The Two Sneaky Numbers of Digitville

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Question

Approach

I scan the array once while keeping a set of numbers I have already seen. Whenever I encounter a number that is already in the set, it must be one of the sneaky repeated numbers, so I add it to the result list. Since there are exactly two repeats, I stop once I found both.

Implementation

class Solution:
    def getSneakyNumbers(self, nums: List[int]) -> List[int]:
        seen = set()
        res = []
        for x in nums:
            if x in seen:
                res.append(x)
                if len(res) == 2:
                    break
            else:
                seen.add(x)
        return res

Complexities

• Time: O(n)
• Space: O(n)