What is the time complexity of Maximum Square Area by Removing Fences From a Field?

The optimal solution for Maximum Square Area by Removing Fences From a Field runs in O(n log n) time and uses O(n) space. The complexity is O(n log n) due to sorting the fences, while the space complexity is O(n) for storing the modified fence arrays.

What is the brute force solution for Maximum Square Area by Removing Fences From a Field?

The brute force approach for Maximum Square Area by Removing Fences From a Field is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible pairs of horizontal and vertical fences to determine the maximum square area that can be formed. This method is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Square Area by Removing Fences From a Field?

Maximum Square Area by Removing Fences From a Field uses the following concepts: Array, Hash Table, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Square Area by Removing Fences From a Field?

Always clarify the problem constraints and edge cases before diving into coding. Think about how to optimize your solution after the brute-force approach. Practice explaining your thought process clearly as you code.

What are common mistakes when solving Maximum Square Area by Removing Fences From a Field?

Not considering the boundaries when calculating gaps. Failing to account for the case where no square can be formed.

#2975

Maximum Square Area by Removing Fences From a Field

Medium

ArrayHash TableEnumerationHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

The optimal solution leverages the differences between the fences to quickly determine the maximum square area without checking every pair. By calculating the maximum gaps between fences, we can directly derive the largest possible square.

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Algorithm

4 steps

1Step 1: Include the boundaries (1 and m for horizontal, 1 and n for vertical) into hFences and vFences.
2Step 2: Sort both hFences and vFences.
3Step 3: Calculate the maximum height and width by finding the maximum differences between consecutive fences.
4Step 4: The maximum square area is the minimum of the maximum height and width squared.

solution.py10 lines

1# Full working Python code
2
3def maxSquareArea(m, n, hFences, vFences):
4    hFences = [1] + sorted(hFences) + [m]
5    vFences = [1] + sorted(vFences) + [n]
6    max_height = max(hFences[i + 1] - hFences[i] for i in range(len(hFences) - 1))
7    max_width = max(vFences[i + 1] - vFences[i] for i in range(len(vFences) - 1))
8    max_side = min(max_height, max_width)
9    return (max_side * max_side) % (10**9 + 7) if max_side > 0 else -1
10

ℹ

Complexity note: The complexity is O(n log n) due to sorting the fences, while the space complexity is O(n) for storing the modified fence arrays.

1Understanding the relationship between the gaps created by fences and the maximum area of squares is crucial.
2Sorting the fences allows for efficient calculation of maximum gaps.

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