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

Code

Time Complexity: $O(k\log k)$, Space Complexity: $O(k)$

類似 k-way merge 的概念。對於 [a, b, c], [d, e, f] 來說，先把 [a, b, c] 和 [d] 配，然後 pop 出來的 [a, d]，因為 a 固定了，所以一定必須把 [a, e] push 進去（下一個順位）。

class Solution {
public:
    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        auto cmp = [&](const pair<int, int>& p1, const pair<int, int>& p2){
            return nums1[p1.first] + nums2[p1.second] > nums1[p2.first] + nums2[p2.second];
        };
        // min heap
        priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(cmp)> pq(cmp);
 
        for(int i = 0; i < nums1.size(); i++) {
            pq.push({i, 0});
        }
        
        vector<vector<int>> res;
        while(k-- && !pq.empty()) {
            auto P = pq.top();
            pq.pop();
            res.push_back({nums1[P.first], nums2[P.second]});
            if(P.second == nums2.size() - 1)
                continue;
            pq.push({P.first, P.second + 1});
        }
        return res;
    }
};

Source

• Find K Pairs with Smallest Sums - LeetCode