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 Elements Appearing More Than n/3 Times in Python
Return all elements that occur more than len(array)/3 times using a simple dictionary counter.
def majority_third(arr):
"""Return elements appearing more than len(arr)/3 times."""
cutoff = len(arr) / 3
counts = {}
for x in arr:
counts[x] = counts.get(x, 0) + 1
return [x for x, c in counts.items() if c > cutoff]
if __name__ == "__main__":
test1 = [3, 2, 3]
test2 = [1, 1, 1, …Find First Duplicate Index in Python
Return the index of the first element that appears more than once in a list, using a dictionary for O(n) time.
def find_first_duplicate(arr):
seen = {}
for index, value in enumerate(arr):
if value in seen:
return index
seen[value] = index
return -1
if __name__ == "__main__":
test_array = [3, 5, 2, 8, 5, 1, 2]
result = find_first_duplicate(test_array)
print(f"Array: {test_arr…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 Count Occurrences of Each Value in Python
Count how many times each value appears in a list using Python's Counter from the collections module.
from collections import Counter
def count_occurrences(values):
"""Return a dictionary mapping each value to its count."""
return dict(Counter(values))
if __name__ == "__main__":
sample_data = ["apple", "banana", "apple", "cherry", "banana", "apple"]
result = count_occurrences(sample_data)
print(r…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 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 Sort Colors (Dutch National Flag) in Python
In-place sorting of a list of 0s, 1s, and 2s using the Dutch National Flag algorithm with O(n) time and O(1) space.
def sort_colors(nums):
low, mid, high = 0, 0, len(nums) - 1
while mid <= high:
if nums[mid] == 0:
nums[low], nums[mid] = nums[mid], nums[low]
low += 1
mid += 1
elif nums[mid] == 1:
mid += 1
else: # nums[mid] == 2
nums[mid], n…How to solve the stock span problem in Python
Calculate the stock span for each day's price using a monotonic stack in O(n) time.
def stock_span(prices):
span = [1] * len(prices)
stack = []
for i in range(len(prices)):
while stack and prices[stack[-1]] <= prices[i]:
stack.pop()
span[i] = i - stack[-1] if stack else i + 1
stack.append(i)
return span
if __name__ == "__main__":
pric…Implement Insert Delete GetRandom O(1) in Python
Build a RandomizedSet class that supports insert, delete, and get_random in average O(1) time using a list and a dictionary mapping values to indices.
import random
class RandomizedSet:
def __init__(self):
self.values = []
self.index_map = {}
def insert(self, val):
if val in self.index_map:
return False
self.index_map[val] = len(self.values)
self.values.append(val)
return True
def delete(self…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 *…Quickselect in Python: Find the kth Smallest Element
Python implementation of the Quickselect algorithm to find the kth smallest element in an unsorted list with average O(n) time complexity.
def quickselect(arr, k):
"""
Returns the k-th smallest element (0-indexed) using Quickselect.
Average: O(n), Worst: O(n^2)
"""
if len(arr) == 1:
return arr[0]
pivot = arr[-1]
left = [x for x in arr[:-1] if x <= pivot]
right = [x for x in arr[:-1] if x > pivot]
if k < len(l…Rearrange array alternately max min in Python
Rearranges a sorted list so its elements alternate between the current maximum and current minimum using two pointers in O(n) time.
def rearrange_alternately(arr):
"""
Rearrange sorted array so elements alternate: max, min, next max, next min...
Returns a new list in O(n) time using O(n) space.
"""
n = len(arr)
result = []
left, right = 0, n - 1
while left <= right:
if left == right:
result.appen…Browse by section
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