Question Link: https://leetcode.com/problems/contains-duplicate/description/

So, I was able to quickly solve this problem with an approach of Sorting. O(n log n) time complexity.

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        nums.sort()
        for i in range(0, len(nums)):
            if (i+1) < len(nums):
                if nums[i] == nums[i+1]:
                    return True
        return False

Now, one point I missed while thinking about the approach was that i+1 can go out of range if there are no duplicates.

There are other possible methods to solve this too. 4 were listed in the solution of this question.

1. Two Loops

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        for i in range(0, len(nums)):
            for j in range(i+1, len(nums)):
                if nums[i] == nums[j]:
                    return True
        return False

1. Sorting

The one I used

1. Hash (set)

This is a better approach it only takes time O(n) however it has O(n) space complexity too.

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        visited = set()
        for i in nums:
            if i in visited:
                return True
            visited.add(i)
        return False

1. Hash Map

This one was similar to the above one time and everything same. Just used a dictionary, i.e., hash map instead of set.

Could be more useful when finding duplicate occurrences.