Algorithms & data structures
Classic patterns — search, sort, stacks, queues, and practical complexity-aware code.
Binary Search for Ship Capacity in Python
Use binary search to find the minimum ship capacity that can transport all packages within a given number of days.
def ship_within_days(weights, days):
def can_ship(capacity):
current = 0
needed_days = 1
for weight in weights:
if current + weight > capacity:
needed_days += 1
current = 0
current += weight
return needed_days <= days
low …Binary Search on Answer in Python: Koko Eating Bananas
Find the minimum eating speed so Koko finishes all banana piles within a given hour limit using binary search on the answer.
import math
def min_eating_speed(piles, h):
"""Return minimum integer eating speed K so Koko finishes within h hours."""
def hours_needed(speed):
return sum(math.ceil(p / speed) for p in piles)
low, high = 1, max(piles)
while low < high:
mid = (low + high) // 2
if hours_needed…Container With Most Water: Two-Pointer Solution in Python
Find the maximum water a container can hold from a list of heights using an efficient two-pointer technique in O(n) time.
from typing import List
def max_water_container(heights: List[int]) -> int:
left, right = 0, len(heights) - 1
max_area = 0
while left < right:
width = right - left
height = min(heights[left], heights[right])
area = width * height
max_area = max(max_area, area)
…Find All Triplets with Sum Zero in Python
This code finds all unique triplets in an array that sum to zero using a sorted array and two-pointer technique.
def find_triplets(nums):
nums.sort()
n = len(nums)
triplets = []
for i in range(n - 2):
if i > 0 and nums[i] == nums[i - 1]:
continue
left, right = i + 1, n - 1
while left < right:
total = nums[i] + nums[left] + nums[right]
if total == 0:
…Find Longest Consecutive Sequence in Python
Find the length of the longest consecutive elements sequence in an unsorted array using a set for O(n) lookups.
def longest_consecutive_length(nums):
num_set = set(nums)
longest = 0
for num in num_set:
if num - 1 not in num_set:
current = num
current_streak = 1
while current + 1 in num_set:
current += 1
current_streak += 1
…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…Find Minimum in Rotated Sorted List in Python
Uses binary search to find the minimum element in a rotated sorted list in O(log n) time.
def find_min(nums):
left, right = 0, len(nums) - 1
while left < right:
mid = (left + right) // 2
if nums[mid] > nums[right]:
left = mid + 1
else:
right = mid
return nums[left]
if __name__ == "__main__":
rotated = [4, 5, 6, 7, 0, 1, 2]
print(f"Minimu…Find Missing Numbers, Duplicates, and Ranges in Python
Analyze a list to identify missing numbers, duplicate values, and contiguous ranges using sets and the Counter class.
def find_missing_duplicates_ranges(numbers):
"""Find missing numbers, duplicates, and ranges in a list."""
from collections import Counter
if not numbers:
return {"missing": [], "duplicates": [], "ranges": []}
full_range = set(range(min(numbers), max(numbers) + 1))
present = set(n…Find Peak Element in Python Using Binary Search
A binary search solution that finds any peak element (an element strictly greater than its neighbors) in an unsorted array in O(log n) time.
def find_peak_element(nums):
left, right = 0, len(nums) - 1
while left < right:
mid = (left + right) // 2
if nums[mid] > nums[mid + 1]:
right = mid
else:
left = mid + 1
return left
if __name__ == "__main__":
test1 = [1, 2, 3, 1]
tes…Find the Duplicate Number in Python Using Floyd's Cycle Detection
Detects the duplicate integer in an array of n+1 numbers (values 1 to n) in O(n) time and O(1) space using Floyd's cycle detection algorithm applied to a linked-list model.
def find_duplicate(nums):
slow = nums[0]
fast = nums[0]
# Phase 1: Find intersection point of the cycle
while True:
slow = nums[slow]
fast = nums[nums[fast]]
if slow == fast:
break
# Phase 2: Find the start of the cycle (the duplicate)
slow = nums[0…Find the Majority Element in Python with Boyer-Moore Vote
Use Boyer-Moore majority vote to find the element appearing more than n/2 times in an array in O(n) time and O(1) space.
def majority_element(nums):
candidate = None
count = 0
for num in nums:
if count == 0:
candidate = num
count += 1 if num == candidate else -1
return candidate
if __name__ == "__main__":
nums = [2, 2, 1, 1, 1, 2, 2]
result = majority_element(nums)
print(f"Major…Find two unique numbers in an array with Python
Returns the two numbers that appear exactly once in a list where every other number appears twice, using XOR bit manipulation.
def find_two_odd(arr):
"""Return the two numbers that appear exactly once, while all others appear twice."""
xor_all = 0
for num in arr:
xor_all ^= num
# xor_all now equals the XOR of the two unique numbers.
# Find a set bit (any bit where they differ).
diff_bit = xor_all & (-xor_all)
…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 Decode a String with Repeated Brackets in Python
Decodes strings with patterns like '3[a]2[bc]' by using a stack to handle nested and repeated bracket groups.
def decode_string(s: str) -> str:
stack = []
current_num = 0
current_str = ""
for ch in s:
if ch.isdigit():
current_num = current_num * 10 + int(ch)
elif ch == "[":
stack.append((current_str, current_num))
current_str = ""
current_num = 0…How to Detect Hardcoded Secrets in Python Source Code
A Python utility that scans source code for common hardcoded secrets like API keys, passwords, tokens, and AWS credentials using regex patterns.
import re
def detect_secrets(text):
"""Detect potential hardcoded secrets in source code."""
patterns = {
'api_key': r'(?i)(api[_-]?key|apikey)\s*[=:]\s*["\']([^"\']+)["\']',
'password': r'(?i)(password|passwd)\s*[=:]\s*["\']([^"\']+)["\']',
'token': r'(?i)(\b(token|secret)\b)\s*[=:]\s…How to Evaluate RPN Expressions in Python
Use a stack to evaluate Reverse Polish Notation token lists with a dictionary of operator lambdas, truncating division toward zero.
def eval_rpn(tokens):
stack = []
ops = {
'+': lambda a, b: a + b,
'-': lambda a, b: a - b,
'*': lambda a, b: a * b,
'/': lambda a, b: int(a / b) # truncate toward zero
}
for token in tokens:
if token in ops:
b = stack.pop()
a = stack.pop(…How to Find Four Sum Quadruplets in Python (Sorted Demo)
Find all unique quadruplets in a sorted array that sum to a target, with duplicate skipping.
def four_sum(nums, target):
nums.sort()
result = []
n = len(nums)
for i in range(n - 3):
if i > 0 and nums[i] == nums[i - 1]:
continue
for j in range(i + 1, n - 2):
if j > i + 1 and nums[j] == nums[j - 1]:
continue
left, right = j + 1…How to Find Intersection of Two Sorted Interval Lists in Python
A two-pointer algorithm that finds all overlapping intervals between two sorted lists of intervals.
def interval_intersection(list1, list2):
i = j = 0
result = []
while i < len(list1) and j < len(list2):
# Find the overlap between current intervals
lo = max(list1[i][0], list2[j][0])
hi = min(list1[i][1], list2[j][1])
# If there's an overlap, add it to result
…How to Find Minimum Swaps to Sort an Array in Python
Calculate the minimum number of adjacent-free swaps needed to sort a permutation array using cycle detection in Python.
def min_swaps_to_sort(arr):
n = len(arr)
arr_pos = sorted((val, idx) for idx, val in enumerate(arr))
visited = [False] * n
swaps = 0
for i in range(n):
if visited[i] or arr_pos[i][1] == i:
continue
cycle_size = 0
j = i
while not visited[j]:
…How to Find the Next Greater Element for Each List Item in Python
Use a monotonic stack to find the next greater element to the right for every item in a list, in O(n) time.
def next_greater_element(nums):
result = [-1] * len(nums)
stack = []
for i in range(len(nums) - 1, -1, -1):
while stack and stack[-1] <= nums[i]:
stack.pop()
result[i] = stack[-1] if stack else -1
stack.append(nums[i])
return result
if __name__ == "__main…How to Find the Previous Smaller Element in Python
Use a monotonic stack to find the nearest smaller element to the left of each item in a list, returning -1 when none exists.
from collections import deque
def previous_smaller_elements(arr):
stack = deque()
result = [-1] * len(arr)
for i in range(len(arr)):
while stack and arr[stack[-1]] >= arr[i]:
stack.pop()
if stack:
result[i] = arr[stack[-1]]
stack.append(i)
return resul…How to Find the n Smallest Items in a Large List with heapq in Python
This code demonstrates how to efficiently extract the n smallest items from a large list using Python's heapq module and a manual max-heap approach.
import heapq
def n_smallest_iterable(data, n):
"""Return the n smallest items without loading the whole list."""
if n <= 0:
return []
return heapq.nsmallest(n, data)
def n_smallest_manual(data, n):
"""Return the n smallest using a heap, O(n log k) time."""
if n <= 0:
return []
…How to Generate a Power Set in Python with Bitmasks
Generate the power set of a small list using a bitmask approach, producing all possible subsets.
def power_set(items):
"""Generate the power set of a list using bitmask approach."""
n = len(items)
result = []
for mask in range(1 << n):
subset = []
for i in range(n):
if mask & (1 << i):
subset.append(items[i])
result.append(subset)
r…How to Search a Rotated Sorted List in Python
Binary search a pivot-rotated sorted list for a target value and return its index in O(log n) time.
from typing import List
def search_rotated(nums: List[int], target: int) -> int:
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return mid
# left half is sorted
if nums[left] <= nums[mid]:
if nums[…Browse by section
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