🔥💥 Easy C++ Solution using HashMap💥🔥
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n1. Create an unordered_
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Create an `unordered_map (indegree)` to represent the directed graph. The keys of the map will be the source cities, and the values will be vectors containing the corresponding destination cities.\n2. Populate the indegree map using the given paths.\n3. Iterate through the paths again, and for each path, check if the source city `(ele[0])` or `(ele[1])` has no outgoing edges (i.e., the size of `indegree[ele[0]] `or `indegree[ele[1]]`  is 0). If it does, return respective `ele[0]` or `ele[1]`.\n# Complexity\n- Time complexity:$$O(n)$$\n    - when `n` is number of paths\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    string destCity(vector<vector<string>>& paths) {\n      int n  = paths.size();\n        unordered_map<string,vector<string>>indegree;\n        for(auto ele:paths){\n          indegree[ele[0]].push_back(ele[1]);\n        }\n        for(auto ele:paths){\n          if(indegree[ele[0]].size() == 0) return ele[0];\n          if(indegree[ele[1]].size() == 0) return ele[1];\n        }\n        return "null";\n    }\n};\n```
['Hash Table', 'String', 'Graph', 'C++']