SRM 521 DIV 1

MySolution  public class MissingParentheses {        int  missingNumber;        public int countCorrections(String str){         int leftCnt = 0;         for(int i=0; i<str.length(); i++){             if(str.charAt(i) == '('){                 leftCnt++;             }else{                 if(leftCnt <=0){                     missingNumber++;                 }else{                     leftCnt--;                 }             }         }         return leftCnt + missingNumber;     } }  Problem Statement
	Given a string of parentheses, you must turn it into a well formed string by inserting as few parentheses as possible, at any position (you cannot delete or change any of the existing parentheses). A well formed string of parentheses is defined by the following rules: The empty string is well formed. If s is a well formed string, (s) is a well formed string. If s and t are well formed strings, their concatenation st is a well formed string. As examples, "(()())", "" and "(())()" are well formed strings and "())(", "()(" and ")" are malformed strings. Given a String par of parentheses, return the minimum number of parentheses that need to be inserted to make it into a well formed string.
Definition
	Class: MissingParentheses Method: countCorrections Parameters: String Returns: int Method signature: int countCorrections(String par) (be sure your method is public)
Limits
	Time limit (s): 2.000 Memory limit (MB): 64
Constraints
-	par will contain between 1 and 50 characters, inclusive.
-	Each character of par will be an opening or closing parenthesis, i.e., '(' or ')'.
Examples
0)
	"(()(()" Returns: 2 The string below is a possible well formed string you can get to by inserting the two closing parentheses marked. (())(()) ^ ^
1)
	"()()(()" Returns: 1 You can fix the string by inserting a single closing parenthesis at the end.
2)
	"(())(()())" Returns: 0 The input string is well formed, so no insertion is needed.
3)
	"())(())((()))))()((())))()())())())()()()" Returns: 7

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