LeetCode - Find k Pairs with Smallest Sums - Day2 (1/2/2020)
Find k Pairs with Smallest Sums
Question
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:
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]
Solution
To be honest, I couldn't solve this question myself. I could come up the brief process but I was not able to turn it to the code.
According to the solution made by someone, we can solve it with the run time O(K logK).
public class Solution { public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) { PriorityQueue<int[]> que = new PriorityQueue<>((a,b)->a[0]+a[1]-b[0]-b[1]); List<int[]> res = new ArrayList<>(); if(nums1.length==0 || nums2.length==0 || k==0) return res; for(int i=0; i<nums1.length && i<k; i++) que.offer(new int[]{nums1[i], nums2[0], 0}); while(k-- > 0 && !que.isEmpty()){ int[] cur = que.poll(); res.add(new int[]{cur[0], cur[1]}); if(cur[2] == nums2.length-1) continue; que.offer(new int[]{cur[0],nums2[cur[2]+1], cur[2]+1}); } return res; } }
He put the diagram, too.
Umm... Idk what's going on lol
Okay, let's generalize the question first. So what we need to do is to pick the combination of nums1[i] & nums2[j] (i and j are indices.). Obviously, nuns1[0] & nums2[0] comes first since the arrays are sorted. Then, we will pick nums1[i + 1] & nums2[j] or nums1[i] & nums2[j + 1].
Now, we know what we have to do next. We will use min-heap to get a min candidate (it takes O(logN) to add and remove.) and then add nums1[i + 1] & nums2[j] or nums1[i] & nums2[j+1] until we get K number of elements.
Wow that's awesome. This concept comes from Shiroyama-san's blog and @kenkoooo -san's gist.
I learned that this is exactly the way to interpreter LeetCode question as a mathematical problem.
public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k) { List<List<Integer>> res = new ArrayList<>(); if (nums1.length == 0 || nums2.length == 0 || k == 0) { return res; } PriorityQueue<List<Integer>> pq = new PriorityQueue<>((o1, o2) -> nums1[o1.get(0)] + nums2[o1.get(1)] - nums1[o2.get(0)] - nums2[o2.get(1)]); TreeSet<List<Integer>> added = new TreeSet<>((o1, o2) -> { if (o1.get(0) != o2.get(0)) { return o1.get(0) - o2.get(0); } return o1.get(1) - o2.get(1); }); pq.add(new ArrayList<>(Arrays.asList(0, 0))); added.add(new ArrayList<>(Arrays.asList(0, 0))); while (res.size() < k && !pq.isEmpty()) { List<Integer> ans = pq.poll(); int i = ans.get(0); int j = ans.get(1); res.add(Arrays.asList(nums1[i], nums2[j])); if (i + 1 < nums1.length && !added.contains(new ArrayList<>(Arrays.asList(i + 1, j)))) { pq.add(new ArrayList<>(Arrays.asList(i + 1, j))); added.add(new ArrayList<>(Arrays.asList(i + 1, j))); } if (j + 1 < nums2.length && !added.contains(new ArrayList<>(Arrays.asList(i, j + 1)))) { pq.add(new ArrayList<>(Arrays.asList(i, j + 1))); added.add(new ArrayList<>(Arrays.asList(i, j + 1))); } } return res; }
Time complexity
O(K log(K)).
Space complexity
O(K).