super easy Intuitive solution
Intuition\n Describe your first thoughts on how to solve this problem. \nIf the total count of any character is not divisible by the number of words, it\'s impo
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf the total count of any character is not divisible by the number of words, it\'s impossible to make the redistribution. \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUse hashmap to get the frequency of each words. If the freq of any character is not divisible by the n.(n=words.size()), then return false, else return true.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n^2)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(n)\n\n# Code\n```\nclass Solution {\npublic:\n    bool makeEqual(vector<string>& words) {\n        // of the freq of any char is not divisible by the no of words, then it cannot be redistributed\n        unordered_map<char,int> mp;\n        int n=words.size();\n        if(n==1) return true;\n        for(auto i:words){\n            string str = i;\n            for(auto itr:str){\n                mp[itr]++;\n            }\n        }\n        for(auto i:mp){\n            if(i.second % n != 0) return false;\n        }\n        return true;\n\n    }\n};\n```
['Hash Table', 'C++']