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基于 SOC 职业分类
| name | python-algorithm-cookbook |
| description | 4 algorithm patterns with full Python implementations for common coding problems |
| version | 1.0.0 |
| category | toolchain |
| author | Claude MPM Team |
| license | MIT |
| progressive_disclosure | {"entry_point":{"summary":"4 algorithm patterns: sliding window, BFS traversal, binary search, and hash map with full implementations","when_to_use":"When implementing algorithms for coding problems or optimizing data processing","quick_start":"Match your problem type to the right pattern: substring constraints -> sliding window, tree/graph -> BFS, sorted data -> binary search, O(1) lookup -> hash map"}} |
| context_limit | 700 |
| tags | ["python","algorithms","sliding-window","bfs","binary-search","hash-map","two-pointers","data-structures","complexity"] |
| requires_tools | [] |
# Pattern: Longest Substring Without Repeating Characters
def length_of_longest_substring(s: str) -> int:
"""Find length of longest substring without repeating characters.
Sliding window technique with hash map to track character positions.
Time: O(n), Space: O(min(n, alphabet_size))
Example: "abcabcbb" -> 3 (substring "abc")
"""
if not s:
return 0
# Track last seen index of each character
char_index: dict[str, int] = {}
max_length = 0
left = 0 # Left pointer of sliding window
for right, char in enumerate(s):
# If character seen AND it's within current window
if char in char_index and char_index[char] >= left:
# Move left pointer past the previous occurrence
# This maintains "no repeating chars" invariant
left = char_index[char] + 1
# Update character's latest position
char_index[char] = right
# Update max length seen so far
# Current window size is (right - left + 1)
max_length = max(max_length, right - left + 1)
return max_length
# Sliding Window Key Principles:
# 1. Two pointers: left (start) and right (end) define window
# 2. Expand window by incrementing right pointer
# 3. Contract window by incrementing left when constraint violated
# Pattern: Binary Tree Level Order Traversal (BFS)
from collections import deque
from typing import Optional
class TreeNode:
def __init__(self, val: int = 0, left: Optional['TreeNode'] = None, right: Optional['TreeNode'] = None):
self.val = val
self.left = left
self.right = right
def level_order_traversal(root: Optional[TreeNode]) -> list[list[int]]:
"""Perform BFS level-order traversal of binary tree.
Returns list of lists where each inner list contains node values at that level.
Time: O(n), Space: O(w) where w is max width of tree
Example:
Input: 3
/ \
9 20
/ \
15 7
Output: [[3], [9, 20], [15, 7]]
"""
if not root:
return []
result: list[list[int]] = []
queue: deque[TreeNode] = deque([root])
while queue:
# CRITICAL: Capture level size BEFORE processing
# This separates current level from next level nodes
level_size = len(queue)
current_level: list[int] = []
# Process exactly level_size nodes (all nodes at current level)
for _ in range(level_size):
node = queue.popleft() # O(1) with deque
current_level.append(node.val)
# Add children for next level processing
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(current_level)
return result
# BFS Key Principles:
# 1. Use collections.deque for O(1) append/popleft operations (NOT list)
# 2. Capture level_size = len(queue) before inner loop to separate levels
# 3. Process entire level before moving to next (prevents mixing levels)
# 4. Add children during current level processing
# Common uses: level order traversal, shortest path, connected components, graph exploration
# Pattern: Median of two sorted arrays
def find_median_sorted_arrays(nums1: list[int], nums2: list[int]) -> float:
"""Find median of two sorted arrays in O(log(min(m,n))) time.
Strategy: Binary search on smaller array to find partition point
"""
# Ensure nums1 is smaller for optimization
if len(nums1) > len(nums2):
nums1, nums2 = nums2, nums1
m, n = len(nums1), len(nums2)
left, right = 0, m
while left <= right:
partition1 = (left + right) // 2
partition2 = (m + n + 1) // 2 - partition1
# Handle edge cases with infinity
max_left1 = float('-inf') if partition1 == 0 else nums1[partition1 - 1]
min_right1 = float('inf') if partition1 == m else nums1[partition1]
max_left2 = float('-inf') if partition2 == 0 else nums2[partition2 - 1]
min_right2 = float('inf') if partition2 == n else nums2[partition2]
# Check if partition is valid
if max_left1 <= min_right2 and max_left2 <= min_right1:
# Found correct partition
if (m + n) % 2 == 0:
return (max(max_left1, max_left2) + min(min_right1, min_right2)) / 2
return max(max_left1, max_left2)
elif max_left1 > min_right2:
right = partition1 - 1
else:
left = partition1 + 1
raise ValueError("Input arrays must be sorted")
# Pattern: Two sum problem
def two_sum(nums: list[int], target: int) -> tuple[int, int] | None:
"""Find indices of two numbers that sum to target.
Time: O(n), Space: O(n)
"""
seen: dict[int, int] = {}
for i, num in enumerate(nums):
complement = target - num
if complement in seen:
return (seen[complement], i)
seen[num] = i
return None