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CheckIfWordEqualsSummationOfTwoWords.cpp
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CheckIfWordEqualsSummationOfTwoWords.cpp
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// Source : https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/
// Author : Hao Chen
// Date : 2021-05-30
/*****************************************************************************************************
*
* 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.
*
* Constraints:
*
* 1 <= firstWord.length, secondWord.length, targetWord.length <= 8
* firstWord, secondWord, and targetWord consist of lowercase English letters from 'a' to 'j'
* inclusive.
******************************************************************************************************/
class Solution {
public:
int strToInt(string& str) {
int n = 0;
for(auto& c :str) {
n = n*10 + c - 'a';
}
return n;
}
bool isSumEqual(string firstWord, string secondWord, string targetWord) {
return strToInt(firstWord) + strToInt(secondWord) == strToInt(targetWord);
}
};