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Solution: Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts

seanpgallivan
Fledgling software developer; the struggle is a Rational Approximation.
Updated on ・4 min read

This is part of a series of Leetcode solution explanations (index). If you liked this solution or found it useful, please like this post and/or upvote my solution post on Leetcode's forums.


Leetcode Problem #1465 (Medium): Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts


Description:


(Jump to: Solution Idea || Code: JavaScript | Python | Java | C++)

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.


Examples:

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.
Visual: Example 1 Visual
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.
Visual: Example 2 Visual
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.

Idea:


(Jump to: Problem Description || Code: JavaScript | Python | Java | C++)

The trick to this problem is realizing that if the horizontal slices and vertical slices are perpendicular, then all vertical slices cross all horizontal slices. This means that we just need to find the largest of each, and the cross-section should be the largest slice.

To find the largest slice of each, we need to first sort the horizontal cuts (hc) and vertical cuts (vc), then iterate through both sets and keep track of the maximum difference found between two consecutive cuts (maxh, maxv). We need to not forget to include the two end cuts, which are found using 0 and h/w, as well.

Once we have the largest difference for both, we can just return the product of these two numbers, modulo 1e9+7.

  • Time Complexity: O(N * log(N) + M * log(M)) where N is the length of hc and M is the length of vc
  • Space Complexity: O(1)

Javascript Code:


(Jump to: Problem Description || Solution Idea)

var maxArea = function(h, w, hc, vc) {
    hc.sort((a,b) => a - b)
    vc.sort((a,b) => a - b)
    let maxh = Math.max(hc[0], h - hc[hc.length-1]),
        maxv = Math.max(vc[0], w - vc[vc.length-1])
    for (let i = 1; i < hc.length; i++)
        maxh = Math.max(maxh, hc[i] - hc[i-1])
    for (let i = 1; i < vc.length; i++)
        maxv = Math.max(maxv, vc[i] - vc[i-1])
    return BigInt(maxh) * BigInt(maxv) % 1000000007n
};
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Python Code:


(Jump to: Problem Description || Solution Idea)

class Solution:
    def maxArea(self, h: int, w: int, hc: List[int], vc: List[int]) -> int:
        hc.sort()
        vc.sort()
        maxh, maxv = max(hc[0], h - hc[-1]), max(vc[0], w - vc[-1])
        for i in range(1, len(hc)):
            maxh = max(maxh, hc[i] - hc[i-1])
        for i in range(1, len(vc)):
            maxv = max(maxv, vc[i] - vc[i-1])
        return (maxh * maxv) % 1000000007
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Java Code:


(Jump to: Problem Description || Solution Idea)

class Solution {
    public int maxArea(int h, int w, int[] hc, int[] vc) {
        Arrays.sort(hc);
        Arrays.sort(vc);
        int maxh = Math.max(hc[0], h - hc[hc.length-1]),
            maxv = Math.max(vc[0], w - vc[vc.length-1]);
        for (int i = 1; i < hc.length; i++)
            maxh = Math.max(maxh, hc[i] - hc[i-1]);
        for (int i = 1; i < vc.length; i++)
            maxv = Math.max(maxv, vc[i] - vc[i-1]);
        return (int)((long)maxh * maxv % 1000000007);
    }
}
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C++ Code:


(Jump to: Problem Description || Solution Idea)

class Solution {
public:
    int maxArea(int h, int w, vector<int>& hc, vector<int>& vc) {
        sort(hc.begin(), hc.end());
        sort(vc.begin(), vc.end());
        int maxh = max(hc[0], h - hc.back()),
            maxv = max(vc[0], w - vc.back());
        for (int i = 1; i < hc.size(); i++)
            maxh = max(maxh, hc[i] - hc[i-1]);
        for (int i = 1; i < vc.size(); i++)
            maxv = max(maxv, vc[i] - vc[i-1]);
        return (int)((long)maxh * maxv % 1000000007);
    }
};
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