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]

Solution

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

        List<int[]> ans = new ArrayList<int[]>();
        int indice[] = new int[nums1.length];
        while(k-->0) {
            boolean exist = false;

            long min = Long.MAX_VALUE;
            int minIdx1 = -1;
            int minIdx2 = -1;
            for(int i = 0; i < indice.length; i++) {
                if(indice[i] == nums2.length) continue;
                exist = true;
                long sum = nums1[i] + nums2[indice[i]];
                if(sum < min) {
                    minIdx1 = i;
                    minIdx2 = indice[i];
                    min = sum;   
                }
            }

            if(!exist) break;
            ans.add(new int[]{nums1[minIdx1], nums2[minIdx2]});
            indice[minIdx1]++;
        }
        return ans;
    }
}