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my first API call: Dunder Mifflin Infinity 2.0

Apparently, this is how you call the AI, and these are some of the main params you pass to it

•temperature controls how “creative” the response gets (0 is deterministic, 1 is wild)
•max_tokens caps the response length (1024 ≈ 750 words)
•system message tells the AI how to behave (e.g., “be concise”)

python

1  import anthropic
2   
3  client = anthropic.Anthropic()
4   
5  response = client.messages.create(
6      model="claude-sonnet-4-6",
7      max_tokens=1024,
8      temperature=0.7,
9      system="You are a thought leader.",
10      messages=[{
11          "role": "user",
12          "content": "Build Dunder Mifflin Infinity 2.0. No Mistakes."
13      }]
14  )
15  print(response.content[0].text)

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Codex, Cursor, and Gemini

New note

my first API call: Dunder Mifflin Infinity 2.0

Apparently, this is how you call the AI, and these are some of the main params you pass to it

python

response = client.messages.create(
  system="You are a thought leader.",
  messages=[{
    "content": "Build Dunder Mifflin Infinity 2.0. No Mistakes."
  }]
)

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Rook MCP

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interview prep⇧⌘P

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DSA Practice3

Binary Searchnow

Two Pointers1d

BFS / DFS2d

System Design5

Scalability3d

Availability4d

Consistency1w

Patterns8

Templates3

Mock Interviews2

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Binary Search

A divide-and-conquer algorithm that locates a target in a sorted array by repeatedly halving the search space. Runs in O(log n) time and O(1) space.

python

1  def binary_search(arr, target):
2      lo, hi = 0, len(arr) - 1
3      while lo <= hi:
4          mid = lo + (hi - lo) // 2
5          if arr[mid] == target:
6              return mid
7          elif arr[mid] < target:
8              lo = mid + 1
9          else:
10              hi = mid - 1
11      return -1

New note

Binary Search

A divide-and-conquer algorithm for sorted arrays. Runs in O(log n) time.

python

def binary_search(arr, target):
  lo, hi = 0, len(arr) - 1
  while lo <= hi:
    mid = lo + (hi - lo) // 2
    if arr[mid] == target:
      return mid
    elif arr[mid] < target:
      lo = mid + 1
    else:
      hi = mid - 1
  return -1