Is Maximum Square Area by Removing Fences From a Field easy or hard?

The problem is typically classified as Medium difficulty. The core challenge is recognizing that a square requires equal horizontal and vertical distances and that these distances come from fence pair differences. Once that insight is clear, a hash set optimization leads to an efficient solution.

Maximum Square Area by Removing Fences From a Field Python/Java solution

Most implementations compute all vertical gaps, store them in a set, and then iterate through horizontal gaps to find the largest matching value. The logic is straightforward to implement in Python using a set, in Java using HashSet, or in C++ using unordered_set. The algorithmic complexity remains O(h^2 + v^2).

How to solve Maximum Square Area by Removing Fences From a Field in O(n)?

A strict O(n) solution is not feasible because the side length candidates come from differences between fence pairs. The practical optimal approach computes all pairwise distances in each direction, which results in O(h^2 + v^2). Using a hash set keeps lookups constant time and avoids O(h^2 * v^2) comparisons.

What is the best approach for Maximum Square Area by Removing Fences From a Field?

The most efficient approach enumerates all vertical fence distances and stores them in a hash set, then checks horizontal fence distances against that set. This reduces comparisons dramatically and runs in O(h^2 + v^2) time. The hash lookup ensures constant‑time checks when verifying whether a square side length exists in both directions.

Is Maximum Square Area by Removing Fences From a Field asked at Google/Amazon/Meta?

Variants of grid geometry and coordinate distance problems appear frequently in interviews at companies like Google, Amazon, and Meta. Problems involving pairwise differences, hash sets, and geometric constraints are common because they test both algorithmic thinking and data structure usage.

What data structure is used in Maximum Square Area by Removing Fences From a Field?

The key data structure is a hash set that stores all vertical fence distance values. This allows constant‑time membership checks when evaluating horizontal distances. Arrays store fence positions, and enumeration generates candidate side lengths.

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

The optimized solution runs in O(h^2 + v^2) time, where h is the number of horizontal fences and v is the number of vertical fences. All pairwise gaps are computed for each direction, and hash set lookups check for matching lengths. Space complexity is O(v^2) for storing vertical distances.

Description Hints 3 Solution Video 32:50

2975. Maximum Square Area by Removing Fences From a Field

Medium49.5% acceptance3 companies

There is a large (m - 1) x (n - 1) rectangular field with corners at (1, 1) and (m, n) containing some horizontal and vertical fences given in arrays hFences and vFences respectively.

Horizontal fences are from the coordinates (hFences[i], 1) to (hFences[i], n) and vertical fences are from the coordinates (1, vFences[i]) to (m, vFences[i]).

Return the maximum area of a square field that can be formed by removing some fences (possibly none) or -1 if it is impossible to make a square field.

Since the answer may be large, return it modulo 10⁹+ 7.

Note: The field is surrounded by two horizontal fences from the coordinates (1, 1) to (1, n) and (m, 1) to (m, n) and two vertical fences from the coordinates (1, 1) to (m, 1) and (1, n) to (m, n). These fences cannot be removed.

Example 1:

Input: m = 4, n = 3, hFences = [2,3], vFences = [2]
Output: 4
Explanation: Removing the horizontal fence at 2 and the vertical fence at 2 will give a square field of area 4.

Example 2:

Input: m = 6, n = 7, hFences = [2], vFences = [4]
Output: -1
Explanation: It can be proved that there is no way to create a square field by removing fences.

Constraints: