How do you solve Number of Pairs of Interchangeable Rectangles on LeetCode?

Number of Pairs of Interchangeable Rectangles (LeetCode #2001) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to reduce ratios using GCD is crucial for this problem.

What is the time complexity of Number of Pairs of Interchangeable Rectangles?

The optimal solution for Number of Pairs of Interchangeable Rectangles runs in O(n) time and uses O(n) space. This complexity is due to iterating through the rectangles once and storing the ratios in a hashmap, which is efficient for counting.

What is the brute force solution for Number of Pairs of Interchangeable Rectangles?

The brute force approach for Number of Pairs of Interchangeable Rectangles is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible pair of rectangles to see if their width-to-height ratios are equal. While straightforward, this method can be slow for larger inputs due to its quadratic time complexity. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Pairs of Interchangeable Rectangles?

Number of Pairs of Interchangeable Rectangles uses the following concepts: Array, Hash Table, Math, Counting, Number Theory. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Pairs of Interchangeable Rectangles?

Always consider edge cases, such as rectangles with the same dimensions. Discuss your thought process and approach before jumping into coding. Be prepared to explain the time and space complexity of your solution.

What are common mistakes when solving Number of Pairs of Interchangeable Rectangles?

Forgetting to reduce the width and height ratio before storing it. Using floating-point division instead of integer operations can lead to precision errors.

#2001

Number of Pairs of Interchangeable Rectangles

Medium

ArrayHash TableMathCountingNumber TheoryHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Instead of checking every pair, we can use a hashmap to count how many rectangles share the same width-to-height ratio. This way, we can calculate the number of interchangeable pairs in linear time.

⚙️

Algorithm

4 steps

1Step 1: Create a hashmap to store the frequency of each ratio in reduced form.
2Step 2: For each rectangle, compute the width and height ratio in reduced form using the greatest common divisor (GCD).
3Step 3: Increment the count in the hashmap for this ratio.
4Step 4: For each unique ratio in the hashmap, calculate the number of interchangeable pairs using the formula: count * (count - 1) / 2.

solution.py14 lines

1# Full working Python code
2from collections import defaultdict
3from math import gcd
4class Solution:
5    def interchangeableRectangles(self, rectangles):
6        ratio_count = defaultdict(int)
7        for width, height in rectangles:
8            g = gcd(width, height)
9            ratio = (width // g, height // g)
10            ratio_count[ratio] += 1
11        count = 0
12        for freq in ratio_count.values():
13            count += freq * (freq - 1) // 2
14        return count

ℹ

Complexity note: This complexity is due to iterating through the rectangles once and storing the ratios in a hashmap, which is efficient for counting.

1Understanding how to reduce ratios using GCD is crucial for this problem.
2Using a hashmap allows for efficient counting of occurrences, avoiding the need for nested loops.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.