contains_dup.py

# 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


# Topics

  

# Recommended Time & Space Complexity
# You should aim for a solution with O(n) time and O(n) space, where n is the size of the input array.


# Hint 1
# A brute force solution would be to check every element against every other element in the array. This would be an O(n^2) solution. Can you think of a better way?


# Hint 2
# Is there a way to check if an element is a duplicate without comparing it to every other element? Maybe there's a data structure that is useful here.


# Hint 3
# We can use a hash data structure like a hash set or hash map to store elements we've already seen. This will allow us to check if an element is a duplicate in constant time.


class Solution:
    def hasDuplicate(self, nums: List[int]) -> bool:
    
        seen = set()

        for num in nums:
            if num in seen:
                return True
            seen.add(num)

        return False