Implementation

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 You are given a board 8 x 8, a position(x,y) as well of a color. You task is to determine if this move is valid. A move is considered valid, if you have can have (vertically, horizontally or in digonal), starting from this position, a structure like so: color given then opposite N colors then color given. For example [“B”,“W”,“W”,“B”] or [“W”,“B”,“W”] is valid in any direction. This [“B”,“B”,“W”] or [“W”,"."] is invalid. ...