Algorithms & data structures
Classic patterns — search, sort, stacks, queues, and practical complexity-aware code.
Find Longest Increasing Subsequence Length in Python
Compute the length of the longest increasing subsequence in an array using dynamic programming.
def longest_increasing_subsequence(nums):
if not nums:
return 0
dp = [1] * len(nums)
for i in range(1, len(nums)):
for j in range(i):
if nums[i] > nums[j]:
dp[i] = max(dp[i], dp[j] + 1)
return max(dp)
if __name__ == "__main__":
# Demo with…
Game of Life Next State Grid in Python
Compute the next generation of Conway's Game of Life from a 2D grid using the standard three rules with neighbor counting.
def next_state(grid):
m, n = len(grid), len(grid[0])
new = [[0] * n for _ in range(m)]
for r in range(m):
for c in range(n):
total = 0
for dr in (-1, 0, 1):
for dc in (-1, 0, 1):
if dr == 0 and dc == 0:
continue
…
How to Solve Daily Temperatures Days Until Warmer in Python
Compute the number of days until a warmer temperature for each day using a monotonic stack.
def daily_temperatures(temps):
n = len(temps)
result = [0] * n
stack = []
for i, temp in enumerate(temps):
while stack and temps[stack[-1]] < temp:
prev_idx = stack.pop()
result[prev_idx] = i - prev_idx
stack.append(i)
return result
if __name__ == …
How to Solve the Trapping Rain Water Problem in Python
Compute the total water trapped between elevation bars using a two-pointer O(n) algorithm.
def trap(height):
if not height:
return 0
left, right = 0, len(height) - 1
left_max, right_max = 0, 0
water = 0
while left < right:
if height[left] < height[right]:
if height[left] >= left_max:
left_max = height[left]
else:
…
Product of Array Except Self in Python Without Division
Compute the product of all array elements except the current one in O(n) time using prefix and suffix products, without using division.
from math import prod
def product_except_self(nums):
n = len(nums)
result = [1] * n
left_product = 1
for i in range(n):
result[i] = left_product
left_product *= nums[i]
right_product = 1
for i in range(n - 1, -1, -1):
result[i] *= right_product
right_product *…
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