C++ (4 ms) with explanation
\nbool canTransform(string start, string end) {\n        int n=start.size();\n        string s1="",s2="";\n\t\t//we take LR from 1st string and LR from 2nd stri
```\nbool canTransform(string start, string end) {\n        int n=start.size();\n        string s1="",s2="";\n\t\t//we take LR from 1st string and LR from 2nd string if it is not equal it means\n\t\t//count of R & l in both the strings is not equal ex- start="X" end="L"\n\t\t//excluding the below checking we get wrong ans.\n        for(int i=0;i<n;i++){\n            if (start[i] != \'X\' ){\n                s1+=start[i];\n            }\n            if(end[i] != \'X\'){\n                s2+=end[i];\n            }\n        }\n        if(s1 != s2){\n            return false;\n        }\n\t\t//then we check if both conditions are true then we return true.\n\t\t//1.if pos of \'L\' of first string is  >= pos of \'L\' in second string \n\t\t//2.pos of \'R\' in first string is <= pos of \'R in second string \n\t\t\n        for(int i=0,j=0;i<n && j<n;){\n            if(start[i] == \'X\'){\n                i++;\n                continue;\n            }\n            else if(end[j] == \'X\'){\n                j++;\n                continue;\n            }\n            else if(start[i] == \'R\' && end[j] == \'R\' && i > j || \n\t\t\tstart[i] == \'L\' && end[j] == \'L\' && i < j ){\n                return false;\n            }\n            i++;\n            j++;\n        }\n        return true;\n    }\n```
['Two Pointers', 'String', 'C++']