Easy C++ Solution || Well-explained
Intuition\n Describe your first thoughts on how to solve this problem. \n- To make all strings (say n) equal means every character should be equal in number in 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- To make all strings *(say n)* equal means every character should be equal in number in all the strings. \n- Since, we need to return if all the strings can be equal after performing given operation any no. of times. \n- Therefore, we assume it as a set of characters(duplicate) given, and now we have to make strings such that they are equal.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Iterate through `words` of size *n* and store the frequency of each character in an unordered map, let\'s say *mp*.\n    ```\n        unordered_map<char, int> mp;\n        for(int i=0; i<n; i++){\n            for(int j=0; j<words[i].size(); j++){\n                mp[words[i][j]]++;\n            }\n        }\n    ```\n- Then, we iterate through the map,\n    ```\n        for(auto ch : mp){\n            if(ch.second % n != 0) return false;\n        }\n    ```\n    doing `(ch.second % n != 0)`, we do this to check\n        - if the number of occurences of any character can be evenly divided into required *n* strings.\n        - if the modulo doesn\'t give 0, implies in no any way, the strings can be equal, hence return false.\n        \n- At last, return true.\n\n# Complexity\n- Time complexity: $$O(n*m)$$, where m is the maximum size of any string in `words`.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- Space complexity: $$O(1)$$, as there can be atmost 26 characters.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    bool makeEqual(vector<string>& words) {\n        int n = words.size();\n        unordered_map<char, int> mp;\n        for(int i=0; i<n; i++){\n            for(int j=0; j<words[i].size(); j++){\n                mp[words[i][j]]++;\n            }\n        }\n        for(auto ch : mp){\n            if(ch.second % n != 0) return false;\n        }\n        return true;\n    }\n};\n```
['Hash Table', 'C++']