Find K Pairs with Smallest Sums
Given two sorted integer arrays `nums1` and `nums2`, and an integer $k$, return the $k$ pairs $(u, v)$ with the smallest sums, where $u$ is from `nums1` and $v$ is from `nums2`. Use a Min-Heap.
Why Interviewers Ask This
Apple interviewers ask this to evaluate a candidate's ability to optimize brute-force solutions using advanced data structures. They specifically test if you can recognize that sorting all pairs is inefficient and instead leverage the sorted nature of input arrays with a Min-Heap for optimal performance. This assesses your grasp of time complexity trade-offs and priority queue manipulation in real-world constraint scenarios.
How to Answer This Question
1. Clarify constraints: Ask about array sizes and whether k exceeds the total possible pairs, as Apple values handling edge cases.
2. Identify the pattern: Explain that since inputs are sorted, the smallest sum must involve nums1[0] and nums2[0], establishing a baseline.
3. Propose the Min-Heap strategy: Describe initializing the heap with (nums1[i] + nums2[0], i, 0) for the first few elements or just the first pair, then iteratively extracting the minimum.
4. Detail the expansion logic: When popping (sum, i, j), push the next potential candidate (i, j+1) into the heap, ensuring you explore neighbors without redundant calculations.
5. Optimize space: Mention limiting the heap size to k or min(len(nums1), k) to save memory, demonstrating awareness of resource efficiency which aligns with Apple's focus on polished, performant code.
Key Points to Cover
- Recognizing that brute force generation of all pairs is too slow for large inputs
- Correctly initializing the Min-Heap with the smallest known candidates
- Understanding the specific expansion rule: moving from (i,j) to (i,j+1)
- Achieving O(K log K) time complexity rather than O(N*M)
- Handling boundary conditions where k is larger than the total number of pairs
Sample Answer
To solve finding the K pairs with the smallest sums efficiently, I would avoid generating all combinations, which would result in O(N*M log N*M) complexity. Instead, I'd leverage the fact that both arrays are already sorted. The core insight is that the smallest sum is always at indices (0,0). From there, any popped element at (i, j) can only be extended to (i, j+1) to find the next smallest candidate involving the same row. I would initialize a Min-Heap with the pair (nums1[0] + nums2[0], 0, 0). In each iteration, I extract the minimum sum from the heap and add it to my result list. Then, I insert the next valid neighbor into the heap. If the current column index is zero, I also consider pushing (i+1, 0) to ensure all rows are explored, though a more optimized approach often initializes the heap with the first element of the first k rows to reduce initial overhead. This approach ensures we visit exactly k elements, resulting in an O(K log K) time complexity and O(K) space complexity. For example, with nums1 = [1,7,11] and nums2 = [2,4,6], the heap guides us to pick (1,2), then (1,4), then (7,2) sequentially. This method is highly scalable and demonstrates how to transform a naive O(N^2) problem into an efficient solution by utilizing the sorted property and priority queues effectively.
Common Mistakes to Avoid
- Generating all N*M pairs first and then sorting them, which fails on large datasets due to excessive memory usage
- Forgetting to check if the next index is within bounds before adding it to the heap, causing runtime errors
- Initializing the heap with only one element and failing to cover multiple starting rows when k is large
- Using a Max-Heap instead of a Min-Heap, which reverses the logic needed to find the smallest sums
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