💡 Problem Formulation: We are set to find the number of rectangles within a collection that can be rearranged to form the largest possible square. Assume we are given a list of tuples with each tuple representing the dimensions of a rectangle (width, height). The goal is to identify how many of these rectangles can be combined to create the largest square, where side lengths are equal. For example, given rectangles [(4,6), (2,3), (5,5), (6,6)], the largest square would be 6×6, and one such rectangle could be used to form this square.
Method 1: Brute Force Approach
This method iterates over each rectangle, finding the largest square that can be formed, and counting the number of rectangles that match this specification. It’s a direct approach but may not be efficient for a large number of rectangles.
Here’s an example:
def find_largest_square(rectangles):
squares = [min(rec) for rec in rectangles]
largest_square = max(squares)
return squares.count(largest_square)
rectangles = [(4, 6), (2, 3), (5, 5), (6, 6)]
print(find_largest_square(rectangles))Output for the provided code snippet: 1
The function find_largest_square calculates the largest square side possible from each rectangle and counts those matching the largest size possible. This example finds that one 6×6 square can be formed from the input list of rectangles.
Method 2: Sorting and Binary Search
Enhancing performance by sorting rectangles to use binary search to quickly find the count of rectangles that can form the largest square. Suitable for larger datasets.
Here’s an example:
def find_largest_square(rectangles):
squares = sorted([min(rec) for rec in rectangles], reverse=True)
largest_square = squares[0]
return squares.count(largest_square)
rectangles = [(4, 6), (2, 3), (5, 5), (6, 6)]
print(find_largest_square(rectangles))Output for the provided code snippet: 1
This optimization sorts the list of possible square sides in descending order. The largest square dimension is at the start of the list, and we count the occurrences of this size. The sorted list can be further leveraged for binary search if needed.
Method 3: Hashing Approach
Optimizing for constant-time lookups by using a hash table to store the counts of each potential square size encountered and directly accessing the largest square’s count.
Here’s an example:
from collections import Counter
def find_largest_square(rectangles):
square_counts = Counter(min(rec) for rec in rectangles)
return square_counts[max(square_counts)]
rectangles = [(4, 6), (2, 3), (5, 5), (6, 6)]
print(find_largest_square(rectangles))Output for the provided code snippet: 1
This approach utilizes the Counter class from Python’s collections module to create a frequency table for each potential square size. The largest key’s value is returned as the result.
Method 4: Functional Programming Approach
Leveraging Python’s functional programming features, this method employs map and max functions for concise and readable code.
Here’s an example:
rectangles = [(4, 6), (2, 3), (5, 5), (6, 6)] largest_square = max(map(min, rectangles)) count_largest_square = list(map(min, rectangles)).count(largest_square) print(count_largest_square)
Output for the provided code snippet: 1
This snippet maps the min function over the list of rectangles to find the largest square possible and then counts how frequently this square size appears.
Bonus One-Liner Method 5: List Comprehension with Lambda
A compact one-liner using list comprehension and a lambda function, which is perfect when looking for a succinct solution.
Here’s an example:
rectangles = [(4, 6), (2, 3), (5, 5), (6, 6)] count_largest_square = (lambda sq: sq.count(max(sq)))([min(r) for r in rectangles]) print(count_largest_square)
Output for the provided code snippet: 1
The lambda function in this one-liner is immediately invoked with a list comprehension that determines the size of the largest square possible, returning the count of rectangles that match this size.
Summary/Discussion
- Method 1: Brute Force Approach. Simple and straightforward. Becomes inefficient with larger datasets due to O(n) time complexity.
- Method 2: Sorting and Binary Search. More efficient with large datasets due to sorted structure. Still operates in O(n log n) time complexity because of sorting.
- Method 3: Hashing Approach. Offers O(n) time complexity with constant-time lookups for better scalability. Python’s Counter class simplifies implementation.
- Method 4: Functional Programming Approach. Provides clarity and conciseness. Relies on the power of
mapandmaxfunctions but is not necessarily faster computationally. - Bonus One-Liner Method 5: Compact code and elegant. However, readability may suffer for those unfamiliar with lambda functions and list comprehensions.