README_EN.md

1880. Check if Word Equals Summation of Two Words

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Description

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.

Solutions

Python3

class Solution:
    def isSumEqual(self, firstWord: str, secondWord: str, targetWord: str) -> bool:
        def convert(word):
            res = 0
            for c in word:
                res *= 10
                res += (ord(c) - ord('a'))
            return res
        return convert(firstWord) + convert(secondWord) == convert(targetWord)

Java

class Solution {
    public boolean isSumEqual(String firstWord, String secondWord, String targetWord) {
        return convert(firstWord) + convert(secondWord) == convert(targetWord);
    }

    private int convert(String word) {
        int res = 0;
        for (char c : word.toCharArray()) {
            res *= 10;
            res += (c - 'a');
        }
        return res;
    }
}

C++

class Solution {
public:
    bool isSumEqual(string firstWord, string secondWord, string targetWord) {
        return convert(firstWord) + convert(secondWord) == convert(targetWord);
    }
private:
    int convert(string word) {
        int res = 0;
        for (char c : word) {
            res *= 10;
            res += (c - 'a');
        }
        return res;
    }
};

JavaScript

/**
 * @param {string} firstWord
 * @param {string} secondWord
 * @param {string} targetWord
 * @return {boolean}
 */
var isSumEqual = function (firstWord, secondWord, targetWord) {
    let carry = 0;
    let n1 = firstWord.length,
        n2 = secondWord.length;
    let n3 = targetWord.length;
    for (let i = 0; i < n3; i++) {
        let num1 = getNum(firstWord.charAt(n1 - 1 - i));
        let num2 = getNum(secondWord.charAt(n2 - 1 - i));
        let sum = carry + num1 + num2;
        if (getNum(targetWord.charAt(n3 - 1 - i)) != sum % 10) return false;
        carry = parseInt(sum / 10);
    }
    return true;
};

function getNum(char) {
    if (!char) return 0;
    return char.charCodeAt() - "a".charCodeAt();
}

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