check_if_word_equals_summation_of_two_words.py

# The letter value of a letter is its position in the alphabet starting from 0 (i.e. 'a' -> 0, 'b' -> 1, 'c' -> 2, etc.).

# The numerical value of some string of lowercase English letters s is the concatenation of the letter values of each letter in s, which is then converted into an integer.

# For example, if s = "acb", we concatenate each letter's letter value, resulting in "021". After converting it, we get 21.
# You are given three strings firstWord, secondWord, and targetWord, each consisting of lowercase English letters 'a' through 'j' inclusive.

# Return true if the summation of the numerical values of firstWord and secondWord equals the numerical value of targetWord, or false otherwise.

 

# Example 1:

# Input: firstWord = "acb", secondWord = "cba", targetWord = "cdb"
# Output: true
# Explanation:
# The numerical value of firstWord is "acb" -> "021" -> 21.
# The numerical value of secondWord is "cba" -> "210" -> 210.
# The numerical value of targetWord is "cdb" -> "231" -> 231.
# We return true because 21 + 210 == 231.
# Example 2:

# Input: firstWord = "aaa", secondWord = "a", targetWord = "aab"
# Output: false
# Explanation: 
# The numerical value of firstWord is "aaa" -> "000" -> 0.
# The numerical value of secondWord is "a" -> "0" -> 0.
# The numerical value of targetWord is "aab" -> "001" -> 1.
# We return false because 0 + 0 != 1.
# Example 3:

# Input: firstWord = "aaa", secondWord = "a", targetWord = "aaaa"
# Output: true
# Explanation: 
# The numerical value of firstWord is "aaa" -> "000" -> 0.
# The numerical value of secondWord is "a" -> "0" -> 0.
# The numerical value of targetWord is "aaaa" -> "0000" -> 0.
# We return true because 0 + 0 == 0.

class Solution:
    def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
        
        def conv_num(word):
            result = ''
            for ch in word:
                result += str(ord(ch) - 97)
                
            return int(result)
        
        return (conv_num(firstWord) + conv_num(secondWord) == conv_num(targetWord))