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373. Find K Pairs with Smallest Sums

Description

You are given two integer arrays nums1 and nums2 sorted in non-decreasing 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.

Return 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]

 

Constraints:

  • 1 <= nums1.length, nums2.length <= 105
  • -109 <= nums1[i], nums2[i] <= 109
  • nums1 and nums2 both are sorted in non-decreasing order.
  • 1 <= k <= 104
  • k <= nums1.length * nums2.length

Solutions

Solution 1

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class Solution:
    def kSmallestPairs(
        self, nums1: List[int], nums2: List[int], k: int
    ) -> List[List[int]]:
        q = [[u + nums2[0], i, 0] for i, u in enumerate(nums1[:k])]
        heapify(q)
        ans = []
        while q and k > 0:
            _, i, j = heappop(q)
            ans.append([nums1[i], nums2[j]])
            k -= 1
            if j + 1 < len(nums2):
                heappush(q, [nums1[i] + nums2[j + 1], i, j + 1])
        return ans

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class Solution {
    public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        for (int i = 0; i < Math.min(nums1.length, k); ++i) {
            q.offer(new int[] {nums1[i] + nums2[0], i, 0});
        }
        List<List<Integer>> ans = new ArrayList<>();
        while (!q.isEmpty() && k > 0) {
            int[] e = q.poll();
            ans.add(Arrays.asList(nums1[e[1]], nums2[e[2]]));
            --k;
            if (e[2] + 1 < nums2.length) {
                q.offer(new int[] {nums1[e[1]] + nums2[e[2] + 1], e[1], e[2] + 1});
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        auto cmp = [&nums1, &nums2](const pair<int, int>& a, const pair<int, int>& b) {
            return nums1[a.first] + nums2[a.second] > nums1[b.first] + nums2[b.second];
        };

        int m = nums1.size();
        int n = nums2.size();
        vector<vector<int>> ans;
        priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(cmp)> pq(cmp);
        for (int i = 0; i < min(k, m); i++)
            pq.emplace(i, 0);
        while (k-- && !pq.empty()) {
            auto [x, y] = pq.top();
            pq.pop();
            ans.emplace_back(initializer_list<int>{nums1[x], nums2[y]});
            if (y + 1 < n)
                pq.emplace(x<