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This approach utilizes a min-heap to efficiently get the smallest sums. We initialize the heap with pairs consisting of the first element from nums1 and each element from nums2. We then extract the minimum sum from the heap, track the index of the element from nums2, and push the next pair from nums1 onto the heap. Repeat the process until we've found k pairs or exhausted possibilities.
Time Complexity: O(k * log(min(k, n))) where n is the length of nums2.
Space Complexity: O(min(k, m*n)) used by the heap where m and n are the lengths of nums1 and nums2, respectively.
1import heapq
2
3def kSmallestPairs(nums1, nums2, k):
4 if not nums1 or not nums2:
5 return []
6 min_heap = []
7 for j in range(min(k, len(nums2))): # Ensure we do not exceed possible pairs
8 heapq.heappush(min_heap, (nums1[0] + nums2[j], 0, j))
9
10 result = []
11 while k > 0 and min_heap:
12 sum, i, j = heapq.heappop(min_heap)
13 result.append([nums1[i], nums2[j]])
14 if i + 1 < len(nums1):
15 heapq.heappush(min_heap, (nums1[i + 1] + nums2[j], i + 1, j))
16 k -= 1
17 return resultThe Python solution initializes a min-heap and then iteratively extracts the smallest elements while maintaining the heap size by considering new potential elements from the arrays. This process continues until we have the k smallest pairs.
In this naive approach, we first generate all possible pairs and their sums, storing them in a list. After generating the pairs, we sort them based on their sums and simply return the first k pairs. This approach, while straightforward, is computationally expensive for large input sizes.
Time Complexity: O(m * n * log(m * n)) where m and n are the lengths of nums1 and nums2, respectively.
Space Complexity: O(m * n) for storing all pairs.
1
This C code handles the generation and sorting of all potential pairs using manual struct management, offering insights into a more verbose but similar straightforward solution.