Two-pass approach: left to right tracking unmatched closing parens, right to left for unmatched opens.
Use a stack to track unmatched opening parentheses and a set to mark invalid closing parentheses. Then build the result by skipping marked characters, ensuring the minimum removals.
Pro tip: Clarify that 'minimum removals' means removing only characters that cause invalidity, and mention that multiple valid answers exist so any is acceptable.
Confirm that we need to remove the fewest characters to make parentheses valid, and that any valid result is acceptable.
Traverse the string, using a stack to match opening and closing parentheses, and mark unmatched closing parentheses and leftover opening parentheses as invalid.
Construct the output string by including only characters that are not marked for removal.
State that the time and space complexity are O(n), where n is the length of the string.
Walk through edge cases like empty string, all invalid parentheses, and nested valid parentheses to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Classic two-pointer trick: walk both nodes up to root simultaneously, swapping to the other node's start when you hit null.
Since each node has a parent pointer, treat the problem as finding the intersection of two linked lists (the paths from each node to the root). Use a two-pointer technique: advance both pointers one step at a time, and when a pointer reaches the root, redirect it to the other node's starting position. They will meet at the LCA after at most O(h) steps, where h is the tree height.
Pro tip: Always clarify with the interviewer whether the nodes are guaranteed to be in the same tree and whether the parent pointers are reliable. Mentioning edge cases like one node being an ancestor of the other shows thoroughness and can lead to a more robust solution.
Confirm that both nodes exist in the same tree, parent pointers are valid, and discuss edge cases such as one node being the root or one node being an ancestor of the other.
Describe how to use two pointers starting at the given nodes, moving up via parent pointers, and redirecting to the other node's start when reaching the root, ensuring they meet at the LCA.
State that the algorithm runs in O(h) time where h is the height of the tree, and uses O(1) extra space, which is optimal.
Mention that using a hash set to store ancestors of one node and then checking the other node's ancestors also works in O(h) time but uses O(h) space, and compare trade-offs.
Walk through edge cases like one node being an ancestor of the other, or nodes at different depths, and confirm the solution handles them correctly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the input structure and depth definition, then choose between recursive DFS and iterative BFS/stack. Implement the solution with careful handling of nested lists and integers, and analyze time and space complexity.
Pro tip: Meta interviewers value clean, bug-free code and clear communication. Start by walking through a simple example to confirm understanding, and discuss trade-offs between recursion and iteration (e.g., recursion depth limits).
Confirm that the input is a nested list where each element is either an integer or a list, and depth starts at 1 for top-level elements. Ask about constraints (e.g., maximum depth, size) and expected output type.
Decide between recursive DFS (simpler, but may hit recursion limit) and iterative BFS/stack (explicit control, avoids recursion depth issues). Consider using a queue for BFS to process level by level.
Write clean code: for DFS, pass depth as parameter and accumulate sum; for BFS, use a queue storing (element, depth) and process each element. Handle both integers and lists correctly.
Walk through a sample input like [1,[4,[6]]] to verify the sum (1*1 + 4*2 + 6*3 = 27). Also test edge cases: empty list, single integer, deeply nested list.
State time complexity O(N) where N is total number of elements (integers and lists), and space complexity O(D) for recursion stack or O(N) for queue in worst case.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Used a max-heap of size K, comparing squared distances to avoid the sqrt.
Clarify the problem constraints (e.g., input size, whether K is valid, and if the output order matters). Then discuss the trade-offs between sorting all points by distance (O(n log n)) and using a max-heap of size K (O(n log K)), and implement the optimal solution with clean code.
Pro tip: Avoid computing square roots by comparing squared distances to prevent floating-point precision issues and improve performance. Also, mention that for very large datasets, a quickselect-based approach can achieve average O(n) time.
Ask about input size, whether K is always valid, if the output needs to be sorted, and if duplicate points are allowed. This shows attention to detail and helps choose the right algorithm.
Explain the brute-force sort approach (O(n log n)) and the heap-based approach (O(n log K)). Mention that for large n and small K, the heap is more efficient.
Select the max-heap of size K as the optimal solution, justifying it by time and space complexity. If K is close to n, sorting might be simpler and equally efficient.
Write clean code using a max-heap (or priority queue) that stores points by their squared distance. Iterate through points, push to heap, and if size exceeds K, pop the farthest. Finally, extract the K points.
State time complexity O(n log K) and space O(K). Test with edge cases like K=0, K=n, duplicate points, and points with same distance.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.