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 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…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 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:
…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…Merge k sorted lists in Python using a heap
Merge k individually sorted lists into one sorted list in Python using a min-heap.
import heapq
def merge_k_sorted_lists(lists):
heap = []
# Push the first element of each list onto the heap
for i, lst in enumerate(lists):
if lst:
heapq.heappush(heap, (lst[0], i, 0))
result = []
while heap:
val, list_idx, elem_idx = heapq.heappop(heap)
re…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]
…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
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