Easy Solution Using Map
Intuition\n Describe your first thoughts on how to solve this problem. \nBy  looking at the example test case w ecan see that the occurence of the characters do
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nBy  looking at the example test case w ecan see that the occurence of the characters does not matter .\n We have to only look at the count of the characters present in the words .\n\n---\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFirstly we have to store the count of the characters in a  ordered map \nwhich will help us to determine how many characters are present in the words\n\n\n\n```\n map<char, int> sa;\n       for(int i =0; i<v.size(); i++){\n           sa[v[i]]++;\n       }\n```\nAfter this the problem looks very easy and can be easily solved .\nif all the occurence of the characters are the multiples of the numbers of strings in the words then we can say YES we can make all the string using each other .\notherwise NO\n```\n bool c = true;\n       for(auto it: sa){\n           if(it.second%n !=0){\n               c = false ;\n               break;\n           }\n       }\n```\n\n---\n\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of the given code is O(n)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of the given code is O(n)\n\n---\n\n\n# Code\n```\nclass Solution {\npublic:\n    bool makeEqual(vector<string>& words) {\n    int n = words.size();\n       if(n==1) return true;\n       string v;\n       for(auto &str :words){\n           v+= str;\n       } \n       map<char, int> sa;\n       for(int i =0; i<v.size(); i++){\n           sa[v[i]]++;\n       }\n       bool c = true;\n       for(auto it: sa){\n           if(it.second%n !=0){\n               c = false ;\n               break;\n           }\n       }\n       return c;\n    }\n};\n```
['String', 'Ordered Map', 'Counting', 'C++']