0ms, 100% faster  || [Java/C++/Python]
Explantation\n\n- l is a count of \'L\'s\n- r is a count of \'R\'s\n- m is a count of \'_\'s\n- n is a length of moves\n\n---\n\n\n\nWith for loop we\'ll find l
# Explantation\n\n- ```l``` is a count of \'L\'s\n- ```r``` is a count of \'R\'s\n- ```m``` is a count of \'_\'s\n- ```n``` is a length of ```moves```\n\n---\n\n\n\nWith for loop we\'ll find ```l``` and ```r```. \n\n*The main point is :* If \'L\'s count is greater than \'R\'s count, then well have to fill the empty places with \'L\', Otherwise \'R\'. \n\nSo, In the first case, we return ```l - r + m```, more clearly, ```l-r + (n-l-r) = n-2r```.\nIn the second case, ```r-l + (n-l-r) = n-2l```\n\n\n---\n\n\n\n**I hope it was clear. Happy coding. Don\'t forget to upvote :)**\n\n---\n\n\n\n\n# Java\n```\nclass Solution {\n    public int furthestDistanceFromOrigin(String moves) {\n        int l = 0, r=0, n = moves.length();\n        for(char c : moves.toCharArray()){\n            if(c==\'L\') l++;\n            else if(c==\'R\') r++;\n        }\n        return (l>r) ? n-2*r : n-2*l;\n    }\n}\n```\n\n---\n\n# Python\n```\nclass Solution(object):\n    def furthestDistanceFromOrigin(self, moves):\n        l = 0\n        r = 0\n        n = len(moves)\n        for c in moves:\n            if c == \'L\':\n                l += 1\n            elif c == \'R\':\n                r += 1\n        return n - 2 * r if l > r else n - 2 * l\n\n```\n\n\n---\n\n# C++\n```\nclass Solution {\npublic:\n    int furthestDistanceFromOrigin(string moves) {\n        int l = 0;\n        int r = 0;\n        int n = moves.length();\n        for (char c : moves) {\n            if (c == \'L\') {\n                l++;\n            } else if (c == \'R\') {\n                r++;\n            }\n        }\n        return (l > r) ? n - 2 * r : n - 2 * l;\n    }\n};\n\n```
['Array', 'Python', 'C++', 'Java']