LeetCode 1: Two Sum — Python Solution

Solve LeetCode 1: Two Sum in Python with a complement lookup approach. The key is to make the state invariant explicit, so the implementation and complexity follow naturally.

This guide paraphrases the task and does not reproduce LeetCode’s prompt. Use the official page for the complete statement, examples, constraints, and submission runner.

What the problem is testing

For each value, check whether its required complement was seen earlier, then record the current index.

Algorithm

1. For each value, check whether its required complement was seen earlier, then record the current index.
2. Maintain this invariant: The map contains one usable prior index for each processed value.
3. Continue until every input item or reachable state has been resolved, then return the accumulated result.

Python solution

from collections import Counter, defaultdict, deque, OrderedDict
import random

class Solution:
    def twoSum(self, nums, target):
        seen = {}
        for i, value in enumerate(nums):
            if target - value in seen: return [seen[target - value], i]
            seen[value] = i

Why this is correct

The proof follows the maintained state: The map contains one usable prior index for each processed value. Each iteration preserves that claim while permanently resolving at least one position, node, interval, or search state. When the loop or recursion ends, every candidate required by the problem has therefore been included or ruled out, so the returned value is correct.

Complexity

O(n) time and O(n) space. The stated auxiliary space excludes the returned output unless the output is the data structure being built.

Edge cases

Duplicate values can form the answer when the target is twice that value.

Tested reference code

This implementation is included in the site’s downloadable 100-solution Python library. The complete suite compiles every solution and runs a behavioral assertion for every problem before publication.

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