786. 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 fori > 0
.- All the numbers of
arr
are unique and sorted in strictly increasing order. 1 <= k <= arr.length * (arr.length - 1) / 2
Rust Solution
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);
}
Having problems with this solution? Click here to submit an issue on github.