Easy Python Solution - 99.9% time, 100% space.
\nclass Solution:\n    def destCity(self, paths: List[List[str]]) -> str:\n        pairs = {}\n        for i in paths:\n            pairs[i[0]] = i[1]\n\t\t\t\n
```\nclass Solution:\n    def destCity(self, paths: List[List[str]]) -> str:\n        pairs = {}\n        for i in paths:\n            pairs[i[0]] = i[1]\n\t\t\t\n        for src, dest in paths:\n            if dest not in pairs:\n                return dest\n```\n\nExplanation:\n1) Create a map where key,value = source, destination.\n2) Iterate over destinations and if it is not a key in map, return destination.
['Python3']