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. 🚀
Earlier in this series: LeetCode #1710: Maximum Units on a Truck.
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, andverticalCuts[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
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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 🧑💻
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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