BEST RECURSIVE CODE, 100% beats
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n    public int minAddToMakeValid(String s) {\n        return helper(s, 0, 0, 0); // Pass initial open, close counts, and index\n    }\n\n    public int helper(String curr, int open, int close, int index) {\n        // Base case: if we\'ve checked all characters in the string\n        if (index == curr.length()) {\n            return open; // At the end, all unmatched open parentheses should be added\n        }\n\n        int count = 0;\n        char c = curr.charAt(index);\n\n        if (c == \'(\') {\n            count += helper(curr, open + 1, close, index + 1); // Increment open count\n        } else {\n            if (open > 0) {\n                count += helper(curr, open - 1, close, index + 1); // Match with open, reduce open\n            } else {\n                count += 1 + helper(curr, open, close, index + 1); // No open to match, add one to count\n            }\n        }\n\n        return count;\n    }\n}\n\n```
['Java']