0373. Find K Pairs with Smallest Sums

https://leetcode.com/problems/find-k-pairs-with-smallest-sums

Description

You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k.

Define a pair (u, v) which consists of one element from the first array and one element from the second array.

Return the k pairs (u1, v1), (u2, v2), ..., (uk, vk) with the smallest sums.

Example 1:

**Input:** nums1 = [1,7,11], nums2 = [2,4,6], k = 3
**Output:** [[1,2],[1,4],[1,6]]
**Explanation:** The first 3 pairs are returned from the sequence: [1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]

Example 2:

**Input:** nums1 = [1,1,2], nums2 = [1,2,3], k = 2
**Output:** [[1,1],[1,1]]
**Explanation:** The first 2 pairs are returned from the sequence: [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]

Example 3:

**Input:** nums1 = [1,2], nums2 = [3], k = 3
**Output:** [[1,3],[2,3]]
**Explanation:** All possible pairs are returned from the sequence: [1,3],[2,3]

Constraints:

• 1 <= nums1.length, nums2.length <= 105

• -109 <= nums1[i], nums2[i] <= 109

• nums1 and nums2 both are sorted in ascending order.

• 1 <= k <= 1000

ac

note: java 8 lambda is much slower than anonymous class.

class Solution {
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        List<int[]> res = new ArrayList<int[]>();
        // edge cases
        if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0 || k <= 0)
            return res;

        // object: [idx1, idx2]
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> {return nums1[a[0]] + nums2[a[1]] - nums1[b[0]] - nums2[b[1]];});

        // init
        for (int i = 0; i < nums1.length && i <= k; i++) {
            pq.add(new int[]{i, 0});
        }
        // get result
        while (!pq.isEmpty() && res.size() < k) {
            int[] curr = pq.poll();
            res.add(new int[]{nums1[curr[0]], nums2[curr[1]]});
            if (curr[1] + 1 < nums2.length) { 
                curr[1]++;
                pq.offer(curr);
            }
        }

        return res;
    }
}

/*
priority queue
add initial object to pq
poll one, add to res
move curr object's idx2 if idx2 < len2 -> add to pq
*/