30. Contains Duplicate

Q: Why early exit?

If the first two elements are duplicates, the early-exit version returns in O(1). The `nums.length !== new Set(nums).size` shortcut forces full iteration even when an early answer exists.

Q: When would sort be preferred?

Memory-constrained contexts where O(n) extra space is too much. Trade-off is O(n log n) time and input mutation.

easyAsked at Vercel

Given an array, return true if any element appears more than once. Vercel asks this as the simplest set-based question; it's the warm-up before they pivot to streaming dedup or cache-key collisions.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-20

Source citations

Public interview reports confirming this problem appears in Vercel loops.

Glassdoor (2025-Q4)— Vercel screen warm-up; expected in 3 minutes.
LeetCode Discuss (2026-Q1)— Listed in Vercel new-grad screen recap.

Problem

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.

Constraints

1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9

Examples

Example 1

Input

nums = [1,2,3,1]

Output

true

Example 2

Input

nums = [1,2,3,4]

Output

false

Example 3

Input

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

Output

true

Approaches

1. Sort then scan

Sort the array; check adjacent elements for equality.

Time: O(n log n)
Space: O(1) or O(n) depending on sort

function containsDuplicate(nums) {
  nums.sort((a, b) => a - b);
  for (let i = 1; i < nums.length; i++) {
    if (nums[i] === nums[i - 1]) return true;
  }
  return false;
}

Tradeoff: Mutates the input. Slower than the Set version, but uses O(1) extra space (depending on sort).

2. Hash set early exit (optimal)

Add each element to a Set; if it's already there, return true.

Time: O(n)
Space: O(n)

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

Tradeoff: O(n) time and space, single pass with early exit on first duplicate. The natural choice.

Vercel-specific tips

Vercel uses this as a baseline; getting it in 3 minutes signals you're moving cleanly. Bonus signal: discussing the trade-off between Set (O(n) space, O(n) time) and sort (O(1) space, O(n log n) time, mutates input), and connecting to streaming dedup with a Bloom filter for huge datasets.

Common mistakes

Using indexOf — O(n^2).
Forgetting early exit — adds every element, slower than necessary on duplicates near the start.
Using `nums.length !== new Set(nums).size` — works but doesn't early-exit and allocates the full Set.

Follow-up questions

An interviewer at Vercel may pivot to one of these next:

Contains Duplicate II (LC 219) — within k indices.
Contains Duplicate III (LC 220) — within k indices and within t value distance.
Streaming dedup — Bloom filter or count-min sketch.

Solve it now

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Output

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FAQ

Why early exit?

If the first two elements are duplicates, the early-exit version returns in O(1). The `nums.length !== new Set(nums).size` shortcut forces full iteration even when an early answer exists.

When would sort be preferred?

Memory-constrained contexts where O(n) extra space is too much. Trade-off is O(n log n) time and input mutation.

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