Leetcode - 217. Contains Duplicate

Easy

Dec 22, 2025

#hashmap

217. Contains Duplicate

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.

Example 1:

Input: nums = [1,2,3,1]

Output: true

Explanation:

The element 1 occurs at the indices 0 and 3.

Example 2:

Input: nums = [1,2,3,4]

Output: false

Explanation:

All elements are distinct.

Example 3:

Input: nums = [1,1,1,3,3,4,3,2,4,2]

Output: true

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

Constraints:

Notes

Intuition

To detect duplicates, we need to remember which numbers we've seen. A hashmap provides O(1) lookup to check if a number has appeared before.

Implementation

Iterate through the array. For each number, check if it's already in the hashmap—if so, we found a duplicate, return True. Otherwise, add the number to the map and continue. If the loop completes, no duplicates exist, return False.

Edge-cases

An empty array or single-element array has no duplicates by definition. The algorithm handles these naturally.

Time: O(n) — single pass with O(1) lookups
Space: O(n) — hashmap stores up to n elements

Complexity

Solution

from typing import List

class Solution:
    def containsDuplicate(self, nums: List[int]) -> bool:
        map = {}
        for num in nums:
            if num in map:
                return True
            map[num] = True

        return False
        



if __name__ == "__main__":
    # Include one-off tests here or debugging logic that can be run by running this file
    # e.g. print(solution.two_sum([1, 2, 3, 4], 3))
    solution = Solution()