The tricky part is handling the length difference case cleanly.
Start by clarifying edge cases (e.g., empty strings, same length, length difference >1) and then propose a linear two-pointer solution that handles insert/delete and replace cases. Walk through the algorithm with examples, analyze time and space complexity, and discuss potential pitfalls like Unicode or case sensitivity.
Pro tip: Mention that the solution can be implemented in O(n) time and O(1) space, and that it's essentially the core of the Levenshtein distance with a threshold of 1. Also, proactively discuss how you would test it, including edge cases like empty strings and strings of very different lengths.
Ask if the strings are ASCII or Unicode, case-sensitive, and if empty strings are allowed. Confirm that 'one edit apart' means exactly one insert, delete, or replace operation.
If the absolute difference in lengths is greater than 1, return false immediately. If lengths are equal, only replacements are possible; if lengths differ by 1, only insertions/deletions are possible.
Iterate through both strings with two pointers. When characters differ, if lengths are equal, skip both characters (replace); if lengths differ, skip the character in the longer string (insert/delete). Count edits and ensure at most one.
State that time complexity is O(n) and space is O(1). Walk through test cases: identical strings (false), one edit apart (true), multiple edits (false), and edge cases like empty strings.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a two-pointer technique to traverse the word and abbreviation simultaneously. When encountering a digit in the abbreviation, parse the full number (ensuring no leading zeros) and advance the word pointer by that many characters. When encountering a letter, check that it matches the current character in the word and advance both pointers. Finally, verify that both pointers have reached the end of their respective strings.
Pro tip: Clarify edge cases upfront, such as empty strings, abbreviations with consecutive numbers, or numbers that skip beyond the word length. This shows thoroughness and helps avoid incorrect assumptions during implementation.
Confirm with the interviewer the exact rules: numbers skip characters, letters must match exactly, leading zeros are invalid, and both strings must be fully consumed. Discuss edge cases like empty strings, numbers at the start/end, and consecutive numbers.
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), check if abbr[j] is a digit. If so, parse the full number (ensuring the first digit is not '0' unless the number is exactly '0'), then advance i by that number. If it's a letter, compare it with word[i]; if they don't match, return false. Advance i and j accordingly.
After the loop, check that both i and j have reached the end of their respective strings. If not, the abbreviation is invalid.
Walk through provided examples and edge cases (e.g., 'word' and 'w0ord' invalid due to leading zero, 'word' and 'w2d' valid) to verify the logic and catch off-by-one errors.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.