Easy C++ Solution
We know that we can move any character, any number of times, to any position in the words array. So, all that we have to check is if all the count of any charac
We know that we can move any character, any number of times, to any position in the words array. So, all that we have to check is if all the count of any character is divisible by n i.e the length of the words array.  For a detailed explanation check out https://helloml.org/redistribute-characters-to-make-all-strings-equal-solution-to-leetcode-problem/. \n\n```cpp\nclass Solution {\npublic:\n    bool makeEqual(vector<string>& words) {\n        int counts[26] = {0};\n        int n = words.size();\n        \n        // Counting the number of times each character appears in words. \n        for(int i = 0; i < words.size(); i++) {\n            string word = words[i];\n            for (int j = 0; j < word.size(); j++) {\n                counts[word[j]-\'a\']++;\n            }\n        }\n        \n        // Now we check if each count is divisible by n\n        for(int i = 0; i < 26; i++) {\n            if(counts[i] % n == 0) {\n                continue;\n            }\n            else {\n                return false;\n            }\n        }\n        \n        return true;\n    }\n};\n```
['C']