What is the time complexity of Number Of Rectangles That Can Form The Largest Square?

The optimal solution for Number Of Rectangles That Can Form The Largest Square runs in O(n) time and uses O(1) space. This complexity is achieved because we only loop through the rectangles once, making it linear in relation to the number of rectangles.

What is the brute force solution for Number Of Rectangles That Can Form The Largest Square?

The brute force approach for Number Of Rectangles That Can Form The Largest Square is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will calculate the maximum square side length for each rectangle and then count how many rectangles can produce that maximum square. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number Of Rectangles That Can Form The Largest Square?

Number Of Rectangles That Can Form The Largest Square uses the following concepts: Array. The recommended patterns to study are: Array, Greedy.

What are the interview tips for Number Of Rectangles That Can Form The Largest Square?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as rectangles with very small dimensions. Practice explaining your thought process as you code, as this helps in interviews.

What are common mistakes when solving Number Of Rectangles That Can Form The Largest Square?

Not understanding that the largest square side is determined by the minimum of length and width. Failing to count rectangles correctly after determining maxLen.

#1725

Number Of Rectangles That Can Form The Largest Square

Easy

ArrayArrayGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution improves efficiency by combining the steps of finding the maximum square side length and counting the rectangles in a single pass. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize maxLen and count to 0.
2Step 2: Iterate through each rectangle, calculating the minimum side length (minSide).
3Step 3: If minSide is greater than maxLen, update maxLen and reset count to 1.
4Step 4: If minSide equals maxLen, increment the count.

solution.py16 lines

1# Full working Python code
2rectangles = [[5,8],[3,9],[5,12],[16,5]]
3
4def countRectangles(rectangles):
5    maxLen = 0
6    count = 0
7    for l, w in rectangles:
8        minSide = min(l, w)
9        if minSide > maxLen:
10            maxLen = minSide
11            count = 1
12        elif minSide == maxLen:
13            count += 1
14    return count
15
16print(countRectangles(rectangles))

ℹ

Complexity note: This complexity is achieved because we only loop through the rectangles once, making it linear in relation to the number of rectangles.

1The largest square side length is determined by the smallest dimension of the rectangles.
2Counting rectangles that can form the largest square can be done in a single pass for efficiency.

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