373. Find K Pairs with Smallest Sums

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.

Solution

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

        int m = nums1.length;
        int n = nums2.length;

        if (k > m * n) {
            k = m * n;
        }
        int[] temp = new int[n];
        int startIndex = 0;

        while (startIndex < k) {
            int min = Integer.MAX_VALUE;
            for (int i = 0; i < n; i++) {
                if (temp[i] < m) {
                    min = Math.min(min, nums1[temp[i]] + nums2[i]);
                }
            }

            int[] current = new int[2];
            for (int i = 0; i < n; i++) {
                if (temp[i] < m && min == nums1[temp[i]] + nums2[i]) {
                    current[0] = nums1[temp[i]];
                    current[1] = nums2[i];

                    temp[i] += 1;
                    break;
                }
            }

            startIndex++;
            result.add(current);
        }

        return result;
    }

}