786. K-th Smallest Prime Fraction

Description

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?

Solutions

Solution 1

Python3JavaC++Go

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]]]

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;
        }
    }
}

class Solution {
public:
    vector<int> kthSmallestPrimeFraction(vector<int>& arr, int k) {
        using pii = pair<int, int>;
        int n = arr.size();
        auto cmp = [&](const pii& a, const pii& b) {
            return arr[a.first] * arr[b.second] > arr[b.first] * arr[a.second];
        };
        priority_queue<pii, vector<pii>, decltype(cmp)> pq(cmp);
        for (int i = 1; i < n; ++i) {
            pq.push({0, i});
        }
        for (int i = 1; i < k; ++i) {
            pii f = pq.top();
            pq.pop();
            if (f.first + 1 < f.second) {
                pq.push({f.first + 1, f.second});
            }
        }
        return {arr[pq.top().first], arr[pq.top().second]};
    }
};

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}
}

786. K-th Smallest Prime Fraction

Description

Solutions

Solution 1

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