2. Contains Duplicate · DSA Practice

Return true if any value appears at least twice in nums, and false if every element is distinct.

Example 1

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

Output: true

Example 2

Input: nums = [7,8,9]

Output: false

Constraints

1 ≤ nums.length ≤ 10⁵

Hint 1

What data structure answers “have I seen this before?” in O(1)?

ArrayHash Table

Approach 1: Sort

Sort the array; duplicates become neighbours.

⏱ Time O(n log n)💾 Space O(1)

Approach 2: Hash set ✅

Insert each value into a set; if it is already there, return true.

⏱ Time O(n)💾 Space O(n)

Reference solution for Hash set (time O(n), space O(n)). Each solution passes every test in all four languages.

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

class Solution {
    public boolean containsDuplicate(int[] nums) {
        Set<Integer> seen = new HashSet<>();
        for (int x : nums)
            if (!seen.add(x)) return true; // add() returns false if x was already there
        return false;
    }
}

class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        unordered_set<int> seen;
        for (int x : nums)
            if (!seen.insert(x).second) return true; // insert fails if x already exists
        return false;
    }
};

function containsDuplicate(nums) {
  const seen = new Set();
  for (const x of nums) {
    if (seen.has(x)) return true;
    seen.add(x);
  }
  return false;
}

Try solving it yourself first. Interviewers care more about how you explain the approach than about the finished code.

This problem is explained in the lesson Arrays, Hashing & Prefix Sums.