## Problem Statement#

### Description#

Given an integer array `nums`, return true if any value appears at least twice in the array, and return `false` if every element is distinct.

### Examples#

#### 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
``````

### Constraints#

• `1 <= nums.length <= 105`
• `-109 <= nums[i] <= 109`

## Key Insights#

This is a very simple problem that introduces the concept of a hash set. As we loop through the array, we check if the current number already exists in the set, if it does, we’re done! We can return `True`. If it doesn’t, we add the number to the set and move on. If we make it through the loop without finding a duplicate, we return `False` at the end.

## Python Solution#

``````class Solution:
def containsDuplicate(self, nums: List[int]) -> bool:
hash_set = set()

# If we find a number in the hash set, return True.
# If we don't find any number in the hash set, we
# know we don't have any duplicates, return False
for num in nums:
if num in hash_set:
return True
else: