From 0391e520514e06a7fd837b595f8e126fe217dfaf Mon Sep 17 00:00:00 2001
From: Hao Chen <haoel@hotmail.com>
Date: Fri, 12 Nov 2021 14:59:12 +0800
Subject: [PATCH] New Problem Solution - "1945. Sum of Digits of String After
 Convert"

---
 README.md                                     |  1 +
 .../SumOfDigitsOfStringAfterConvert.cpp       | 77 +++++++++++++++++++
 2 files changed, 78 insertions(+)
 create mode 100644 sumOfDigitsOfStringAfterConvert/SumOfDigitsOfStringAfterConvert.cpp

diff --git a/README.md b/README.md
index 20a9c568..4345c749 100644
--- a/README.md
+++ b/README.md
@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1945|[Sum of Digits of String After Convert](https://leetcode.com/problems/sum-of-digits-of-string-after-convert/) | [C++](/Users/chenhao/GitHub/leetcode/sumOfDigitsOfStringAfterConvert/SumOfDigitsOfStringAfterConvert.cpp)|Easy|
 |1935|[Maximum Number of Words You Can Type](https://leetcode.com/problems/maximum-number-of-words-you-can-type/) | [C++](./algorithms/cpp/maximumNumberOfWordsYouCanType/MaximumNumberOfWordsYouCanType.cpp)|Easy|
 |1882|[Process Tasks Using Servers](https://leetcode.com/problems/process-tasks-using-servers/) | [C++](./algorithms/cpp/processTasksUsingServers/ProcessTasksUsingServers.cpp)|Medium|
 |1881|[Maximum Value after Insertion](https://leetcode.com/problems/maximum-value-after-insertion/) | [C++](./algorithms/cpp/maximumValueAfterInsertion/MaximumValueAfterInsertion.cpp)|Medium|
diff --git a/sumOfDigitsOfStringAfterConvert/SumOfDigitsOfStringAfterConvert.cpp b/sumOfDigitsOfStringAfterConvert/SumOfDigitsOfStringAfterConvert.cpp
new file mode 100644
index 00000000..d2dcfe72
--- /dev/null
+++ b/sumOfDigitsOfStringAfterConvert/SumOfDigitsOfStringAfterConvert.cpp
@@ -0,0 +1,77 @@
+// Source : https://leetcode.com/problems/sum-of-digits-of-string-after-convert/
+// Author : Hao Chen
+// Date   : 2021-11-12
+
+/***************************************************************************************************** 
+ *
+ * You are given a string s consisting of lowercase English letters, and an integer k.
+ * 
+ * First, convert s into an integer by replacing each letter with its position in the alphabet (i.e., 
+ * replace 'a' with 1, 'b' with 2, ..., 'z' with 26). Then, transform the integer by replacing it with 
+ * the sum of its digits. Repeat the transform operation k times in total.
+ * 
+ * For example, if s = "zbax" and k = 2, then the resulting integer would be 8 by the following 
+ * operations:
+ * 
+ * 	Convert: "zbax" ➝ "(26)(2)(1)(24)" ➝ "262124" ➝ 262124
+ * 	Transform #1: 262124 ➝ 2 + 6 + 2 + 1 + 2 + 4 ➝ 17
+ * 	Transform #2: 17 ➝ 1 + 7 ➝ 8
+ * 
+ * Return the resulting integer after performing the operations described above.
+ * 
+ * Example 1:
+ * 
+ * Input: s = "iiii", k = 1
+ * Output: 36
+ * Explanation: The operations are as follows:
+ * - Convert: "iiii" ➝ "(9)(9)(9)(9)" ➝ "9999" ➝ 9999
+ * - Transform #1: 9999 ➝ 9 + 9 + 9 + 9 ➝ 36
+ * Thus the resulting integer is 36.
+ * 
+ * Example 2:
+ * 
+ * Input: s = "leetcode", k = 2
+ * Output: 6
+ * Explanation: The operations are as follows:
+ * - Convert: "leetcode" ➝ "(12)(5)(5)(20)(3)(15)(4)(5)" ➝ "12552031545" ➝ 12552031545
+ * - Transform #1: 12552031545 ➝ 1 + 2 + 5 + 5 + 2 + 0 + 3 + 1 + 5 + 4 + 5 ➝ 33
+ * - Transform #2: 33 ➝ 3 + 3 ➝ 6
+ * Thus the resulting integer is 6.
+ * 
+ * Example 3:
+ * 
+ * Input: s = "zbax", k = 2
+ * Output: 8
+ * 
+ * Constraints:
+ * 
+ * 	1 <= s.length <= 100
+ * 	1 <= k <= 10
+ * 	s consists of lowercase English letters.
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    int sumChar(char c) {
+        int x = c - 'a' + 1;
+        return x < 10 ? x : x / 10 + x % 10;
+    }
+    int sumInt(int x) {
+        int s = 0;
+        while( x > 0 ) {
+            s += x % 10;
+            x /= 10;
+        }
+        return s;
+    }
+    int getLucky(string s, int k) {
+        int result = 0;
+        for (auto c : s) {
+            result += sumChar(c);
+        }
+        for (; k > 1 ; k--) {
+            result = sumInt(result);
+        }
+        return result;
+    }
+};