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Contains Duplicate

Hash Set Membership Check — Early Exit

LeetCode 217 • Easy • Arrays & Hashing

Input: nums = [5, 3, 8, 1, 3, 9, 2] → Output: true (3 appears at index 1 and index 4)

TimeO(n)single pass

SpaceO(n)hash set

Checked: 0/7 Set size: 0 Lookups: 0

Current element

In set (safe)

Duplicate found!

All unique

Current —

Set

{ }

Result —

Speed:

Ready

Press Play to watch the algorithm, or Step to advance one operation at a time.
For each element, check: is it already in the set? If yes → duplicate. If no → add it and move on.

✎ Whiteboard

Size: 3

⌨ Type It — Build Muscle Memory

Practice until you don't need to look. Use the guide comments below as scaffolding. Type the full implementation in the editor. The green highlights are the nuances to burn into memory.

# ─── CONTAINS DUPLICATE (LeetCode 217) ───

# Pattern: Hash set membership check

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

# 1. Define: containsDuplicate(self, nums)

# 2. Init empty set:

# seen = set() ← tracks values we've visited

# 3. Loop through array:

# for num in nums:

# 4. Check if already seen:

# if num in seen: ← O(1) lookup!

# return True ← early exit

# 5. Not seen yet — remember it:

# seen.add(num)

# 6. No duplicates: return False

# Vars: seen (set), num (current element)

# Key: set() gives O(1) lookup vs O(n) list scan

# Why not sort? Sorting is O(n log n) — set is O(n)

▼ your implementation ▼

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