In detailed explained..
\n# Code\n\nclass Solution {\npublic:\n    bool isTransformable(string s, string t) {\n        // we have to check whether a number is less than the current num
\n# Code\n```\nclass Solution {\npublic:\n    bool isTransformable(string s, string t) {\n        // we have to check whether a number is less than the current number in left side or not..\n        // if we found the occurence of number less than current in left than return false\n        //for example..\n        // 34521\n        /*  \n        3 -> 0\n        4 -> 1 \n        5 -> 2 \n        2 -> 3 \n        1 -> 4\n        traverse in t now 23415\n        step 1-> encounter 2 no element less than 2 in left..\n        step 2-> encounter 3 no element less than 3  in left..\n        step 3-> encounter 4 no element less than 4  in left\n         we have already encountered 3 and decreased its occurence..\n        step 4-> encounter 1 no element less than 1  in left..\n        step 5-> encounter 5 no element less than 5  in left..\n        return true.. \n        */\n        \n\n        vector<vector<int>>v(10);\n        for (int i = s.size() - 1; i >= 0; i--)\n        {\n            v[s[i] - \'0\'].push_back(i);\n        }\n        \n        for (int i = 0; i < t.size(); i++) {\n            int n = t[i] - \'0\';\n            \n            if (v[n].size()==0)\n            { return false;\n            }\n            for (int j = 0; j < n; j++) {\n                if (v[j].size() and v[j].back() < v[n].back()) return false;\n            }\n            \n            v[n].pop_back();\n        }\n        \n        return true;\n    }\n};\n```
['C++']