Most frequent element

Counter.most_common pattern.

counter dict

Implement `most_frequent(nums: list) -> any`.

Return the element with the highest frequency. If tie, return the smallest.

Constraints

1 ≤ len(nums) ≤ 10^5

Example

>>> most_frequent([1,2,2,3,3,3])
3

8 points ~8 min

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Hints

collections.Counter(nums).most_common(1)[0][0].

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