Validate word abbreviation and reconcile two abbreviations

Implement a function isValidAbbreviation(word: string, abbr: string) that returns true if abbr is a valid abbreviation of word, where a positive integer in abbr represents the number of letters skipped and leading zeros are invalid. Follow-ups: (a) Given two abbreviations abbr1 and abbr2, determine whether there exists at least one original word (over lowercase letters) that both could represent; if so, return true and output one possible alignment of kept/skipped positions, otherwise return false. (b) Analyze time and space complexity, and discuss how you would handle very long inputs or streaming validation.

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Valid Word Abbreviation

Implement isValidAbbreviation(word, abbr) that returns true if abbr is a valid abbreviation of word.

An abbreviation replaces any number of consecutive characters with the count of characters replaced. A positive integer in abbr represents how many letters are skipped; the remaining characters in abbr must match word exactly, in order. Numbers must not contain leading zeros (e.g. "01", "s010n" are invalid abbreviations), and the abbreviation is valid only if the two strings are fully consumed at the same time.

Example: isValidAbbreviation("internationalization", "i12iz4n") = true because i + skip 12 + iz + skip 4 + n reconstructs the word. isValidAbbreviation("apple", "a2e") = false because a + skip 2 lands on the 4th char (l), not e.

Use two pointers: walk word with i and abbr with j. On a letter in abbr, it must equal word[i] (advance both). On a digit, reject a leading 0, parse the full number, and advance i by that count. Return true iff both pointers finish exactly at the end.

Constraints

• 1 <= word.length, but word and abbr may also be empty in edge cases.
• word consists of lowercase English letters.
• abbr consists of lowercase English letters and digits.
• Numbers in abbr have no leading zeros (a leading zero makes the abbreviation invalid).
• Parsed skip counts fit in a 32-bit / 64-bit integer.

Hints

1. Use two pointers, one over word and one over abbr, advancing them independently.
2. When you hit a digit, parse the entire consecutive run of digits as one number, then jump that many positions in word.
3. Reject the abbreviation immediately if a number starts with '0' (leading zeros are invalid).
4. The abbreviation is valid only if BOTH pointers reach the end together — a number that runs past the end of word, or leftover characters, means invalid.

Reconcile Two Word Abbreviations

Follow-up to Valid Word Abbreviation. Given two abbreviation strings abbr1 and abbr2, determine whether there exists at least one original word over lowercase letters that BOTH abbreviations could represent. Return true if such a word exists, otherwise false.

Each abbreviation uses the same rules as before: a positive integer (no leading zeros) means "skip that many letters" (a wildcard run that can stand for any letters), and a literal letter must appear at that exact position. Two abbreviations describe the same word iff they have the same total length and, at every position where BOTH pin down a concrete letter, those letters are equal. A skip in either abbreviation acts as a wildcard at that position.

Example: canBothRepresentSameWord("a2c", "ab1c") = true — both have length 4, abbr1 = a,,,c and abbr2 = a,b,_,c; no conflicting fixed letters, so a word like "abxc" works. canBothRepresentSameWord("a2c", "a3c") = false — lengths 4 vs 5 can never describe the same word.

Approach: expand each abbreviation into a per-position list where a literal contributes its character and a skip of k contributes k wildcard slots. If lengths differ, return false; otherwise return false only if some position has two differing fixed letters. Reject any abbreviation containing a leading-zero number.

Constraints

• abbr1 and abbr2 each follow the abbreviation grammar: lowercase letters and positive integers with no leading zeros.
• An abbreviation containing a leading-zero number is invalid and yields false.
• The original word, if it exists, is over lowercase English letters.
• Either abbreviation may be empty; two empty abbreviations both represent the empty word.

Hints

1. Think of each digit-run as a sequence of wildcard slots and each letter as a fixed slot, then line the two sequences up position by position.
2. If the two abbreviations expand to different total lengths, no single word can satisfy both — return false.
3. A conflict only happens when both abbreviations fix a concrete (and different) letter at the same position; a skip on either side is a wildcard that matches anything.
4. Don't forget to reject abbreviations with leading zeros before comparing.

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

Implement a function isValidAbbreviation(word: string, abbr: string) that returns true if abbr is a valid abbreviation of word, where a positive integer in abbr represents the number of letters skipped and leading zeros are invalid. Follow-ups: (a) Given two abbreviations abbr1 and abbr2, determine whether there exists at least one original word (over lowercase letters) that both could represent; if so, return true and output one possible alignment of kept/skipped positions, otherwise return false. (b) Analyze time and space complexity, and discuss how you would handle very long inputs or streaming validation.

Recommended Approach

Use two pointers over word and abbr. When abbr[j] is a letter, it must equal word[i]. When abbr[j] is a digit, parse the full positive number, reject leading zero, and advance i by that count. The abbreviation is valid only if both pointers finish exactly at the end.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

Time: O(len(word) + len(abbr)). Space: O(1).

Edge Cases and Tests

Leading zeros, abbreviation skipping past the word, empty abbreviation, exact match, adjacent numeric chunks, and case sensitivity.