LeetCode 217 – Python 3 (Week 06 – 05)

Solution 1

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        nums.sort()
        for i in range(len(nums)-1):
            if nums[i] == nums[i+1]:
                return True
            
        return False

Time complexity is O(nlog⁡n). Space complexity is O(1).

Solution 2

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        s = set()
        for num in nums:
            if num in s: return True
            else: s.add(num)
            
        return False

Time complexity is O(n). Space complexity is O(1).

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