0373.Find_K_Pairs_with_Smallest_Sums.py

"""
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.

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]
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]
 

Constraints:

1 <= nums1.length, nums2.length <= 105
-109 <= nums1[i], nums2[i] <= 109
nums1 and nums2 both are sorted in ascending order.
1 <= k <= 104
"""

class Solution:
    def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
        hq = []
        heappush(hq, (nums1[0] + nums2[0], 0, 0))
        visited = set()
        visited.add((0, 0))
        
        res = []
        m, n = len(nums1), len(nums2)
        for _ in range(min(m*n, k)):
            curr, i, j = heappop(hq)
            res.append([nums1[i], nums2[j]])
            if i + 1 < m and (i + 1, j) not in visited:
                heappush(hq, (nums1[i + 1] + nums2[j], i + 1, j))
                visited.add((i + 1, j))
            if j + 1 < n and (i, j + 1) not in visited:
                heappush(hq, (nums1[i] + nums2[j + 1], i, j + 1))
                visited.add((i, j + 1))
        return res