New Problem Solution - "1880. Check if Word Equals Summation of Two Words"

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|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|
 |1871|[Jump Game VII](https://leetcode.com/problems/jump-game-vii/) | [C++](./algorithms/cpp/jumpGame/jumpGame.VII.cpp)|Medium|
 |1870|[Minimum Speed to Arrive on Time](https://leetcode.com/problems/minimum-speed-to-arrive-on-time/) | [C++](./algorithms/cpp/minimumSpeedToArriveOnTime/MinimumSpeedToArriveOnTime.cpp)|Medium|
 |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

@@ -0,0 +1,71 @@
+// 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);
+    }
+};