2022-07-02 Daily-Challenge

Today I have done leetcode's July LeetCoding Challenge with cpp.

July LeetCoding Challenge 2

Description

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.

 

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.

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.

Example 3:

Input: h = 5, w = 4, horizontalCuts = [3], verticalCuts = [3]
Output: 9

 

Constraints:

  • 2 <= h, w <= 109
  • 1 <= horizontalCuts.length <= min(h - 1, 105)
  • 1 <= verticalCuts.length <= min(w - 1, 105)
  • 1 <= horizontalCuts[i] < h
  • 1 <= verticalCuts[i] < w
  • All the elements in horizontalCuts are distinct.
  • All the elements in verticalCuts are distinct.

Solution

const int MOD = 1e9 + 7;
class Solution {
public:
  int maxArea(int h, int w, vector<int>& horizontalCuts, vector<int>& verticalCuts) {
    int maxH = 0;
    int maxW = 0;
    int curH = 0;
    int curW = 0;
    sort(horizontalCuts.begin(), horizontalCuts.end());
    sort(verticalCuts.begin(), verticalCuts.end());
    for(auto h : horizontalCuts) {
      maxH = max(maxH, h - curH);
      curH = h;
    }
    maxH = max(maxH, h - curH);
    for(auto w : verticalCuts) {
      maxW = max(maxW, w - curW);
      curW = w;
    }
    maxW = max(maxW, w - curW);
    return 1LL * maxH * maxW % MOD;
  }
};

// Accepted
// 56/56 cases passed (104 ms)
// Your runtime beats 48.69 % of cpp submissions
// Your memory usage beats 76.96 % of cpp submissions (32.2 MB)