✅ C++ | With inline explanation | Easy to understand
\n\nclass Solution {\npublic:\n   \n    int longestPalindrome(vector<string>& words)\n    {\n        int n = words.size() ;\n        \n        map<string,int> m
```\n\nclass Solution {\npublic:\n   \n    int longestPalindrome(vector<string>& words)\n    {\n        int n = words.size() ;\n        \n        map<string,int> mp ; // map to store frequency of each word\n        \n        for(string x : words)\n        {\n            mp[x]++ ;\n        }\n        \n        int count = 0 ;\n        bool flag = false ; // assume that we have no suitable word w such that z = awb is a palindrome where b = reverse(a)\n        \n        for(auto p : mp)\n        {\n            string x = p.first ; // x ->current string\n            \n            string y = x ;\n            reverse(y.begin(),y.end()) ; // y -> reverse of current string\n            \n            if(x!=y) // if x is not a palindrome\n            {\n                // let the final string(palindrome) is z = awb , where w can be a empty string\n                //then\n                // if we want to place x in a , we should have one y (=reverse(x)) to place it in \n\t\t\t\t//corresponding position in b inorder to maintain palindrome property\n                // so if frequency of x is 10 and frequency of y is 5,\n                // then we can only use 5 x and 5 y\n                int c = min(mp[x],mp[y]) ;\n                count+=c ; // count stores number of words present in a (where final answer z = awb)\n                \n                mp[y]-=c ; mp[x]-=c ; // we have used c occurances of x and y , so decrease its frequency\n            \n            }\n            else\n            {   // if x is a palindrome\n                // let frequency of x be 10, then\n                // we can use only 5 x in a, because we need 5 x in b\n                count+=mp[x]/2 ;\n                \n                // if we have 11 x (odd number of x)\n                // we use 5 x in a, 5 x in b\n                // and the remaining x can be used in w \n                // so make flag = true \n                if(mp[x]%2 && x[0]==x[1])\n                    flag = true ;\n                \n                // we have used all the x, so reduce the frequency\n                mp[x] = 0  ;\n            }\n                \n        }\n        \n        // count is the number of words in a\n        // number of words in b = number of words in a = count\n        // number of words in a and b = 2*count\n        // each word has 2 characters\n        // number of characters in a and b = 2*2*count = 4*count\n        int ans = count*4 ;\n        \n        // if we had got some word that can be placed in w\n        // then length of final answer will be increased by 2\n        if(flag)\n            ans+=2 ;\n        \n        return ans ;\n        \n    }\n};\n\n```\n\nUpdate : \nas pointed out by @hardik2001\nNo need to check x[0]==x[1], as x==y so it will be x[0]==x[1]\n
['C']