373. Find K Pairs with Smallest Sums

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373. Find K Pairs with Smallest Sums

Description

Difficulty: Medium

Related Topics: Array, Heap (Priority Queue)

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:

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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:

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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:

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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]

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

Hints/Notes

  • Priority queue
  • Don’t think about moving 2 needles together, pair the first item in 1st list with all items in the 2nd list

Solution

Language: C++

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class Solution {
public:
    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        int size1 = nums1.size(), size2 = nums2.size();
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
        for (int i = 0; i < size1; i++) {
            pq.push({nums1[i] + nums2[0], i, 0});
        }
        vector<vector<int>> res;
        while (k > 0 && !pq.empty()) {
            vector<int> top = pq.top();
            pq.pop();
            int i1 = top[1];
            int i2 = top[2];
            res.push_back({nums1[i1], nums2[i2]});
            if (i2 < size2 - 1) {
                i2++;
                pq.push({nums1[i1] + nums2[i2], i1, i2});
            }
            k--;
        }
        return res;
    }
};