You are also given an integer dist. You can only climb to the next highest rung if the distance between where you are currently at (the floor or on a rung) and the next rung is at most dist. You are able to insert rungs at any positive integer height if a rung is not already there.

Return the minimum number of rungs that must be added to the ladder in order for you to climb to the last rung.

Example 1:

Input: rungs = [1,3,5,10], dist = 2
Output: 2
Explanation:
You currently cannot reach the last rung.
Add rungs at heights 7 and 8 to climb this ladder. 
The ladder will now have rungs at [1,3,5,7,8,10].

Example 2:

Input: rungs = [3,6,8,10], dist = 3
Output: 0
Explanation:
This ladder can be climbed without adding additional rungs.

Example 3:

Input: rungs = [3,4,6,7], dist = 2
Output: 1
Explanation:
You currently cannot reach the first rung from the ground.
Add a rung at height 1 to climb this ladder.
The ladder will now have rungs at [1,3,4,6,7].

Constraints:

• 1 <= rungs.length <= 105
• 1 <= rungs[i] <= 109
• 1 <= dist <= 109
• rungs is strictly increasing.

Solution: Math

Check two consecutive rungs, if their diff is > dist, we need insert (diff – 1) / dist rungs in between.
ex1 5 -> 11, diff = 6, dist = 2, (diff – 1) / dist = (6 – 1) / 2 = 2. => 5, 7, 9, 11.
ex2 0 -> 3, diff = 3, dist = 1, (diff – 1) / dist = (3 – 1) / 1 = 2 => 0, 1, 2, 3

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int addRungs(vector<int>& rungs, int dist) {
    int ans = 0;
    for (size_t i = 0; i < rungs.size(); ++i)
      ans += (rungs[i] - (i ? rungs[i - 1] : 0) - 1) / dist;    
    return ans;
  }
};

# Author: Huahua
class Solution:
  def addRungs(self, rungs: List[int], dist: int) -> int:
    return sum((r - 1 - (rungs[i - 1] if i else 0)) // dist for i, r in enumerate(rungs))

The post 花花酱 LeetCode 1936. Add Minimum Number of Rungs appeared first on Huahua's Tech Road.

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.

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 <= 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.

Solution: Geometry

Find the max gap between vertical cuts mx and max gap between horizontal cuts my. ans = mx * my

Time complexity: O(nlogn)
Space complexity: O(1) if sort in place otherweise O(n)

// Author: Huahua
class Solution {
public:
  int maxArea(int h, int w, vector<int>& horizontalCuts, vector<int>& verticalCuts) {
    constexpr int kMod = 1e9 + 7;
    sort(begin(verticalCuts), end(verticalCuts));
    sort(begin(horizontalCuts), end(horizontalCuts));
    int mx = max(verticalCuts[0], w - verticalCuts.back());
    int my = max(horizontalCuts[0], h - horizontalCuts.back());
    for (int i = 1; i < verticalCuts.size(); ++i)
      mx = max(mx, verticalCuts[i] - verticalCuts[i - 1]);
    for (int i = 1; i < horizontalCuts.size(); ++i)
      my = max(my, horizontalCuts[i] - horizontalCuts[i - 1]);               
    return static_cast<long>(mx) * my % kMod;
  }
};

The post 花花酱 LeetCode 1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts appeared first on Huahua's Tech Road.