## LeetCode Question - 1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts π°

### About the Series

Problem-solving is a key skill set for any tech-related stuff you might be working on.

When it comes to developers it's one of the most crucial skills which is needed in almost any day-to-day code you might be writing.

So, this series of blogs is all about practicing Daily LeetCode Challenges & Problem-solving. π

### Problem Statement

**Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts π°**

You are given a rectangular cake of size

`h x 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, and`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 large number, return this`modulo 109 + 7`

### Video Explanation

### Solution

**Algorithm**

- Push
`0`

and`h`

in the`horizontalCuts`

. - Sort the
`horizontalCuts`

array in Ascending Order. - Take the maximum difference between the cuts by iterating on the list of
`horizontalCuts`

- Push
`0`

and`h`

in the`horizontalCuts`

. - Sort the
`horizontalCuts`

array in Ascending Order. - Take the maximum difference between the cuts by iterating on the list of
`horizontalCuts`

- Multiply both the max values and return the modulo
`10^9 + 7`

- Now, we got the largest piece of Cake π°

**Code in JS π§βπ»**

```
/**
* @param {number} h
* @param {number} w
* @param {number[]} horizontalCuts
* @param {number[]} verticalCuts
* @return {number}
*/
var maxArea = function (h, w, horizontalCuts, verticalCuts) {
horizontalCuts.push(0, h);
horizontalCuts.sort((a, b) => a - b);
var maxH = 0;
verticalCuts.push(0, w);
verticalCuts.sort((a, b) => a - b);
var maxW = 0;
for (var i = 1; i < horizontalCuts.length; i++) {
maxH = Math.max(maxH, horizontalCuts[i] - horizontalCuts[i - 1]);
}
for (var j = 1; j < verticalCuts.length; j++) {
maxW = Math.max(maxW, verticalCuts[j] - verticalCuts[j - 1]);
}
return (BigInt(maxH) * BigInt(maxW)) % BigInt(1e9 + 7);
};
```

**Time Complexity : O(nlogn)**

**Space Complexity: O(1)**

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