[Python , Javascript]  Easy solution with very clear Explanation
Please dont downvote guys if cannot support,We are putting lot of effort in it\uD83D\uDE42\n\n\nWhat the Question asking us to do \uD83E\uDD14 ?\nYour task is t
**Please dont downvote guys if cannot support,We are putting lot of effort in it\uD83D\uDE42**\n\n```\nWhat the Question asking us to do \uD83E\uDD14 ?\nYour task is to remove the minimum number of parentheses ( \'(\' or \')\', in any positions )\nso that the resulting parentheses string is valid and return any valid string\n```\n\n![image](https://assets.leetcode.com/users/images/bd148e51-991a-4cea-9683-831cc402b696_1647314863.6571114.gif)\n\n![image](https://assets.leetcode.com/users/images/a56af075-c77d-476b-9105-996c19fe3d8e_1647314916.9967928.gif)\n\n\n![image](https://assets.leetcode.com/users/images/d8f0c7fd-88a3-467a-aded-c698cb73b595_1647320162.3985708.gif)\n\n\n```\nBig o:\n    n-->size of the s\n    Time: O(n)\n    Space: O(n)\n```\n\n`Javascript`\n\n```\nconst minRemoveToMakeValid = (str)=> {\n     const stack = [];\n  const splitted_str = str.split("");\n  for (let i = 0; i < str.length; i++) {\n    const char = str[i];\n    if (char === "(") stack.push(i); // if curr char is (  then we will push into our stack\n    else if (char === ")") {\n      if (stack.length === 0) {\n          // if out stack is empty then we will make ) as \'\'\n        splitted_str[i] = "";\n      } else {\n          //! if stack is not empty then we will pop top of the stack\n        stack.pop();\n      }\n    }\n  }\n    // if we have extra ( bracket we will remove it by making it as \'\'\n  for (let i = 0; i < stack.length; i++) {\n    const char = stack[i];\n    splitted_str[char] = "";\n  }\n\n  return splitted_str.join(""); // at last we will join the splitted_str\n};\n```\n\n`Python`\n\n```\nclass Solution:\n    def minRemoveToMakeValid(self, s) :\n        stack=[]\n        split_str=list(s)\n        for i in range(len(s)):\n            if s[i]==\'(\':\n                # if current char is \'(\' then push it to stack\n                stack.append(i)\n            elif s[i]==\')\':\n                # if current char is \')\' then pop top of the stack\n                if len(stack) !=0:\n                    stack.pop()\n                else:\n                    # if our stack is empty then we can\'t pop so make  current char as \'\'\n                    split_str[i]=""\n        for i in stack:\n            split_str[i]=""\n        return \'\' .join(split_str)\n```\n\n```\nUPVOTE if you like \uD83D\uDE03 , If you have any question, feel free to ask.\n```
['Python', 'JavaScript']