1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts

# 1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts#

## 题目 #

Given a rectangular cake with height `h` and width `w`, and two arrays of integers `horizontalCuts` and `verticalCuts` where `horizontalCuts[i]` is the distance from the top of the rectangular cake to the `ith` horizontal cut and similarly, `verticalCuts[j]` is the distance from the left of the rectangular cake to the `jth` vertical cut.

Return the maximum area of a piece of cake after you cut at each horizontal and vertical position provided in the arrays `horizontalCuts` and `verticalCuts`. Since the answer can be a huge number, return this modulo 10^9 + 7.

Example 1:

``````Input: h = 5, w = 4, horizontalCuts = [1,2,4], verticalCuts = [1,3]
Output: 4
Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green piece of cake has the maximum area.

``````

Example 2:

``````Input: h = 5, w = 4, horizontalCuts = [3,1], verticalCuts = [1]
Output: 6
Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green and yellow pieces of cake have the maximum area.

``````

Example 3:

``````Input: h = 5, w = 4, horizontalCuts = [3], verticalCuts = [3]
Output: 9

``````

Constraints:

• `2 <= h, w <= 10^9`
• `1 <= horizontalCuts.length < min(h, 10^5)`
• `1 <= verticalCuts.length < min(w, 10^5)`
• `1 <= horizontalCuts[i] < h`
• `1 <= verticalCuts[i] < w`
• It is guaranteed that all elements in `horizontalCuts` are distinct.
• It is guaranteed that all elements in `verticalCuts` are distinct.

## 解题思路 #

• 读完题比较容易想到解题思路。找到水平切口最大的差值和竖直切口最大的差值，这 4 条边构成的矩形即为最大矩形。不过有特殊情况需要判断，切口除了题目给的切口坐标以外，默认还有 4 个切口，即蛋糕原始的 4 条边。如下图二，最大的矩形其实在切口之外。所以找水平切口最大差值和竖直切口最大差值需要考虑到蛋糕原始的 4 条边。

## 代码 #

``````package leetcode

import "sort"

func maxArea(h int, w int, hcuts []int, vcuts []int) int {
sort.Ints(hcuts)
sort.Ints(vcuts)
maxw, maxl := hcuts[0], vcuts[0]
for i, c := range hcuts[1:] {
if c-hcuts[i] > maxw {
maxw = c - hcuts[i]
}
}
if h-hcuts[len(hcuts)-1] > maxw {
maxw = h - hcuts[len(hcuts)-1]
}
for i, c := range vcuts[1:] {
if c-vcuts[i] > maxl {
maxl = c - vcuts[i]
}
}
if w-vcuts[len(vcuts)-1] > maxl {
maxl = w - vcuts[len(vcuts)-1]
}
return (maxw * maxl) % (1000000007)
}
``````

Apr 8, 2023