Contains duplicate

Time: O(N); Space: O(N); easy

Given an array of integers, find if the array contains any duplicates.

Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.

Example 1:

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

Output: True

Example 2:

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

Output: False

Example 3:

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

Output: True

1. Naive Linear Search [O(N^2), O(N^2) - Time Limit Exceeded]

2. Sorting [O(NLogN), O(1)]

3.Hash Table [O(N), O(N)]

4. Using Set [O(N), O(N)]

[1]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(N)
    """
    def containsDuplicate(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        return len(nums) > len(set(nums))

[2]:

s = Solution1()

nums = [1,2,3,1]
assert s.containsDuplicate(nums) == True

nums = [1,2,3,4]
assert s.containsDuplicate(nums) == False

nums = [1,1,1,3,3,4,3,2,4,2]
assert s.containsDuplicate(nums) == True