SIMPLE SOLUTION ✅✅
Intuition\nCount frequency of each letter and see if each string can have same number of that letter or not.\n\n# Complexity\n- Time complexity: O(n*w)\n- Space
# Intuition\nCount frequency of each letter and see if each string can have same number of that letter or not.\n\n# Complexity\n- Time complexity: $$O(n*w)$$\n- Space complexity: $$O(26)$$\n\n# Code\n```\nclass Solution {\npublic:\n    bool makeEqual(vector<string>& words) {\n        int n = words.size();\n        vector<int>freq(26, 0);\n        //count frequency of each alphabet\n        for(auto s : words){\n            for(auto c : s){\n                freq[c-\'a\']++;\n            }\n        }\n        //try to divide the alphabet into n given strings equally.\n        for(int i = 0;i<26;i++){\n            if(freq[i]%n!=0) return false;\n        }\n        return true;\n    }\n};\n```
['C++']