Two Sum - Visual Walkthrough & Python Solution

Why it works

A brute-force solution checks every pair of numbers, which is O(n²). The hash map turns the inner loop into a constant-time lookup: instead of searching for a partner, you ask memory directly. One pass, O(n) time, O(n) space for the map.

The problem

Given an array of integers nums and an integer target, return the indices of the two numbers that add up to target. Each input has exactly one solution. You may not use the same element twice.

Hints

• A brute-force O(n^2) solution checks every pair. Can you do better?
• Store each number's index in a hash map as you iterate.
• For each element x, check if (target - x) is already in the map.

Test cases

Reference starter code

def two_sum(nums, target):
    # Your solution here
    pass

# Test
print(two_sum([2, 7, 11, 15], 9))   # [0, 1]
print(two_sum([3, 2, 4], 6))        # [1, 2]
print(two_sum([3, 3], 6))           # [0, 1]

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