easy c++||  no queues just vectors ||explained
```\n//Logic is store the number of L and R in with respect to their positions so that we can find if we can shift L or R accordingly \n\nclass Solution {\npubl
```\n//Logic is store the number of L and R in with respect to their positions so that we can find if we can shift L or R accordingly \n\nclass Solution {\npublic:\n    bool canChange(string start, string target) {\n    vector<pair<char, int>> v;\n        vector<pair<char, int>> m;\n    int a , b =0;\n        for(int i =0;i<start.size();i++){\n        if(start[i]==\'L\'){\n            v.push_back({\'L\' , i});\n        }\n        else if ( start[i]==\'R\'){\n            v.push_back({\'R\', i});\n        }\n        else{\n            a++;\n        }\n    }\n        \n         for(int i =0;i<target.size();i++){\n        if(target[i]==\'L\'){\n            m.push_back({\'L\' , i});\n        }\n        else if ( target[i]==\'R\'){\n            m.push_back({\'R\', i});\n        }\n        else{\n            b++;\n        }\n    }\n\t// to check if the number of  blanks are extra then we cannot obtain target \n        if(a!= b){\n            \n            return false;\n        }\n\t\t\n\t\t\n\t\t// if the element is R and it is appearing after in start with respect to target then we return false as we cannot move it left to make target .\n        for(int  i =0;i<v.size();i++){\n            if(v[i].first == m[i].first and v[i].first == \'R\'){\n                \n                if(v[i].second > m[i].second){\n                    return false;\n                }\n            }\n\t\t\t// if the element is L and it is appearing before in  start with respect to target then return false as we  cannot move it right to make target \n            else if (v[i].first == m[i].first and v[i].first == \'L\'){\n                if(v[i].second <m[i].second){\n                    return false;\n                }\n            }\n\t\t\t// if the arrangement is disordered eg - LR and RL \n            else{\n                return false;\n            }\n        }\n        \n        return true;\n    }\n\t//  please upvote if you understood  :)) happy coding\n};
['C']