0217.Contains Duplicate

• Link: 217. Contains Duplicate
• Track: NeetCode150

Question

Example 1

• Input: nums = [1,2,3,1]
• Output: true
• Explanation: The element 1 occurs at the indices 0 and 3.

Restate the problem

• Given an integer array nums
• Return True if any value appears at least twice, otherwise return False.

Edge Case

• [], return False
• len(nums) < 2, return False

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Method	Approach	Time Complexity	Space Complexity
Method 1 ⭐	Hashmap	O(n)	O(n)
Method 2	Sort	O($nlogn$)	O(1)

Method 1: HashSet

Approach

• Initial an empty set seen(only stores unique values)
• Scan the nums
• If num in seen, return True
  • Add num into set

Complexity

• Time Complexity: O(n)
• Space Complexity: O(n)

Code

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        seen = set()
        for num in nums:
            if num not in seen:
                seen.add(num)
            else:
                return True
        
        return False

Method 2: Sort + Scan

Approach

• Sort nums
• Scan sorted nums, start from 1
  • If the current number equals the previous number, return True

Complexity

• Time Complexity: O($nlogn$)
  • Sorting: $nlogn$
• Space Complexity: O(1)

Code

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

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History

• Feb-27-2026 Solved
• Jan-15-2026 Solved