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 get21
.
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.
Approach
C++
#include <bits/stdc++.h>using namespace std;bool isSumEqual(string firstWord,string secondWord, string targetWord){string sum1 = "", sum2 = "", sum3 = "";for (int i = 0; i < firstWord.size(); i++){sum1 += to_string(firstWord[i] - 'a');}for (int i = 0; i < secondWord.size(); i++){sum2 += to_string(secondWord[i] - 'a');}for (int i = 0; i < targetWord.size(); i++){sum3 += to_string(targetWord[i] - 'a');}if (stoi(sum1) + stoi(sum2) == stoi(sum3))return true;return false;}int main(){string firstWord = "acb";string secondWord = "cba", targetWord = "cdb";if (isSumEqual(firstWord, secondWord, targetWord))cout << "true\n";elsecout << "false\n";return 0;}
No comments:
Post a Comment