LeetCode 217 - Contains Duplicate

2026-04-01

easy　array

Given an integer array nums, return true if any value appears more than once in the array, otherwise return false.

Example 1:

Input: nums = [1, 2, 3, 3]

Output: true

Example 2:

Input: nums = [1, 2, 3, 4]

Output: false

Solution 1:

class Solution:
    def hasDuplicate(self, nums: List[int]) -> bool:
        # we can use set and check the length, because set don't allow duplicate
        # time complexity is O(n) since it need to transfer to set
        # space complexity also is O(n) because it need space for set
        return len(set(nums)) != len(nums)

Solution 2:

class Solution:
    def hasDuplicate(self, nums: List[int]) -> bool:
        # Hash Set Solution
        # Time Complexity: O(n) in the worst case. However, it leverages an "early exit"
        # (short-circuiting) mechanism, returning True immediately upon finding a match.
        # Space Complexity: O(n) in the worst case, as the set will store all elements
        # if the array contains entirely unique integers.

        duplicate_set = set()
        for i in nums:
            if i in duplicate_set:
                return True
            else:
                duplicate_set.add(i)
        return False