# Blind 75 - Contains Duplicate

Create a set and put items in one by one while checking if `num` is already there.

## Video

https://youtu.be/7KHKFqLPMbs

## Problem and Constraints

If the array contains any duplicate, return true. Else false.

## All Approaches and Explanations in English

### O(n^2) solution. O(1) space complexity

Double loop. And check if the elements are equals.

Bad. Solution.

### O(n) time complexity. O(n) space complexity

Create a set and put items in one by one while checking if `num`

is already there.

### Code, if any

```
class Solution {
public boolean containsDuplicate(int[] nums) {
Set<Integer> visited = new HashSet<>();
for(int num: nums){
if(visited.contains(num)){
return true;
}
visited.add(num);
}
return false;
}
}
```