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

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

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. 🚀

Problem Statement

Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts 🍰

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 🍰

Code in JS 🧑‍💻

/**
 * @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)

You can find me on the web 🌍

• Github
• LinkedIn
• Twitter

Add your solution or approach in the comments below.

Show your love by Sharing the blog. 🤗

“The best way to predict the future is to create it.”

~ Peter Drucker