x402 Gateway – OpenAI o4-mini (Fast Reasoning) is a paid API for AI agents from x402-gateway-production.up.railway.app, paid per call via x402, $0.03/call, status unknown (last checked 2026-09-15).
Runs OpenAI's o4-mini small reasoning model via the x402 payment gateway for fast chain-of-thought inference on math, coding, and analysis tasks
OpenAI's latest small reasoning model — fast chain-of-thought for math, coding, and analysis
A text completion from OpenAI's o4-mini model containing the model's chain-of-thought reasoning and final answer, covering the math, code, or analytical task submitted in the prompt.
POSThttps://x402-gateway-production.up.railway.app/api/llm/o4-miniChoose this endpoint when you need fast, affordable chain-of-thought reasoning — especially for math, coding, logic, or structured analysis — and don't need the full power of a larger frontier model. Ideal for agent pipelines where cost efficiency matters and the task benefits from step-by-step reasoning over raw fluency.
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"content": "Given a list of integers [3, 1, 4, 1, 5, 9], write a Python function that returns all pairs summing to 10 and explain the time complexity."
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"content": "Here’s a simple O(n)-time, O(n)-space solution using a hash set to find all unique pairs that sum to a given target (10 in your example):\n\n```python\ndef find_pairs(arr, target):\n seen = set() # numbers we’ve seen so far\n pairs = set() # store result pairs as (small, large) to avoid dupes\n\n for x in arr:\n y = target - x\n if y in seen:\n # order the pair so (1,9) and (9,1) aren’t counted separately\n pairs.add((min(x, y), max(x, y)))\n seen.add(x)\n\n # convert set of tuples to a list if you like\n return list(pairs)\n\n# Example\nnums = [3, 1, 4, 1, 5, 9]\nprint(find_pairs(nums, 10)) # [(1, 9)]\n```\n\nExplanation of time complexity:\n\n- We loop over the list once: O(n) iterations.\n- Each iteration does at most two set operations (`in`-check and `add`), each O(1) on average.\n- So total time is O(n). \n- We use O(n) extra space for the `seen` set (and up to O(n) for the result set in the worst"
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