Solution: Contains Duplicate

I appreciate that 3 example solutions are given but they are not given in order from least to most optimal. It would be nice to at least callout which solution for problems is the most optimal solution.

class Solution:
    def containsDuplicate(self, nums):
      numsSet = set(nums)
      return not (len(nums) == len(numsSet))

Is there a reason why this would not be a good solution? Maybe I am overlooking something here

const uniqueSet = new Set(nums);

if (uniqueSet.size !== nums.length) {
	return true; 
}

I just found another solution using Set(): by copying each element into a new set. Since a set only stores distinct elements, if the number of elements in the input matches the size of the Set, it means there are no duplicates

function containsDuplicate(nums) {
    if(!nums || nums.length == 0){
      return false;
    }
    
    let nums_copy = new Set();

    for(let i = 0; i<nums.length;i++){
      nums_copy.add(nums[i]);
    }

    return nums_copy.size == nums.length ? false:true;
  }
 

I have another thought of how to proceed with this problem, but the problem is, I do not know the space and time complexity of this.

My thought process is as follows. Add all the elements in the array into a hash set. Count the number of existing elements in the two collections and compare to see whether there are any duplicates. The list collection should be equal to the set collection if there are no duplicates and return true.

I.e.

def containsDuplicate(nums):

`from collections import Counter`

`return Counter(nums) == Counter(set(nums))`

What if we need to use O(1) space and O(N) time is it possible?

For approach 2, what is the difference between using set and using list since the if condition will check if the number is unique or not.

def containsDuplicate(nums):
    for i in range(len(nums)):
      if nums[i] in nums[1+i:]:
        return True
    return False