← Back to problems Solve on LeetCode → Next: Reverse Linked List →

Two Sum

Hash Map Complement Lookup — One Pass

LeetCode 1 • Easy • Arrays & Hashing

Input: nums = [3, 8, 1, 5, 11, 7], target = 12 → Output: [2, 4] (nums[2]+nums[4] = 1+11 = 12)

TimeO(n)single pass

SpaceO(n)hash map

Checked: 0/6 Map size: 0 Lookups: 0

Current element

Complement

In hash map

Pair found

Complement —

Map

{ }

Result [ ]

Speed:

Ready

Press Play to watch the algorithm, or Step to advance one operation at a time.
For each element, compute complement = target − nums[x]. If the complement is already in the hash map, we found our pair.

✎ 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.

# ─── TWO SUM (LeetCode 1) ───

# Pattern: Hash map complement lookup

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

# 1. Define: twoSum(self, nums, target)

# 2. Init hash map:

# map = {} ← stores value → index

# 3. Loop: for x in range(len(nums)):

# 4. Calculate what you need:

# complement = target - nums[x]

# 5. Check if complement already seen:

# if complement in map: ← O(1) lookup!

# return [map[complement], x]

# 6. Store current for future lookups:

# map[nums[x]] = x ← value as key, index as val

# 7. No pair: return [-1, -1]

# Vars: map (dict), complement, x

# Key: target - nums[x] gives what you need

# Map stores values you've ALREADY SEEN

▼ your implementation ▼

Verify your solution: