# 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.

```java
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
*/
```