✅ C++ Solution 🔥 Fastest Solution , Beats 100%🔥
\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nTo cheak by swifting any charecters we can able to make all strings equal or not.
![100% beats 2.png](https://assets.leetcode.com/users/images/0b445d58-f3b0-48d4-9735-20bf52b36fba_1703947560.89523.png)\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo cheak by swifting any charecters we can able to make all strings equal or not.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Calculate total charecters occurance.\n   Calculate Total number of charecter Occurance in given string array using unordered_map. \n2. mp[i] % world.size() != 0  return false.\n3. else return true.\n\n# Complexity\n- Time complexity: $$O(N*N)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(N)$$  where N = number of charecters in Total string\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    bool makeEqual(vector<string>& words) {\n        unordered_map<char,int>mp;\n        // count all occurance of charecters using map\n        for(auto it:words){\n            for(auto it2:it){\n                mp[it2]++;\n            }\n        }\n        // cheak mp[it2]%words.size()==0 is true then return true else false\n        for(auto it:mp){\n            if(it.second%words.size()!=0)\n              return false; \n        }\n        return true;\n    }\n};\n```\n![UPvote2.jpg](https://assets.leetcode.com/users/images/15b56310-0e3e-4dbf-a0f6-c64fe9a9903f_1703947577.6097116.jpeg)\n
['Hash Function', 'C++']