Easy 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. \nUse Hash Map and count 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUse Hash Map and count the all character in words and check is it divisible by size of words or not . if it not divisible then return false otherwise return true\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n)\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        unordered_map<char,int> mp;\n        int n=words.size();\n        for(int i=0;i<n;i++){\n            int m=words[i].size();\n            for(int j=0;j<m;j++){\n                mp[words[i][j]]++;\n            }\n        }\n        int flg=1;\n        for(auto i:mp){\n            if(i.second%n!=0){\n                flg=0;\n                break;\n            }\n        }\n\n        if(flg){\n            return true;\n        }\n        else{\n            return false;\n        }\n    }\n};\n```
['C++']