### Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts LeetCode Solution - The Coding Shala

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In this post, we will learn how to solve LeetCode's Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts Problem and will implement its solution in Java.

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

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

Practice this problem on LeetCode.

## LeetCode - Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts Java Solution

Approach 1

Here we need to find the maximum gap between two cuts both horizontally and vertically then return their product. The steps are as below.

Step 1. Sort both the arrays.

Step 2. Find the maximum difference between the two cuts.

Step 3. Return the product of maximum differences with modulo 1000000007

Java Program:

```class Solution {
public int maxArea(int h, int w, int[] horizontalCuts, int[] verticalCuts) {
Arrays.sort(horizontalCuts);
Arrays.sort(verticalCuts);

long maxH = 0;
long maxV = 0;

// maximum horizontal difference
maxH = Math.max(horizontalCuts[0] - 0, h - horizontalCuts[horizontalCuts.length-1]);
for (int i = 1; i < horizontalCuts.length; i++) {
maxH = Math.max(maxH, horizontalCuts[i] - horizontalCuts[i-1]);
}

// maximum veritcal difference
maxV = Math.max(verticalCuts[0] - 0, w - verticalCuts[verticalCuts.length-1]);
for (int i = 1; i < verticalCuts.length; i++) {
maxV = Math.max(maxV, verticalCuts[i] - verticalCuts[i-1]);
}

long res = (maxH * maxV) % 1000000007;

return (int) res;
}
}
```

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