Find K Pairs with Smallest Sums

Problem:

ou 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.

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

Example 1:

Given nums1 = [1,7,11], nums2 = [2,4,6],  k = 3

Return: [1,2],[1,4],[1,6]

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:

Given nums1 = [1,1,2], nums2 = [1,2,3],  k = 2

Return: [1,1],[1,1]

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:

Given nums1 = [1,2], nums2 = [3],  k = 3 

Return: [1,3],[2,3]

All possible pairs are returned from the sequence:
[1,3],[2,3]

Analysis:
Similar to kth smallest element in a sorted matrix. We can treat the number in arrays as coordinates, and sum as value in matrix.

Solution:

class Solution {
    class Tuple {
        int x;
        int y;
        int sum;
        public Tuple(int x, int y, int sum) {
            this.x = x;
            this.y = y;
            this.sum = sum;
        }
    }
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        List<int[]> res = new ArrayList<>();
        if (nums1 == null || nums2 == null) return res;
        if (nums1.length == 0 || nums2.length == 0) return res;
        PriorityQueue<Tuple> queue = new PriorityQueue<>(k, new Comparator<Tuple>() {
            public int compare(Tuple a, Tuple b) {
                return a.sum - b.sum;
            }
        });
        for (int i = 0; i < nums1.length; i++) {
            queue.offer(new Tuple(i, 0, nums1[i] + nums2[0]));
        }
        for (int i = 0; i < k; i++) {
            if (queue.isEmpty())
                break;
            Tuple cur = queue.poll();
            res.add(new int[]{nums1[cur.x], nums2[cur.y]});
            if (cur.y + 1 < nums2.length) {
                queue.add(new Tuple(cur.x, cur.y + 1, nums1[cur.x] + nums2[cur.y + 1]));    
            }
        }
        return res;
    }
}