Minimum Remove to make Valid parentheses
//Input: s = "lee(t(c)o)de)"\nOutput: "lee(t(c)o)de"\nExplanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.//\n******Explanations*****\n-->In the
//Input: s = "lee(t(c)o)de)"\nOutput: "lee(t(c)o)de"\nExplanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.//\n*********************************************Explanations************************************\n-->In the given que we need to remove the extra \'(\' & \')\' (brackets).\n-->first apporach in my mind that we use string , stack , map  and so on.\n-->so , I m using the string .\n-->firstly  iterate the loop from the starting and the secondly iterate the loop from the ending.\n--> so , my apporach is that we iterate the loop from the starting  and if we find the "("(opening bracket) , then increment it , and if we find the ")"(closing bracket) is more then the "(" then we need to decrement it .(no. closing> no.opening)\n--> thats why we are using the s[i]==\'@\', and decrement the count --;\n--> second loop started from the end  . \n--> if no. opening > the no. closing  , then decrement the \'(\' (opening)\n--> at last run the  loop \n\n```\nclass Solution {\npublic:\n    string minRemoveToMakeValid(string s) {\n        int n=s.length();\n        int count=0;\n\t\t// first loop from the starting \n        for(int i=0;i<n;i++){\n            if(s[i]==\'(\'){\n                count++;\n            }\n            else if(s[i]==\')\'){\n                if(count==0){\n                    s[i]=\'@\';\n                }\n                else{\n                    count--;\n                }\n            }\n        }\n\t\t// second loop from the ending\n        count=0;\n        for(int i=n-1;i>=0;i--){\n            if(s[i]==\')\'){\n                count++;\n            }\n            else if(s[i]==\'(\'){\n                if(count==0){\n                    s[i]=\'@\';\n                }\n                else{\n                    count--;\n                }\n            }\n        }\n        string ans="";\n        for(int i=0;i<n;i++){\n            if(s[i]!=\'@\'){\n                ans.push_back(s[i]);\n            }\n        }\n        return ans;\n    }\n};\n```\n**If this code is helpful  then please upvote my solution and if any query then feel free to ask \nTHANKU**
['String', 'C', 'C++']