New Problem Solution - "Minimum Length of String After Deleting Similar Ends"

Author: haoel

README.md

@@ -12,6 +12,7 @@ LeetCode
 |1754|[Largest Merge Of Two Strings](https://leetcode.com/problems/largest-merge-of-two-strings/) | [C++](./algorithms/cpp/largestMergeOfTwoStrings/LargestMergeOfTwoStrings.cpp)|Medium|
 |1753|[Maximum Score From Removing Stones](https://leetcode.com/problems/maximum-score-from-removing-stones/) | [C++](./algorithms/cpp/maximumScoreFromRemovingStones/MaximumScoreFromRemovingStones.cpp)|Medium|
 |1752|[Check if Array Is Sorted and Rotated](https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/) | [C++](./algorithms/cpp/checkIfArrayIsSortedAndRotated/CheckIfArrayIsSortedAndRotated.cpp)|Easy|
+|1750|[Minimum Length of String After Deleting Similar Ends](https://leetcode.com/problems/minimum-length-of-string-after-deleting-similar-ends/) | [C++](./algorithms/cpp/minimumLengthOfStringAfterDeletingSimilarEnds/MinimumLengthOfStringAfterDeletingSimilarEnds.cpp)|Medium|
 |1749|[Maximum Absolute Sum of Any Subarray](https://leetcode.com/problems/maximum-absolute-sum-of-any-subarray/) | [C++](./algorithms/cpp/maximumAbsoluteSumOfAnySubarray/MaximumAbsoluteSumOfAnySubarray.cpp)|Medium|
 |1748|[Sum of Unique Elements](https://leetcode.com/problems/sum-of-unique-elements/) | [C++](./algorithms/cpp/sumOfUniqueElements/SumOfUniqueElements.cpp)|Easy|
 |1605|[Find Valid Matrix Given Row and Column Sums](https://leetcode.com/problems/find-valid-matrix-given-row-and-column-sums/) | [C++](./algorithms/cpp/FindValidMatrixGivenRowAndColumnSums/FindValidMatrixGivenRowAndColumnSums.cpp)|Medium|

algorithms/cpp/minimumLengthOfStringAfterDeletingSimilarEnds/MinimumLengthOfStringAfterDeletingSimilarEnds.cpp

@@ -0,0 +1,68 @@
+// Source : https://leetcode.com/problems/minimum-length-of-string-after-deleting-similar-ends/
+// Author : Hao Chen
+// Date   : 2021-02-12
+
+/***************************************************************************************************** 
+ *
+ * Given a string s consisting only of characters 'a', 'b', and 'c'. You are asked to apply the 
+ * following algorithm on the string any number of times:
+ * 
+ * 	Pick a non-empty prefix from the string s where all the characters in the prefix are equal.
+ * 	Pick a non-empty suffix from the string s where all the characters in this suffix are equal.
+ * 	The prefix and the suffix should not intersect at any index.
+ * 	The characters from the prefix and suffix must be the same.
+ * 	Delete both the prefix and the suffix.
+ * 
+ * Return the minimum length of s after performing the above operation any number of times (possibly 
+ * zero times).
+ * 
+ * Example 1:
+ * 
+ * Input: s = "ca"
+ * Output: 2
+ * Explanation: You can't remove any characters, so the string stays as is.
+ * 
+ * Example 2:
+ * 
+ * Input: s = "cabaabac"
+ * Output: 0
+ * Explanation: An optimal sequence of operations is:
+ * - Take prefix = "c" and suffix = "c" and remove them, s = "abaaba".
+ * - Take prefix = "a" and suffix = "a" and remove them, s = "baab".
+ * - Take prefix = "b" and suffix = "b" and remove them, s = "aa".
+ * - Take prefix = "a" and suffix = "a" and remove them, s = "".
+ * 
+ * Example 3:
+ * 
+ * Input: s = "aabccabba"
+ * Output: 3
+ * Explanation: An optimal sequence of operations is:
+ * - Take prefix = "aa" and suffix = "a" and remove them, s = "bccabb".
+ * - Take prefix = "b" and suffix = "bb" and remove them, s = "cca".
+ * 
+ * Constraints:
+ * 
+ * 	1 <= s.length <= 105
+ * 	s only consists of characters 'a', 'b', and 'c'.
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    int minimumLength(string s) {
+        char ch; 
+        int left=0, right=s.size()-1;
+        
+        while(left < right) {
+            ch = s[left];
+            if (s[right] != ch) break;
+            
+            while (s[left] == ch) left++;
+    
+            if (left >= right ) return 0;
+            while (s[right] == ch) right--;
+        }
+        
+        return right - left + 1;
+        
+    }
+};