8000 New Problem Solution - "1880. Check if Word Equals Summation of Two W… · royeLin/leetcode_cpp@d8e8f74 · GitHub
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New Problem Solution - "1880. Check if Word Equals Summation of Two Words"
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‎README.md

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| # | Title | Solution | Difficulty |
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|---| ----- | -------- | ---------- |
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|1880|[Check if Word Equals Summation of Two Words](https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/) | [C++](./algorithms/cpp/checkIfWordEqualsSummationOfTwoWords/CheckIfWordEqualsSummationOfTwoWords.cpp)|Easy|
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|1871|[Jump Game VII](https://leetcode.com/problems/jump-game-vii/) | [C++](./algorithms/cpp/jumpGame/jumpGame.VII.cpp)|Medium|
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|1870|[Minimum Speed to Arrive on Time](https://leetcode.com/problems/minimum-speed-to-arrive-on-time/) | [C++](./algorithms/cpp/minimumSpeedToArriveOnTime/MinimumSpeedToArriveOnTime.cpp)|Medium|
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|1869|[Longer Contiguous Segments of Ones than Zeros](https://leetcode.com/problems/longer-contiguous-segments-of-ones-than-zeros/) | [C++](./algorithms/cpp/longerContiguousSegmentsOfOnesThanZeros/LongerContiguousSegmentsOfOnesThanZeros.cpp)|Easy|

‎algorithms/cpp/checkIfWordEqualsSummationOfTwoWords/CheckIfWordEqualsSummationOfTwoWords.cpp

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// Source : https://leetcode.com/problems/check-if-word-equals-summation-of-two-words/
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// Author : Hao Chen
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// Date : 2021-05-30
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/*****************************************************************************************************
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*
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* The letter value of a letter is its position in the alphabet starting from 0 (i.e. 'a' -> 0, 'b' ->
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* 1, 'c' -> 2, etc.).
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*
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* The numerical value of some string of lowercase English letters s is the concatenation of the
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* letter values of each letter in s, which is then converted into an integer.
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*
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* For example, if s = "acb", we concatenate each letter's letter value, resulting in "021".
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* After converting it, we get 21.
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*
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* You are given three strings firstWord, secondWord, and targetWord, each consisting of lowercase
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* English letters 'a' through 'j' inclusive.
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*
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* Return true if the summation of the numerical values of firstWord and secondWord equals the
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* numerical value of targetWord, or false otherwise.
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*
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* Example 1:
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*
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* Input: firstWord = "acb", secondWord = "cba", targetWord = "cdb"
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* Output: true
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* Explanation:
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* The numerical value of firstWord is "acb" -> "021" -> 21.
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* The numerical value of secondWord is "cba" -> "210" -> 210.
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* The numerical value of targetWord is "cdb" -> "231" -> 231.
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* We return true because 21 + 210 == 231.
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*
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* Example 2:
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*
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* Input: firstWord = "aaa", secondWord = "a", targetWord = "aab"
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* Output: false
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* Explanation:
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* The numerical value of firstWord is "aaa" -> "000" -> 0.
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* The numerical value of secondWord is "a" -> "0" -> 0.
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* The numerical value of targetWord is "aab" -> "001" -> 1.
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* We return false because 0 + 0 != 1.
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*
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* Example 3:
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*
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* Input: firstWord = "aaa", secondWord = "a", targetWord = "aaaa"
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* Output: true
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* Explanation:
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* The numerical value of firstWord is "aaa" -> "000" -> 0.
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* The numerical value of secondWord is "a" -> "0" -> 0.
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* The numerical value of targetWord is "aaaa" -> "0000" -> 0.
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* We return true because 0 + 0 == 0.
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*
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* Constraints:
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*
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* 1 <= firstWord.length, secondWord.length, targetWord.length <= 8
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* firstWord, secondWord, and targetWord consist of lowercase English letters from 'a' to 'j'
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* inclusive.
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******************************************************************************************************/
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class Solution {
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public:
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int strToInt(string& str) {
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int n = 0;
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for(auto& c :str) {
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n = n*10 + c - 'a';
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}
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return n;
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}
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bool isSumEqual(string firstWord, string secondWord, string targetWord) {
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return strToInt(firstWord) + strToInt(secondWord) == strToInt(targetWord);
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}
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};

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