Two Sum in Python — Codekilla

Problem

Given nums = [2, 7, 11, 15] and target = 9, return [0, 1] because nums[0] + nums[1] == 9.

Each input has exactly one solution; you may not use the same element twice.

Naive — O(n²)

python

def two_sum_brute(nums, target):
    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            if nums[i] + nums[j] == target:
                return [i, j]

Quadratic time; works but slow for len(nums) > 10_000.

Optimal — O(n) with a hash map

python

def two_sum(nums, target):
    seen = {}                        # value -> index
    for i, n in enumerate(nums):
        if (need := target - n) in seen:
            return [seen[need], i]
        seen[n] = i

Walk-through for nums = [2, 7, 11, 15], target = 9:

i	n	need	seen before?	action
0	2	7	no	store `2 → 0`
1	7	2	yes (index 0)	return `[0, 1]`

Tests

python

assert two_sum([2,7,11,15], 9)  == [0, 1]
assert two_sum([3,2,4],    6)  == [1, 2]
assert two_sum([3,3],      6)  == [0, 1]

What interviewers look for

Can you reach O(n) from O(n²) by trading space for time?
Walrus operator := for "compute once, test, reuse" (Python 3.8+).
Edge cases — negative numbers, duplicates, the same index twice (not allowed).

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