)
	h := make(hp, 0, n-1)
	for i := 1; i < n; i++ {
		h = append(h, frac{1, arr[i], 0, i})
	}
	heap.Init(&h)
	for i := 1; i < k; i++ {
		f := heap.Pop(&h).(frac)
		if f.i+1 < f.j {
			heap.Push(&h, frac{arr[f.i+1], arr[f.j], f.i + 1, f.j})
		}
	}
	return []int{h[0].x, h[0].y}
}

# Accepted solution for LeetCode #786: K-th Smallest Prime Fraction
class Solution:
    def kthSmallestPrimeFraction(self, arr: List[int], k: int) -> List[int]:
        h = [(1 / y, 0, j + 1) for j, y in enumerate(arr[1:])]
        heapify(h)
        for _ in range(k - 1):
            _, i, j = heappop(h)
            if i + 1 < j:
                heappush(h, (arr[i + 1] / arr[j], i + 1, j))
        return [arr[h[0][1]], arr[h[0][2]]]

// Accepted solution for LeetCode #786: K-th Smallest Prime Fraction
struct Solution;
use std::cmp::Ord;
use std::cmp::Ordering;

use std::collections::BinaryHeap;

struct Fraction(i32, i32, usize, usize);

impl PartialEq for Fraction {
    fn eq(&self, other: &Self) -> bool {
        self.0 * other.1 == self.1 * other.0
    }
}

impl Eq for Fraction {}

impl PartialOrd for Fraction {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

impl Ord for Fraction {
    fn cmp(&self, other: &Self) -> Ordering {
        (self.1 * other.0).cmp(&(self.0 * other.1))
    }
}

impl Solution {
    fn kth_smallest_prime_fraction(a: Vec<i32>, k: i32) -> Vec<i32> {
        let mut queue: BinaryHeap<Fraction> = BinaryHeap::new();
        let n = a.len();
        let k = k as usize;
        for i in 0..n {
            queue.push(Fraction(a[i], a[n - 1], i, n - 1));
        }
        for _ in 0..k - 1 {
            let f = queue.pop().unwrap();
            if f.3 - 1 > f.2 {
                queue.push(Fraction(a[f.2], a[f.3 - 1], f.2, f.3 - 1));
            }
        }
        let f = queue.pop().unwrap();
        vec![f.0, f.1]
    }
}

#[test]
fn test() {
    let a = vec![1, 2, 3, 5];
    let k = 3;
    let res = vec![2, 5];
    assert_eq!(Solution::kth_smallest_prime_fraction(a, k), res);
    let a = vec![1, 7];
    let k = 1;
    let res = vec![1, 7];
    assert_eq!(Solution::kth_smallest_prime_fraction(a, k), res);
}

// Accepted solution for LeetCode #786: K-th Smallest Prime Fraction
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #786: K-th Smallest Prime Fraction
// class Solution {
//     public int[] kthSmallestPrimeFraction(int[] arr, int k) {
//         int n = arr.length;
//         Queue<Frac> pq = new PriorityQueue<>();
//         for (int i = 1; i < n; i++) {
//             pq.offer(new Frac(1, arr[i], 0, i));
//         }
//         for (int i = 1; i < k; i++) {
//             Frac f = pq.poll();
//             if (f.i + 1 < f.j) {
//                 pq.offer(new Frac(arr[f.i + 1], arr[f.j], f.i + 1, f.j));
//             }
//         }
//         Frac f = pq.peek();
//         return new int[] {f.x, f.y};
//     }
// 
//     static class Frac implements Comparable {
//         int x, y, i, j;
// 
//         public Frac(int x, int y, int i, int j) {
//             this.x = x;
//             this.y = y;
//             this.i = i;
//             this.j = j;
//         }
// 
//         @Override
//         public int compareTo(Object o) {
//             return x * ((Frac) o).y - ((Frac) o).x * y;
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.