Algorithms & data structures
Classic patterns — search, sort, stacks, queues, and practical complexity-aware code.
Binary Search for Ship Capacity in Python
Use binary search to find the minimum ship capacity that can transport all packages within a given number of days.
def ship_within_days(weights, days):
def can_ship(capacity):
current = 0
needed_days = 1
for weight in weights:
if current + weight > capacity:
needed_days += 1
current = 0
current += weight
return needed_days <= days
low …Binary Search on Answer in Python: Koko Eating Bananas
Find the minimum eating speed so Koko finishes all banana piles within a given hour limit using binary search on the answer.
import math
def min_eating_speed(piles, h):
"""Return minimum integer eating speed K so Koko finishes within h hours."""
def hours_needed(speed):
return sum(math.ceil(p / speed) for p in piles)
low, high = 1, max(piles)
while low < high:
mid = (low + high) // 2
if hours_needed…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)
…Drop Elements From Start While Condition Is True in Python
This generator function drops elements from the beginning of an iterable while a predicate returns true, then yields the rest.
def drop_while(predicate, iterable):
"""Drop elements from the start while predicate is true."""
it = iter(iterable)
for item in it:
if not predicate(item):
yield item
break
yield from it
if __name__ == "__main__":
numbers = [1, 2, 3, 4, 1, 2, 5]
result = list(d…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 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 Missing Number in Python Sequence 1 to N
Find the missing number from a list containing numbers 1 to N using the arithmetic sum formula.
def find_missing_number(nums, n):
expected_sum = n * (n + 1) // 2
actual_sum = sum(nums)
return expected_sum - actual_sum
if __name__ == "__main__":
n = 10
numbers = [1, 2, 3, 4, 5, 6, 7, 9, 10]
missing = find_missing_number(numbers, n)
print(f"The missing number is: {missing}")Find Missing Numbers, Duplicates, and Ranges in Python
Analyze a list to identify missing numbers, duplicate values, and contiguous ranges using sets and the Counter class.
def find_missing_duplicates_ranges(numbers):
"""Find missing numbers, duplicates, and ranges in a list."""
from collections import Counter
if not numbers:
return {"missing": [], "duplicates": [], "ranges": []}
full_range = set(range(min(numbers), max(numbers) + 1))
present = set(n…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 k Closest Points to Origin in Python
Sorts a list of (x, y) point tuples by their Euclidean distance from the origin and returns the k nearest points.
import math
def k_closest(points, k):
points.sort(key=lambda p: math.sqrt(p[0]**2 + p[1]**2))
return points[:k]
if __name__ == "__main__":
points = [(1, 2), (3, 4), (-1, 0), (5, 5), (0, 1)]
k = 3
result = k_closest(points, k)
print(f"Original points: {points}")
print(f"K closest points (k…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 Last Index Where a Condition Is True in Python
This code scans a sequence from the end and returns the index of the last element that satisfies a given condition, or -1 if none do.
def last_index_where(sequence, condition):
"""Return the index of the last element in sequence that satisfies condition."""
for i in range(len(sequence) - 1, -1, -1):
if condition(sequence[i]):
return i
return -1
if __name__ == "__main__":
numbers = [1, 4, 7, 2, 9, 5, 8, 3]
is_…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 the Second Largest Unique Number in a Python List
This Python function finds the second largest unique number from a list by converting it to a set, removing the maximum, and returning the new maximum.
def second_largest_unique(numbers):
unique_numbers = set(numbers)
if len(unique_numbers) < 2:
return None
unique_numbers.remove(max(unique_numbers))
return max(unique_numbers)
if __name__ == "__main__":
test_list = [4, 2, 9, 5, 2, 9, 1, 5]
result = second_largest_unique(test_list)
…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)
…Game of Life Next State Grid in Python
Compute the next generation of Conway's Game of Life from a 2D grid using the standard three rules with neighbor counting.
def next_state(grid):
m, n = len(grid), len(grid[0])
new = [[0] * n for _ in range(m)]
for r in range(m):
for c in range(n):
total = 0
for dr in (-1, 0, 1):
for dc in (-1, 0, 1):
if dr == 0 and dc == 0:
continue
…Generate Pascal's Triangle Rows in Python
Builds Pascal's triangle as a list of rows, where each inner value is the sum of the two values above it.
def generate_pascals_triangle(rows):
triangle = []
for row_num in range(rows):
row = [1] * (row_num + 1)
for col in range(1, row_num):
row[col] = triangle[row_num - 1][col - 1] + triangle[row_num - 1][col]
triangle.append(row)
return triangle
if __name__ == "__main__":
…How to Apply a Function to Sliding Window Slices in Python
This Python code applies a given function to every contiguous window of a specified size in a list, returning a list of results.
def apply_to_sliding_windows(data, window_size, func):
return [func(data[i:i + window_size]) for i in range(len(data) - window_size + 1)]
if __name__ == "__main__":
numbers = [1, 2, 3, 4, 5, 6]
window_size = 3
results = apply_to_sliding_windows(numbers, window_size, sum)
print(results)
results…How to Compare Two Lists Elementwise for Greater Flags in Python
Compare two equal-length lists element by element and return a list of booleans marking where list_a values are greater than list_b values.
def compare_lists_greater(list_a, list_b):
"""
Compare two lists elementwise and return a list of booleans
indicating whether each element in list_a is greater than the
corresponding element in list_b.
"""
if len(list_a) != len(list_b):
raise ValueError("Lists must have the same length"…How to Compute Jaccard Similarity in Python
Compute the Jaccard similarity between two lists by converting them to sets and dividing the intersection size by the union size.
def jaccard_similarity(list1, list2):
set1 = set(list1)
set2 = set(list2)
intersection = set1 & set2
union = set1 | set2
if not union:
return 0.0
return len(intersection) / len(union)
if __name__ == "__main__":
a = [1, 2, 3, 4, 5]
b = [3, 4, 5, 6, 7]
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