wont get a more straight forward solution than this.
Intuition\n Describe your first thoughts on how to solve this problem. \nguys please upvote , I work so hard explaining things and you potatoes dont even upvote
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nguys please upvote , I work so hard explaining things and you potatoes dont even upvote the post . SHAME on you. NOW UPVOTE\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n\n1. **Initializing Variables:**\n   - `StringBuilder sb = new StringBuilder("");`: Creates a `StringBuilder` object named `sb` to efficiently build the resulting string.\n   - `int count = 0`: Initializes a variable `count` to 0 to keep track of the nesting depth of parentheses encountered while iterating through the string.\n\n2. **Iterating Through the String:**\n   - The `for` loop iterates through each character (`s.charAt(i)`) in the `s` string.\n     - **Handling Opening Parentheses (\'(\'):**\n       - `if (s.charAt(i) == \'(\')`: Checks if the current character is an opening parenthesis.\n         - `count++`: Increments the `count` to indicate entering a new level of nesting.\n         - `if (count > 1)`: If the current nesting depth (`count`) is greater than 1 (meaning we\'re inside an existing parenthetical group), the character is appended to the `StringBuilder` (`sb.append(s.charAt(i))`) as it\'s not part of the outermost parentheses.\n     - **Handling Closing Parentheses (\')\'):**\n       - `else`: If the current character is a closing parenthesis.\n         - `count--`: Decrements the `count` to indicate exiting a level of nesting.\n         - `if (count > 0)`: If the current nesting depth (`count`) is still greater than 0 (meaning we\'re still within a parenthetical group), the character is appended to the `StringBuilder` (`sb.append(s.charAt(i))`) as it\'s not part of the outermost parentheses.\n\n3. **Building the Result:**\n   - After iterating through the entire string, the `StringBuilder` `sb` contains the characters that are not part of the outermost parentheses in the original string.\n\n4. **Returning the Result:**\n   - `return sb.toString()`: Converts the `StringBuilder` `sb` to a String object and returns it.\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- The code iterates through the string once, resulting in a time complexity of O(n), where n is the length of the string `s`.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n- The space complexity is O(n) due to the use of the `StringBuilder` object, which creates extra space in proportion to the length of the resulting string. However, this is generally considered more efficient than creating a new String object for each character appended during modification.\n# Code\n```\nclass Solution {\n    public String removeOuterParentheses(String s) {\n        StringBuilder sb=new StringBuilder("");\n        int count=0;\n        for(int i=0;i<s.length();i++){\n            if(s.charAt(i)==\'(\'){\n                count++;\n                if(count>1){\n                    sb.append(s.charAt(i));\n                }\n            }\n            else{\n                count--;\n                if(count>0){\n                    sb.append(s.charAt(i));\n                }\n            }\n        }\n        return sb.toString();\n    }\n}\n```
['String', 'Java']