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Algorithms & data structures

Classic patterns — search, sort, stacks, queues, and practical complexity-aware code.

14 matches
Algorithms & data structures medium

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.

two-pointer array algorithm
Python
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)
        …
15 0 Open
Algorithms & data structures medium

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.

binary-search minimum rotated-array
Python
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…
12 0 Open
Algorithms & data structures medium

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.

binary-search peak array
Python
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…
17 0 Open
Algorithms & data structures medium

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.

floyd-cycle duplicate-number two-pointers
Python
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…
14 0 Open
Algorithms & data structures medium

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.

boyer-moore majority-element array
Python
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…
14 0 Open
Algorithms & data structures medium

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.

bit-manipulation xor arrays
Python
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)
…
13 0 Open
Algorithms & data structures medium

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.

stack monotonic stack algorithm
Python
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…
13 0 Open
Algorithms & data structures medium

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.

binary-search rotated-array search-algorithm
Python
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[…
12 0 Open
Algorithms & data structures medium

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.

algorithm sorting two-pointers
Python
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…
14 0 Open
Algorithms & data structures medium

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.

stack monotonic-stack algorithm
Python
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…
13 0 Open
Algorithms & data structures medium

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.

randomized-set o1-lookup hash-map
Python
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…
12 0 Open
Algorithms & data structures medium

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.

array prefix suffix
Python
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]
    
…
14 0 Open
Algorithms & data structures medium

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.

arrays prefix-product suffix-product
Python
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 *…
14 0 Open
Algorithms & data structures medium

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.

quickselect selection algorithm
Python
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…
16 0 Open

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