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.

#2975

Medium

auto_awesome数组·哈希·扫描

LeetCode 题解工作台

移除栅栏得到的正方形田地的最大面积

有一个大型的 (m - 1) x (n - 1) 矩形田地，其两个对角分别是 (1, 1) 和 (m, n) ，田地内部有一些水平栅栏和垂直栅栏，分别由数组 hFences 和 vFences 给出。水平栅栏为坐标 (hFences[i], 1) 到 (hFences[i], n) ，垂直栅栏为坐…

数组哈希表枚举

题目描述

有一个大型的 (m - 1) x (n - 1) 矩形田地，其两个对角分别是 (1, 1) 和 (m, n) ，田地内部有一些水平栅栏和垂直栅栏，分别由数组 hFences 和 vFences 给出。

水平栅栏为坐标 (hFences[i], 1) 到 (hFences[i], n)，垂直栅栏为坐标 (1, vFences[i]) 到 (m, vFences[i]) 。

返回通过移除一些栅栏（可能不移除）所能形成的最大面积的 正方形 田地的面积，或者如果无法形成正方形田地则返回 -1。

由于答案可能很大，所以请返回结果对 10⁹ + 7 取余后的值。

注意：田地外围两个水平栅栏（坐标 (1, 1) 到 (1, n) 和坐标 (m, 1) 到 (m, n) ）以及两个垂直栅栏（坐标 (1, 1) 到 (m, 1) 和坐标 (1, n) 到 (m, n) ）所包围。这些栅栏不能被移除。

示例 1：

输入：m = 4, n = 3, hFences = [2,3], vFences = [2]
输出：4
解释：移除位于 2 的水平栅栏和位于 2 的垂直栅栏将得到一个面积为 4 的正方形田地。

示例 2：

输入：m = 6, n = 7, hFences = [2], vFences = [4]
输出：-1
解释：可以证明无法通过移除栅栏形成正方形田地。

提示：

3 <= m, n <= 10⁹
1 <= hFences.length, vFences.length <= 600
1 < hFences[i] < m
1 < vFences[i] < n
hFences 和 vFences 中的元素是唯一的。

lightbulb

解题思路

方法一：枚举

我们可以枚举 $\textit{hFences}$ 中的任意两条水平栅栏 $a$ 和 $b$ ，计算 $a$ 和 $b$ 之间的距离 $d$ ，记录在哈希表 $hs$ 中，然后枚举 $\textit{vFences}$ 中的任意两条垂直栅栏 $c$ 和 $d$ ，计算 $c$ 和 $d$ 之间的距离 $d$ ，记录在哈希表 $vs$ 中，最后遍历哈希表 $hs$ ，如果 $hs$ 中的某个距离 $d$ 在哈希表 $vs$ 中也存在，那么说明存在一个正方形田地，其边长为 $d$ ，面积为 $d^2$ ，我们只需要取最大的 $d$ ，求 $d^2 \bmod 10^9 + 7$ 即可。

时间复杂度 $O(h^2 + v^2)$ ，空间复杂度 $O(h^2 + v^2)$ 。其中 $h$ 和 $v$ 分别是 $\textit{hFences}$ 和 $\textit{vFences}$ 的长度。

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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

speed

复杂度分析

指标	值
时间	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.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

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