Very easy......🔥🔥🔥🔥
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: 100%\u2705\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```c []\n\nint minAddToMakeValid(char *s) {\n    int open = 0, close = 0;\n\n    for (int i = 0; s[i] != \'\\0\'; i++) {\n        if (s[i] == \'(\') {\n            open++;  // Count of unmatched opening brackets\n        } else if (s[i] == \')\') {\n            if (open > 0) {\n                open--;  // Match with an opening bracket\n            } else {\n                close++;  // Unmatched closing brackets\n            }\n        }\n    }\n\n    // The result is the sum of unmatched open and close brackets\n    return open + close;\n}\n\n```
['String', 'Stack', 'Greedy', 'C']