In computer science and coding interviews, problems related to arrays and geometry frequently arise. One such problem is the “Number Of Rectangles That Can Form The Largest Square” on Leetcode. This problem tests your ability to handle array manipulations, simple geometry, and optimizing algorithms. In this article, we will dive deep into various ways of solving this problem in Python, covering every little detail from understanding the problem to optimizing the algorithm.
Table of Contents
- Problem Statement
- Understanding the Problem
- Brute-Force Approach
- Optimized Algorithm
- Python Code Walkthrough
- Time and Space Complexity Analysis
- Test Cases
- Common Pitfalls and How to Avoid Them
1. Problem Statement
You are given a list of rectangles represented by their widths and heights. Your task is to find the maximum square side length that can be formed using these rectangles and return the number of rectangles that can be used to form a square with this maximum side length.
rectangles.lengthis in the range [1, 1000].
1 <= rectangles[i], rectangles[i] <= 10^9
2. Understanding the Problem
Each rectangle is represented by its width and height, both of which are integers. You need to identify the maximum square length that can be formed and the number of rectangles that can contribute to forming a square with that length.
3. Brute-Force Approach
The simplest approach is to go through each rectangle one by one and calculate the minimum of the width and height for each, treating it as the maximum side length for a square that can be formed using that rectangle. Then, you could find the maximum of these and count how many rectangles have this maximum value.
4. Optimized Algorithm
The brute-force approach is not efficient, especially when the number of rectangles increases. We can optimize this by performing a single pass through the rectangles array, updating the maximum square side length and the corresponding count in one go.
5. Python Code Walkthrough
Let’s examine a Python function that uses this optimized approach.
def countGoodRectangles(rectangles): max_len = 0 count = 0 for rec in rectangles: square_len = min(rec) if square_len > max_len: max_len = square_len count = 1 elif square_len == max_len: count += 1 return count
max_len keeps track of the maximum square side length, and
count keeps track of the number of rectangles that can form a square with side length
6. Time and Space Complexity Analysis
The time complexity of this optimized algorithm is O(n), where n is the number of rectangles. It only involves a single pass through the array, making it quite efficient. The space complexity is O(1), as we only use a constant amount of extra space.
7. Test Cases
Test your function with a variety of test cases to ensure that it works as expected.
print(countGoodRectangles([[5, 8], [3, 9], [5, 12], [16, 5]])) # Should return 3 print(countGoodRectangles([[2, 3], [3, 7], [4, 3], [3, 7]])) # Should return 3
8. Common Pitfalls and How to Avoid Them
- Ignoring Edge Cases: Always remember to consider edge cases, such as when all rectangles are squares, or when there’s only one rectangle.
- Forgetting to Update Count: Make sure to update the
countvariable correctly, both when you find a new maximum and when you find another rectangle that matches the current maximum.
The “Number Of Rectangles That Can Form The Largest Square” problem tests your understanding of array manipulations and basic geometry while challenging you to optimize your code. The problem is simple enough to be solved with a straightforward algorithm, but it offers a good opportunity to practice optimization techniques and think critically about edge cases and constraints.