C++ easy to understand solution
Intuition\nTwo strings are similar if they contain the same set of characters(order and frequency of characters is not important)\n# Approach\nDeclare a map of 
# Intuition\nTwo strings are similar if they contain the same set of characters(order and frequency of characters is not important)\n# Approach\nDeclare a map of set of characters and store the frequency of the  set of characters obtained from the given vector of strings. Then count the identical pairs in a single iteration\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(N)\n\n# Code\n```\nclass Solution {\npublic:\n    int similarPairs(vector<string>& words) {\n        map<set<char>,int> mp;int count=0;  map<set<char>,int>::iterator it;\n        for(int i=0;i<words.size();i++){\n            set<char> st;\n            for(int j=0;j<words[i].size();j++){\n            st.insert(words[i][j]);\n            }\n            mp[st]++;\n        }\n        for(it=mp.begin();it!=mp.end();it++){\n            count+=(((it->second)*(it->second-1))/2);\n        }\n        return count;\n    }\n};\n```
['Ordered Map', 'Counting', 'Ordered Set', 'C++']