Super Easy using One HashSet 5 line solution . Beats 100%;
Intuition\n Describe your first thoughts on how to solve this problem. \nif we put all source in a hashset, it is sure that there will not be one destination . 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nif we put all source in a hashset, it is sure that there will not be one destination . \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nLike    A->B\n        B->C\n        Z->D\n        C->Z\nthen if set contains A,B,C,Z then D is not present . So D will be answer.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(N)\n\n# Code\n```\nclass Solution {\n    public String destCity(List<List<String>> paths) {\n        HashSet<String> set1 = new HashSet<>();\n        \n        for(int i=0;i<paths.size();++i){ \n            set1.add(paths.get(i).get(0));\n        }\n        \n        for(int i=0;i<paths.size();++i){ \n            if(set1.contains(paths.get(i).get(1))==false){\n                return paths.get(i).get(1);\n            }\n          \n        }\n       \n        return "Please upvote";\n    }\n}\n```
['Java']