💀NO MAP , NO SET || Using  Find( ) ||🔥 Easy C++
Intuition\nSimply Pick each words Reverse it and check by using find().\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n- Dry run t
# Intuition\nSimply Pick each words Reverse it and check by using find().\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n- Dry run the 1st Test Case.\n- pick the 1st string of the words and reverse it .\n- Befor finding the reversed string first check either 1st and 2nd letter are same or not (since there is only 2 char) \n- NOW using find().. we search for str(ie reversed string) in words[]\n- if found then increase the count \n- AND Here the important stepwe have to make words[i]=str ?? think u will get if not then....\n\n---> because if we futher check we will again reach to element whose reversed is present before ie increase the count 2 times .\n\n---> Eg ; 1St TC : ["cd","ac","dc",...]  for "cd" we get "dc" in words[] and again for "dc" we get "cd" means extra count so replace with reversed(ie str itself) simpply...\n\ndry run more .... :)\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:O(N^2)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int maximumNumberOfStringPairs(vector<string>& words) {\n        int cnt=0;\n        for(int i=0;i<words.size();i++)\n        {\n            string str=words[i];\n            reverse(str.begin(),str.end());\n            if(str[0]==str[1]) continue; \n\n            auto it = find(words.begin(),words.end(),str);\n            if(it!=words.end())\n            {\n                cnt++;\n            }\n            words[i]=str;\n        }\n        return cnt;\n    }\n};\n```
['String', 'String Matching', 'C++']