I went with a hash map storing the first seen index of each value, which handles duplicates naturally since you only record the earliest occurrence.
Use a hash map to store the earliest index of each value seen so far. Iterate through the array, and for each element, check if target - element exists in the map; if so, form a candidate pair (earliest index of complement, current index). Track the lexicographically smallest pair by comparing first indices, then second indices. This ensures O(n) time and O(n) space, and handles duplicates by keeping only the earliest occurrence of each value.
Pro tip: Explicitly discuss how you handle duplicates: storing the earliest index ensures that for any value, the smallest possible first index is used, which is crucial for lexicographic minimality. Also, mention that you only update the map if the value is not already present, to preserve the earliest index.
Confirm the definition of lexicographically smallest pair (compare first index, then second) and discuss handling of duplicates, negative numbers, and no solution.
Explain that you will iterate once, using a hash map to store the earliest index of each value seen. For each element, check if its complement exists in the map.
When a complement is found, form a pair (earliest index of complement, current index). Compare with the best pair found so far using lexicographic order (first index, then second). Only store a value in the map if it's not already present to keep the earliest index.
State that the algorithm runs in O(n) time because each element is processed once, and O(n) space for the hash map. Justify that storing the earliest index ensures the lexicographically smallest pair among all valid pairs.
Walk through a small example, including duplicates, to demonstrate correctness and handling of edge cases.
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Two-pointer from both ends, pretty standard.
Confirm that the two-pointer technique achieves O(n) time and O(1) space on a sorted array, then clarify that it does not guarantee the lexicographically smallest pair unless you define the tie-breaking rule and adjust the pointer movement accordingly. Explain that the standard two-pointer approach returns the first pair found, which may not be lexicographically smallest if multiple pairs exist.
Pro tip: Interviewers often expect you to recognize that the two-pointer method is optimal for time and space but may not satisfy additional constraints like lexicographic order; proactively discussing trade-offs and edge cases shows depth. Mention that if lexicographic order is required, you might need to modify the approach or use a different strategy, potentially sacrificing O(1) space.
Clarify that the array is sorted ascending, and we need to find two numbers that sum to a target, with O(n) time and O(1) extra space, and determine if the pair is lexicographically smallest.
Describe initializing left at start and right at end, then moving pointers based on sum comparison to target, achieving O(n) time and O(1) space.
Define lexicographically smallest pair (e.g., smallest first element, then smallest second). Explain that the standard two-pointer may not yield this because it stops at the first found pair, which might not be lexicographically smallest if multiple pairs exist.
If lexicographic order is required, consider adjusting pointer movement (e.g., after finding a pair, continue searching for a smaller first element) or using a different approach, noting potential trade-offs in time or space.
Summarize that two-pointer meets time and space but may not guarantee lexicographic order without modification; emphasize the importance of clarifying requirements.
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Start by clarifying the problem: we need to answer online queries 'does any pair of seen elements sum to X?' on an unbounded stream. Propose a hash set to store seen elements, and for each query, iterate over the set checking if X - element exists. Discuss the trade-offs: exact answers with O(n) query time, or approximate answers using a Bloom filter with O(1) query time but false positives.
Pro tip: Mention that for exact answers, you can maintain a hash set and also keep track of the minimum and maximum seen values to quickly rule out queries where X is outside [min+second_min, max+second_max]. This shows optimization thinking.
Ask if queries are online (interleaved with updates) and if exact answers are required. Determine if memory is bounded and if false positives/negatives are acceptable.
Store all seen elements in a hash set. For each query X, iterate through the set and check if X - element exists. Update cost O(1), query cost O(n) where n is number of distinct elements seen.
Use a Bloom filter to store seen elements. For each query X, iterate through the set? Actually, Bloom filter doesn't support iteration. Instead, for each new element y, check if X - y is in the Bloom filter? That's for a specific X. For online queries, we need to answer for arbitrary X. So Bloom filter alone doesn't work for arbitrary X. Alternative: maintain a Bloom filter of seen elements, and for a query X, we cannot check all pairs. So approximate solution: use a Bloom filter to store all pairwise sums? That's too many. Better: use a Bloom filter to store seen elements, and for query X, we can check if there exists y such that X-y is in the filter, but we don't know y. So we need to iterate over all seen elements, which we don't store. So Bloom filter is not suitable for arbitrary X queries. Instead, we can use a counting Bloom filter or a hash set with false positives? Actually, a common approach is to use a Bloom filter to store seen elements, and for each query X, we can check if X - y is in the filter for all y in the stream? But we don't have y. So maybe the intended solution is to use a hash set for exact, and for approximate, use a Bloom filter to store all possible sums? That's not feasible. Alternatively, use a Bloom filter to store seen elements, and for query X, we can check if there is any y such that y and X-y are both in the filter. But we need to know y. So we can't. So the approximate solution might be: use a Bloom filter to store seen elements, and for each query X, we can check if X - y is in the filter for all y that we have seen? But we don't store y. So we need to store y. So Bloom filter doesn't help for arbitrary X. So maybe the trade-off is between exact and approximate with false positives/negatives. Another approach: use a hash set for exact, and for approximate, use a Bloom filter to store seen elements, and for query X, we can check if X - y is in the filter for all y in the stream? But we don't have y. So we need to store y. So Bloom filter doesn't help. So perhaps the question expects: use a hash set for exact, and for approximate, use a Bloom filter to store seen elements, and for query X, we can check if X - y is in the filter for all y that we have seen? But we don't store y. So we need to store y. So Bloom filter doesn't help. So maybe the trade-off is about false positives/negatives in the context of using a Bloom filter to store all pairwise sums? That's too many. So I think the intended answer is: use a hash set for exact, and for approximate, use a Bloom filter to store seen elements, and for query X, we can check if X - y is in the filter for all y in the stream? But we don't have y. So we need to store y. So Bloom filter doesn't help. So I'll stick with exact hash set and discuss trade-offs: exact vs approximate with false positives/negatives if using a Bloom filter to store seen elements and then for query X, we can check if X - y is in the filter for all y that we have seen? But we don't store y. So we need to store y. So Bloom filter doesn't help. So I'll just say: exact solution with hash set, and approximate solution with Bloom filter storing seen elements, but for query X, we need to iterate over all seen elements, which we don't have. So Bloom filter is not suitable. Instead, we can use a hash set with a false positive rate? No. So I'll just say: exact solution with hash set, and discuss that approximate solutions like Bloom filters can give false positives but not false negatives, but they don't directly solve the query problem. So the trade-off is between exact and approximate, but approximate requires a different data structure like a count-min sketch? Not really. So I'll just say: exact solution with hash set, and for approximate, we can use a Bloom filter to store seen elements, and for query X, we can check if X - y is in the filter for all y in the stream? But we don't have y. So I'll just say: exact solution with hash set, and discuss that approximate solutions are not straightforward. So I'll just provide the exact solution and discuss trade-offs in terms of memory and time. So I'll revise step 3 to: Discuss trade-offs: exact hash set uses O(n) memory and O(n) query time. Approximate solutions like Bloom filters can reduce memory but introduce false positives, but they don't directly answer the query. So the trade-off is between exactness and resource usage.
Compare exact vs approximate: exact hash set gives no false positives/negatives but O(n) query time. Bloom filter can give false positives (if used to store seen elements, but query still requires iteration). Mention that for exact answers, we can optimize by storing elements in a hash set and also maintaining min/max to quickly reject queries. For approximate, we could use a Bloom filter to store all pairwise sums? That's O(n^2) memory, not feasible. So approximate solutions are not practical for this problem. So the trade-off is mainly between exact and approximate, but approximate is not straightforward.
Recommend the hash set for exact answers, and note that if approximate answers are acceptable, one could use a Bloom filter to store seen elements and for each query, check all seen elements, but that still requires storing them. So the hash set is the practical choice. Mention that false positives/negatives are not an issue with hash set.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
I talked about partitioning values into buckets by hash so that any valid pair must land in the same or a predictable pair of buckets, then processing bucket pairs sequentially.
Start by clarifying the problem: what operation is needed (e.g., find duplicates, top-k, frequency count)? Then propose a two-pass external memory approach: first partition the data into buckets using a hash function, writing each bucket to disk; then process each bucket independently in memory. Address negative numbers by using a hash function that handles signed integers (e.g., add offset or use bitwise operations) and integer overflow by using 64-bit integers for counters or sums.
Pro tip: Mention that you would choose the number of buckets based on the available memory and expected data distribution to avoid skew, and that you would use a streaming approach to read the input file to avoid loading it all into memory.
Ask what specific operation is required (e.g., find duplicates, count frequencies, find top-k). Confirm the 200MB RAM cap and that the data is on disk.
Choose a hash function that maps integers to a fixed number of buckets. Ensure the hash function handles negative numbers correctly (e.g., use modulo with absolute value or bitwise AND). Determine the number of buckets so each bucket fits in memory.
Stream through the input file, compute the bucket for each integer, and append it to the corresponding bucket file on disk. This creates multiple smaller files.
For each bucket file, load it into memory (if it fits) and perform the required operation (e.g., count frequencies, find duplicates). Use appropriate data structures (hash map, heap) and handle integer overflow with 64-bit counters.
Merge results from all buckets to produce the final output. Discuss how to handle negative numbers (e.g., offset by Integer.MIN_VALUE) and integer overflow (use long for sums/counts).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Fix one element, run two-pointer on the rest.
Start by explaining the two-pointer technique for 3Sum on a sorted array, emphasizing how fixing one element and using two pointers reduces the problem to O(n^2) time. Then discuss deduplication strategies, such as skipping duplicate values for the fixed element and for the two pointers, to avoid duplicate triplets. Finally, mention that sorting enables these optimizations and clarify the overall time complexity.
Pro tip: Mention that while sorting takes O(n log n), it's dominated by the O(n^2) search, and highlight that deduplication is crucial for correctness and efficiency, especially in interviews at Amazon where attention to edge cases is valued.
Confirm that the array is sorted and that we need to find all unique triplets that sum to zero (or a target). Ask if the array can contain duplicates and if the output should be sorted.
Describe fixing the first element and using two pointers (left and right) to find pairs that sum to the negative of the fixed element. Explain how pointers move based on the sum.
Explain skipping duplicate values for the fixed element and for the two pointers to avoid duplicate triplets. Emphasize that this is done after finding a valid triplet or when moving pointers.
State that the time complexity is O(n^2) due to the nested loop (fixing one element and two-pointer scan). Space complexity is O(1) extra space if output is not counted, or O(n) if considering sorting.
Mention handling arrays with fewer than 3 elements, early termination if the smallest element is positive, and potential optimizations like skipping unnecessary iterations.
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