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)
…Count Smaller Elements to the Right in Python
Return a list where each index counts how many elements to its right are smaller than that element using a clean O(n²) nested-loop approach.
def count_smaller_elements(arr):
"""
Return a list where result[i] is the number of elements
to the right of arr[i] that are smaller than arr[i].
"""
result = []
for i in range(len(arr)):
count = 0
for j in range(i + 1, len(arr)):
if arr[j] < arr[i]:
…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 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 Longest Consecutive Run in an Unsorted List in Python
Find the length of the longest sequence of consecutive integers in an unsorted list using a set and a linear scan.
def longest_run(nums):
if not nums:
return 0
num_set = set(nums)
longest = 0
for num in num_set:
# Only start counting from the smallest number in a sequence
if num - 1 not in num_set:
current = num
length = 1
while current + 1 in num_set:
…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 Maximum Distance Between Identical Elements in Python
Compute the maximum index distance between any two identical elements in a list using a dictionary to track first occurrences.
from collections import defaultdict
def max_distance_between_identical(nums):
first_occurrence = {}
max_dist = 0
for i, num in enumerate(nums):
if num in first_occurrence:
dist = i - first_occurrence[num]
max_dist = max(max_dist, dist)
else:
first_occur…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 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 Pivot Index in Python
Locate the index where the sum of elements to the left equals the sum to the right, using a single pass with prefix sums.
def find_pivot_index(nums):
total = sum(nums)
left_sum = 0
for i, num in enumerate(nums):
if left_sum == total - left_sum - num:
return i
left_sum += num
return -1
if __name__ == "__main__":
test_cases = [
[1, 7, 3, 6, 5, 6],
[1, 2, 3],
[2, 1, -…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 =…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 Rotate an Array by k Steps in Python
This code rotates a list to the right by k positions using modulo arithmetic to handle k larger than the list length.
def rotate_array(nums, k):
if not nums:
return []
n = len(nums)
k = k % n
return nums[-k:] + nums[:-k] if k else nums[:]
if __name__ == "__main__":
arr = [1, 2, 3, 4, 5, 6]
k = 2
result = rotate_array(arr, k)
print(f"Original: {arr}")
print(f"Rotated by {k}: {result}")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
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