New Solution "1541. Minimum Insertions to Balance a Parentheses String"

README.md

@@ -10,6 +10,7 @@ LeetCode
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
 |1573|[Number of Ways to Split a String](https://leetcode.com/problems/number-of-ways-to-split-a-string/) | [C++](./algorithms/cpp/NumberOfWaysToSplitString/NumberOfWaysToSplitString.cpp)|Medium|
+|1541|[Minimum Insertions to Balance a Parentheses String](https://leetcode.com/problems/minimum-insertions-to-balance-a-parentheses-string/) | [C++](./algorithms/cpp/minimumInsertionsToBalanceAParenthesesString/MinimumInsertionsToBalanceAParenthesesString.cpp)|Medium|
 |1535|[Find the Winner of an Array Game](https://leetcode.com/problems/find-the-winner-of-an-array-game/) | [C++](./algorithms/cpp/findTheWinnerOfAnArrayGame/FindTheWinnerOfAnArrayGame.cpp)|Medium|
 |1470|[Shuffle the Array](https://leetcode.com/problems/shuffle-the-array/) | [C++](./algorithms/cpp/shuffleTheArray/ShuffleTheArray.cpp)|Easy|
 |1464|[Maximum Product of Two Elements in an Array](https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/) | [C++](./algorithms/cpp/maximumProductOfTwoElementsInAnArray/MaximumProductOfTwoElementsInAnArray.cpp)|Easy|

algorithms/cpp/minimumInsertionsToBalanceAParenthesesString/MinimumInsertionsToBalanceAParenthesesString.cpp

@@ -0,0 +1,92 @@
+// Source : https://leetcode.com/problems/minimum-insertions-to-balance-a-parentheses-string/
+// Author : Hao Chen
+// Date   : 2020-10-02
+
+/***************************************************************************************************** 
+ *
+ * Given a parentheses string s containing only the characters '(' and ')'. A parentheses string is 
+ * balanced if:
+ * 
+ * 	Any left parenthesis '(' must have a corresponding two consecutive right parenthesis '))'.
+ * 	Left parenthesis '(' must go before the corresponding two consecutive right parenthesis 
+ * '))'.
+ * 
+ * In other words, we treat '(' as openning parenthesis and '))' as closing parenthesis.
+ * 
+ * For example, "())", "())(())))" and "(())())))" are balanced, ")()", "()))" and "(()))" are not 
+ * balanced.
+ * 
+ * You can insert the characters '(' and ')' at any position of the string to balance it if needed.
+ * 
+ * Return the minimum number of insertions needed to make s balanced.
+ * 
+ * Example 1:
+ * 
+ * Input: s = "(()))"
+ * Output: 1
+ * Explanation: The second '(' has two matching '))', but the first '(' has only ')' matching. We need 
+ * to to add one more ')' at the end of the string to be "(())))" which is balanced.
+ * 
+ * Example 2:
+ * 
+ * Input: s = "())"
+ * Output: 0
+ * Explanation: The string is already balanced.
+ * 
+ * Example 3:
+ * 
+ * Input: s = "))())("
+ * Output: 3
+ * Explanation: Add '(' to match the first '))', Add '))' to match the last '('.
+ * 
+ * Example 4:
+ * 
+ * Input: s = "(((((("
+ * Output: 12
+ * Explanation: Add 12 ')' to balance the string.
+ * 
+ * Example 5:
+ * 
+ * Input: s = ")))))))"
+ * Output: 5
+ * Explanation: Add 4 '(' at the beginning of the string and one ')' at the end. The string becomes 
+ * "(((())))))))".
+ * 
+ * Constraints:
+ * 
+ * 	1 <= s.length <= 10^5
+ * 	s consists of '(' and ')' only.
+ ******************************************************************************************************/
+class Solution {
+public:
+    int minInsertions(string s) {
+        vector<char> stack;
+        
+        int cnt = 0;
+        int len = s.size();
+        for (int i=0; i<len; i++) {
+            
+            if ( s[i] == '(' ) {
+                stack.push_back( s[i] );
+                continue;
+            }
+            // if s[i] is ')'
+            if (stack.size() > 0)  {
+                stack.pop_back();
+            } else {
+                cnt++; // missed the '('
+            }
+            
+            // if  s[i+1] is ')', need to skip
+            if ( i < len -1 && s[i+1] == ')' ) {
+                i++;
+            }else{
+                cnt++; //missed the ')'
+            }
+         
+        }
+        
+        // if the stack still has '(', which means need double of ')'
+        return cnt + stack.size()*2;
+    }
+};