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Discussion on: Are these lists equal?

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Hong Jiang

You could build a map with list 1, and then iterate over list 2 to compare each element with its count in the map. The asymptotic complexity is the same, but the code would be more concise and efficient (2 loops instead of 3).

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Carlos Roso Author

That's right. That's the kind of response I was expecting from this post. My O(n) solution is clear enough to lay a foundation for non-asymptotic optimizations like yours or that of the first comment. Thanks for contributing!