Valid Word Abbreviation

Challenge

Given two params a word and it’s abbreviation return True if the abbreviation is valid.

Examples:

"kubernetes" "k8s" => True
"apple" "a2e" => False

Solution

This is the kind of implementation challenges easy to get wrong and/or forgot edge cases…

class Solution:
    def validWordAbbreviation(self, word: str, abbr: str) -> bool:
        i = 0
        tmp = ""
        if len(abbr) > 1 and abbr.isdecimal(): return False
        index = 0
        while i < len(abbr):
            number = ""
            while i < len(abbr) and abbr[i].isdigit():
                number += abbr[i]
                i += 1
            if number:
                if number[0] == "0": return False
                if index + int(number) > len(word): return False
                tmp += word[index:index+ int(number)] 
                index = len(tmp)
            elif i < len(abbr):
                tmp += abbr[i]
                i += 1
            index = len(tmp)
        return tmp == word

Time Complexity: O(n)
Space Complexity: O(n)

Challenge#

Solution#

Challenge

Solution