373. Find K Pairs with Smallest Sums

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.

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]

Credits:
Special thanks to @elmirap and @StefanPochmann for adding this problem and creating all test cases.

通过这个题熟悉PriorityQueue用法。

Basic idea: Use min_heap to keep track on next minimum pair sum, and we only need to maintain K possible candidates in the data structure.

Some observations: For every numbers in nums1, its best partner(yields min sum) always strats from nums2[0] since arrays are all sorted; And for a specific number in nums1, its next candidate sould be [this specific number] + nums2[current_associated_index + 1], unless out of boundary;)

Here is a simple example demonstrate how this algorithm works.

The run time complexity is O(kLogk) since que.size <= k and we do at most k loop.

代码如下：

public class Solution {
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        List<int[]> res = new ArrayList<int[]>();
        PriorityQueue<int[]> queue = new PriorityQueue<int[]>((a, b) -> a[0] + a[1] - b[0] - b[1]);
        if (nums1.length == 0 || nums2.length == 0 || k == 0) {
            return res;
        }
        for (int i = 0; i < nums1.length && i < k; i++) {
            queue.offer(new int[]{nums1[i], nums2[0], 0});
        }
        while (k > 0 && !queue.isEmpty()) {
            int[] curr = queue.poll();
            res.add(new int[]{curr[0], curr[1]});
            k --;
            if (curr[2] == nums2.length - 1) {
                continue;
            }
            queue.offer(new int[]{curr[0], nums2[curr[2] + 1], curr[2] + 1});
        }
        return res;
    }
}