The base version was fine, stack-based, explained complexity, tested it.
Start by clarifying the exact definition of a 'run' (e.g., pairs vs. any length) and whether removal is iterative or single-pass. Then propose a stack-based solution that processes characters in one pass, and discuss how the follow-up changes the logic (e.g., using a stack of characters and counts).
Pro tip: Mention that the follow-up (runs of any length) is equivalent to the 'remove all adjacent duplicates' problem, which can be solved with a stack in O(n) time; also note that if removal is iterative (like Candy Crush), a stack of (char, count) is needed to handle cascading removals.
Ask whether a run is exactly two characters or any number, and whether removal is single-pass or iterative (cascading). Confirm input/output format and constraints.
For single-pass removal of any-length runs, use a stack of characters. For iterative removal, use a stack of (character, count) pairs to track runs and trigger removals.
Iterate through the string, pushing characters onto the stack. If the top of the stack matches the current character, increment the count (or pop for pairs). If a run reaches the removal threshold, pop it.
State that the time complexity is O(n) because each character is pushed and popped at most once, and space is O(n) for the stack in the worst case.
Walk through edge cases: empty string, all same characters, alternating characters, and cascading removals (e.g., 'abba' -> '' if iterative).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First clarify that the arrays are sorted and we need to return a new sorted array of squares. Then present a two-pointer approach from the end of the merged array, comparing absolute values to place the largest square at the end, achieving O(n+m) time and O(1) extra space (excluding output).
Pro tip: Explicitly call out that squaring can destroy the sorted order when negatives are present, and that comparing absolute values is the key insight. This shows you anticipate edge cases and think about invariants, which interviewers value.
Confirm that the input arrays are sorted, can contain negative numbers, and that the output should be a new sorted array of squared values. Ask about duplicates, empty arrays, and whether in-place modification is allowed.
Mention that merging then squaring then sorting is O((n+m) log(n+m)) and simple, but not optimal. Explain that squaring first and then merging is tricky because squares of negatives can be out of order.
Use two pointers starting at the end of each array, compare absolute values, and place the larger square at the end of the result array. Move the pointer of the chosen element inward and repeat.
State time complexity O(n+m) and space O(n+m) for the output (or O(1) extra if output is pre-allocated). Walk through edge cases: one array empty, all negatives, all positives, zeros, and duplicates.
Write clean code with meaningful variable names, then test with examples like [-4,-1,0,3,10] and [-7,-3,2,3,11]. Verify the output is sorted and correct.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.