• AutoBuilder
• @github@anaclumos@rsc-swr

alias cc="CLAUDE_CODE_NO_FLICKER=1 claude --dangerously-skip-permissions --teammate-mode auto"alias ccl="CLAUDE_CODE_NO_FLICKER=1 CLAUDE_CODE_USE_VERTEX=1 CLOUD_ML_REGION=global ANTHROPIC_VERTEX_PROJECT_ID=tech-innovation-prod ANTHROPIC_MODEL=claude-opus-4-6[1m] ANTHROPIC_DEFAULT_OPUS_MODEL=claude-opus-4-6[1m] claude --dangerously-skip-permissions --teammate-mode auto"

It's surprisingly easy

Solved at: 220710

Question

• Two Sum - LeetCode

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.

Solution

So the first obvious answer is to iterate twice. This finishes calculations in $O(n^2)$ time.

class Solution:    def twoSum(self, nums: List[int], target: int) -> List[int]:        for idx1, val1 in enumerate(nums):            for idx2, val2 in enumerate(nums):                if idx1 == idx2:                    continue                if val1 + val2 == target:                    return [idx1, idx2]

However, this gives a timeout.

Improved

I used Python Dictionary to store complementing values. Python Dictionary will have $O(1)$ access time for most cases. This solution will run in $O(n)$ time.

• One caveat: depending on the hash function, it can go as bad as $O(n^2)$.

class Solution:    def twoSum(self, nums: List[int], target: int) -> List[int]:
        # map for complementing elements: complementary-idx        complementing_map = {}
        for idx, val in enumerate(nums):            if val in complementing_map:                return [complementing_map[val], idx]            complementing_map[target - val] = idx

Results

Runtime

• 60 ms, faster than 97.16% of Python3 online submissions for Two Sum.

Memory Usage

• 15.4 MB, less than 14.24% of Python3 online submissions for Two Sum.

Other Answers Online

• Sort first, $O(n \log n)$
• For all elements, $O(n)$
  • Perform binary search $O(\log n)$
• In total: $O(n \log n)$