Algorithms & data structures
Classic patterns — search, sort, stacks, queues, and practical complexity-aware code.
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 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)
…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 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 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[…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 All Elements Except Self in Python
Given a list of integers, return a list where each element is the product of all other elements except itself, using prefix and suffix products in O(n) time and O(1) extra space.
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 *= nums[i]
…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 *…Set Matrix Zeroes in Python: Markers List Grid Demo
Given a matrix, this code finds all rows and columns that contain a zero and sets every element in those rows and columns to zero, using boolean marker arrays.
def set_zeroes(matrix):
rows, cols = len(matrix), len(matrix[0])
row_markers = [False] * rows
col_markers = [False] * cols
# First pass: record which rows and columns contain zeros
for i in range(rows):
for j in range(cols):
if matrix[i][j] == 0:
row_markers[i] …Split Array Largest Sum in Python (Minimize Largest Subarray Sum)
Binary search + greedy check to split an array into k subarrays while minimizing the largest subarray sum.
def can_split(nums, k, max_sum):
subarrays = 1
current_sum = 0
for num in nums:
if current_sum + num <= max_sum:
current_sum += num
else:
subarrays += 1
current_sum = num
if subarrays > k:
return False
return True
def spli…Browse by section
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