One of the Best Digit DP + Bitmask question solution(c++ solution)
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n    int dp[70][2][1025][2][2];\n    int f(int i,int t,int flag,int flag2,int flag3,string &s){\n\n        if(i == s.size() && flag2 == true)return 1;\n        else if(i == s.size())return 0;\n        \n        if(dp[i][t][flag][flag2][flag3]!=-1)return dp[i][t][flag][flag2][flag3];\n\n        int ans = 0;\n        int high = t ? s[i]-\'0\' : 9;\n        int low = 0;\n        for(int k=low;k<=high;k++){\n            if(flag3 == false){//we have not yet started \n                if(k == 0)ans+=f(i+1,t && high == k,flag,flag2,0,s);\n                else ans+= f(i+1,t && high == k,flag|(1<<k),0,1,s);\n            }\n            else if((flag&(1<<k))){//this digit is already taken previously\n                ans+= f(i+1,t && high == k,flag|(1<<k),1,1,s);//i have taken repeated digit\n            }\n            else ans+=f(i+1,t && high == k,flag|(1<<k),flag2,1,s);\n        }\n        return dp[i][t][flag][flag2][flag3] = ans;\n    }\npublic:\n    int numDupDigitsAtMostN(int n) {\n        \n        string res = "";\n        while(n>0){\n            res.push_back(n%10+\'0\');\n            n = n/10;\n        }\n        reverse(res.begin(),res.end());\n        memset(dp,-1,sizeof(dp));\n        return f(0,1,0,0,0,res);\n    }\n};\n```
['C++']