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Copy-ready Python snippets by topic and difficulty — short, focused, and runnable in the browser editor.
How to Merge Two Sorted Lists in Python
Merge two sorted lists into one sorted list using a two-pointer loop, then extend with remaining elements.
def merge_sorted_lists(list1, list2):
merged = []
i = j = 0
while i < len(list1) and j < len(list2):
if list1[i] <= list2[j]:
merged.append(list1[i])
i += 1
else:
merged.append(list2[j])
j += 1
merged.extend(list1[i:])
merged…How to Reverse a List in Place Without Using reverse() in Python
A two-pointer while loop swaps elements from both ends toward the center to reverse a list in place without creating a copy.
def reverse_list_in_place(lst):
left = 0
right = len(lst) - 1
while left < right:
lst[left], lst[right] = lst[right], lst[left]
left += 1
right -= 1
if __name__ == "__main__":
my_list = [1, 2, 3, 4, 5]
print("Original:", my_list)
reverse_list_in_place(my_list)
prin…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 Median of Two Sorted Arrays in Python
Merges two sorted arrays with a two-pointer walk and returns the median of the combined sorted sequence.
def median_of_two_sorted_arrays(nums1, nums2):
merged = []
i = j = 0
while i < len(nums1) and j < len(nums2):
if nums1[i] <= nums2[j]:
merged.append(nums1[i])
i += 1
else:
merged.append(nums2[j])
j += 1
merged.extend(nums1[i:])
merged.…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 Equilibrium Index of a List in Python
Find every index in a list where the sum of elements to its left equals the sum to its right, using a single pass.
def find_equilibrium_indexes(arr):
total = sum(arr)
left_sum = 0
indexes = []
for i, num in enumerate(arr):
total -= num
if left_sum == total:
indexes.append(i)
left_sum += num
return indexes
if __name__ == "__main__":
test = [1, 2, 3, -1, 2, 3]
result =…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 Array by Parity (Even Before Odd) in Python
Rearrange an array so all even numbers appear before all odd numbers using a simple two-list partition approach.
def sort_array_by_parity(nums):
"""
Rearrange the array so that all even integers come first,
followed by all odd integers. The order within even and odd
groups is not required to be sorted.
"""
even = []
odd = []
for num in nums:
if num % 2 == 0:
even.append(nu…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 Two Sorted Arrays Without Extra Space in Python
Merge two sorted arrays in-place from the end, using the trailing zeros in the first array to avoid extra space.
def merge_sorted(arr1, arr2):
m, n = len(arr1), len(arr2)
i, j = m - 1, n - 1
while j >= 0:
if i >= 0 and arr1[i] > arr2[j]:
arr1[i + j + 1] = arr1[i]
i -= 1
else:
arr1[i + j + 1] = arr2[j]
j -= 1
return arr1
if __name__ == "__main__":
…Move Zeroes to End in Python Maintaining Order
In-place algorithm that moves all zeroes to the end of a list while preserving the relative order of non-zero elements.
def move_zeroes(nums):
non_zero_index = 0
for i in range(len(nums)):
if nums[i] != 0:
nums[non_zero_index], nums[i] = nums[i], nums[non_zero_index]
non_zero_index += 1
return nums
if __name__ == "__main__":
example = [0, 1, 0, 3, 12]
result = move_zeroes(example)
…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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