Pretty approachable if you've done merge sort variants before.
Use a min-heap to perform a k-way merge across the three arrays, tracking the last added value to skip duplicates. Alternatively, use three pointers to merge sequentially while deduplicating. Discuss time and space complexity and handle edge cases like empty arrays.
Pro tip: Clarify whether the input arrays can contain duplicates and whether the output must be sorted in ascending order. Also, mention that if the arrays are very large, a heap-based approach is more efficient than merging two at a time.
Ask about input sizes, whether arrays can be empty, if duplicates exist within each array, and if the output should be a new array or in-place. Confirm the expected time/space complexity.
Decide between a heap-based k-way merge (O(N log k) time) or iterative two-array merge (O(N * k) time). Explain the trade-offs and pick the most efficient for the given constraints.
For heap approach: initialize a min-heap with the first element of each non-empty array, along with array index and element index. Repeatedly extract the minimum, add to result if different from last added, and push the next element from the same array. For pointer approach: merge arrays one by one while skipping duplicates.
During merge, compare the current element with the last added element to skip duplicates. Handle empty arrays by ignoring them in the heap or pointer initialization. Ensure the result is sorted and contains no duplicates.
State time complexity: O(N log k) for heap, where N is total elements and k=3; space O(N) for output plus O(k) for heap. Walk through a small example to verify correctness, including duplicates and empty arrays.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one took me a minute to even figure out the right approach.
Use BFS level-by-level to remove one parenthesis at a time, checking validity at each level. Stop at the first level where valid strings are found, as that represents the minimum removals. Use a set to avoid duplicates and return all valid strings at that level.
Pro tip: Mention that BFS guarantees minimum removals because it explores all strings with k removals before k+1 removals. Also, deduplicate at each level to avoid exponential blowup, and note that the maximum removals is n, so the algorithm is O(2^n) in the worst case but often much faster.
Confirm that a valid string has balanced parentheses and no unmatched closing before opening. Define a helper function to check validity in O(n) time.
Start with the original string in a queue. For each level, process all strings, and if any is valid, return all valid strings at that level. Otherwise, generate next level by removing one parenthesis from each string.
Use a set to store strings at each level, ensuring no duplicates are processed. This is crucial for efficiency and to return distinct results.
For each string, iterate through characters; if a character is '(' or ')', create a new string by removing it. Add to next level set.
When valid strings are found at a level, return them as a list. Discuss time and space complexity: worst-case O(2^n) but often better due to pruning.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.