Pretty approachable once you see it as a counting problem.
Clarify that the custom order defines a total order for characters present in it, and characters not in the order can be placed arbitrarily. Propose a solution that counts frequencies of each character in the second string, then outputs characters in the custom order first, followed by the remaining characters in any order. Discuss time and space complexity, and consider edge cases like empty strings or characters outside the order.
Pro tip: Mention that you can achieve O(n) time by using a hash map for character frequencies and iterating through the custom order once, then through the remaining characters. This avoids sorting and shows you optimize for efficiency.
Confirm that the custom order string contains unique characters and that characters not in it can be placed anywhere. Ask if the output should preserve any specific order for those extra characters.
Decide between sorting with a custom comparator (O(n log n)) or frequency counting (O(n)). The frequency approach is optimal and straightforward.
Count occurrences of each character in the second string using a hash map. Then iterate through the custom order, appending each character repeated by its count. Finally, append any remaining characters.
State that time complexity is O(n + m) where n is length of second string and m is length of custom order, and space is O(k) for distinct characters. Discuss edge cases like empty strings, all characters in order, or none in order.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This is the one that made me sweat a little.
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 skip that many characters in the word. For non-digit characters, check for an exact match and advance both pointers.
Pro tip: Clarify edge cases upfront, such as empty strings, abbreviations with only digits, and words with digits. Also, mention that you'll handle leading zeros by checking if the first digit is '0' and the number has more than one digit.
Set pointers i for word and j for abbreviation, both starting at 0.
While j < len(abbr), check if abbr[j] is a digit. If so, parse the number, ensuring no leading zeros, and increment i by that number. If not, compare abbr[j] with word[i] and increment both if they match.
When parsing a number, if the first digit is '0' and the number has more than one digit, return false. Also, ensure the parsed number doesn't exceed the remaining length of word.
For non-digit characters, if they don't match or i is out of bounds, return false.
After processing all characters in abbreviation, ensure i has reached the end of word. If so, return true; otherwise, false.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.