Deque Solution | TC: O(n)
\nstring removeDuplicates(string s) {\n        deque<char> q;\n        string res;\n        for(int i=0;i<s.length();i++) {\n            if(!q.empty() && q.back
```\nstring removeDuplicates(string s) {\n        deque<char> q;\n        string res;\n        for(int i=0;i<s.length();i++) {\n            if(!q.empty() && q.back()==s[i])  //when q.back is same as the current char\n                q.pop_back();\n            else\n                q.push_back(s[i]);  //when q is empty or when char at q back !=s[i]\n        }\n\t\t//Concatenate chars in q from front to back\n        while(!q.empty()){\n            res+=q.front();\n            q.pop_front();\n        }\n        return res;\n    }\n```
['Queue', 'C', 'C++']