#2975

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Maximum Square Area by Removing Fences From a Field

This problem challenges you to calculate the largest square area possible by removing fences from a given rectangular field.

array hash table enumeration

Problem Statement

Given a rectangular field of dimensions (m - 1) x (n - 1), defined by horizontal and vertical fences, determine the largest square area that can be formed by removing some or all of the fences. Horizontal fences are described by the array hFences, where each fence spans across the entire row at a given y-coordinate. Similarly, vertical fences are represented by the array vFences, spanning the entire column at each x-coordinate.

Your task is to return the maximum area of a square that can be formed in the field after removing some fences. If no square can be formed, return -1. The solution must leverage the differences between adjacent fences to identify potential square dimensions.

Examples

Example 1

Input: m = 4, n = 3, hFences = [2,3], vFences = [2]

Output: 4

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

It can be proved that there is no way to create a square field by removing fences.

Constraints

3 <= m, n <= 109
1 <= hFences.length, vFences.length <= 600
1 < hFences[i] < m
1 < vFences[i] < n
hFences and vFences are unique.

Solution Approach

Array Scanning

Start by transforming the arrays of horizontal and vertical fences. Add 1 to the beginning and m (for hFences), and 1 to the beginning and n (for vFences). Then, calculate the difference between each consecutive pair of values in the transformed arrays. These differences represent potential side lengths for squares.

Hash Lookup for Optimal Squares

Use hash tables to efficiently check the presence of horizontal and vertical gaps. For each potential square size, confirm if the square can be formed by checking if the corresponding rows and columns are free of fences. This step ensures that you do not repeatedly scan the entire grid.

Edge Case Handling

Consider edge cases where it may not be possible to form a square, such as when the field is too small or the fences block all possible square areas. Return -1 in these cases to reflect the impossibility of forming a valid square.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the final approach, particularly the number of fences and the efficiency of the array scanning and hash lookup steps. Space complexity depends on the storage needed for the transformed fence arrays and hash table lookups.

What Interviewers Usually Probe

The candidate should focus on efficiently scanning the arrays and utilizing hash lookups to avoid brute force solutions.
Look for the candidate to demonstrate how to handle the large constraints effectively, especially with hash tables and efficient array manipulation.
The candidate should consider and discuss edge cases, especially situations where no square can be formed.

Common Pitfalls or Variants

Common pitfalls

Failing to handle edge cases where no square can be formed, leading to incorrect or missing outputs.
Not efficiently calculating the differences between consecutive fences, which could result in slow performance for larger inputs.
Overcomplicating the approach with unnecessary nested loops or brute force methods instead of leveraging array scanning and hash lookup.

Follow-up variants

What if you were given the ability to remove only specific fences, not all fences?
How would the approach change if the dimensions of the field were not fixed and could vary during the problem?
What if there were additional constraints on the positioning of the fences or the size of the field?

FAQ

What is the main approach to solve the Maximum Square Area by Removing Fences From a Field problem?

The problem is best solved by scanning the arrays of fences, calculating the gaps between consecutive fences, and using hash lookups to confirm possible square areas.

What pattern is most applicable to this problem?

The problem follows the 'Array scanning plus hash lookup' pattern, where efficiency is key to handling the large input size.

Can the Maximum Square Area by Removing Fences From a Field problem be solved with a brute force approach?

While a brute force approach is possible, it would not be efficient enough for large inputs, making array scanning and hash lookup the optimal approach.

How does handling edge cases affect the solution to the Maximum Square Area by Removing Fences From a Field problem?

Edge cases, like when no square can be formed, are crucial to address in order to avoid incorrect outputs or missed solutions. Always consider them in the design.

What makes this problem a good example of array and hash table usage?

This problem relies heavily on efficiently scanning arrays and leveraging hash tables for quick lookups, making it a great example of using these techniques in combination.

terminal

Solution

Solution 1: Enumeration

We can enumerate any two horizontal fences $a$ and $b$ in $\textit{hFences}$, calculate the distance $d$ between $a$ and $b$, and record it in the hash table $hs$. Then, we enumerate any two vertical fences $c$ and $d$ in $\textit{vFences}$, calculate the distance $d$ between $c$ and $d$, and record it in the hash table $vs$. Finally, we traverse the hash table $hs$. If a certain distance $d$ in $hs$ also exists in the hash table $vs$, it indicates that there exists a square field with a side length of $d$, and the area is $d^2$. We just need to take the largest $d$ and calculate $d^2 \bmod 10^9 + 7$.

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class Solution:
    def maximizeSquareArea(
        self, m: int, n: int, hFences: List[int], vFences: List[int]
    ) -> int:
        def f(nums: List[int], k: int) -> Set[int]:
            nums.extend([1, k])
            nums.sort()
            return {b - a for a, b in combinations(nums, 2)}

        mod = 10**9 + 7
        hs = f(hFences, m)
        vs = f(vFences, n)
        ans = max(hs & vs, default=0)
        return ans**2 % mod if ans else -1

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