The leading zeros edge case is the one that'll get you if you're moving fast.
Use a two-pointer technique to traverse the word and abbreviation simultaneously. When encountering a digit, parse the full number (ensuring no leading zeros) and skip that many characters in the word. At the end, verify both pointers have reached the end of their respective strings.
Pro tip: Clarify edge cases upfront, such as empty strings, abbreviations with only digits, and words with uppercase/lowercase sensitivity. This shows attention to detail and prevents incorrect assumptions.
Ask about case sensitivity, empty inputs, and whether the abbreviation can contain other characters. Confirm that digits represent skips and no leading zeros are allowed.
Set two pointers, i for the word and j for the abbreviation, both starting at 0. These will track the current position in each string.
While j < len(abbr), if abbr[j] is a letter, compare it with word[i] and advance both pointers. If it's a digit, parse the number (checking for leading zeros) and advance i by that number.
After the loop, ensure both i and j have reached the end of their strings. If not, the abbreviation is invalid.
Walk through provided examples and edge cases (e.g., 'a' and '1', 'ab' and 'a1', 'ab' and '2') to verify correctness and discuss time/space complexity.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the problem constraints (e.g., which bracket types, empty string, invalid characters) and then propose a stack-based solution. Walk through the algorithm step-by-step, analyze time and space complexity, and test with edge cases.
Pro tip: Mention that you can early-return if the string length is odd, and discuss how to handle multiple bracket types by using a mapping. This shows attention to optimization and robustness.
Ask about the types of brackets allowed (e.g., (), [], {}), whether the string can be empty, and if there are any invalid characters. Confirm the expected return type (boolean).
Explain that you will iterate through the string, pushing opening brackets onto a stack and popping when encountering a closing bracket, checking for a match. If the stack is empty at the end, the brackets are balanced.
Trace through a sample input like '{[()]}' to demonstrate how the stack operates, and also show a failing case like '([)]' to highlight the importance of order.
State that time complexity is O(n) and space complexity is O(n) in the worst case. Discuss edge cases: empty string, single bracket, odd length, and mismatched types.
Mention that early termination on odd length can save time, and briefly note that a counter-based approach works only for a single bracket type, so a stack is more general.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.