算法: 超级落蛋计算第一次蛋碎的楼层887. Super Egg Drop

Posted 2022-01-06 AI架构师易筋

887. Super Egg Drop

You are given k identical eggs and you have access to a building with n floors labeled from 1 to n.

You know that there exists a floor f where 0 <= f <= n such that any egg dropped at a floor higher than f will break, and any egg dropped at or below floor f will not break.

Each move, you may take an unbroken egg and drop it from any floor x (where 1 <= x <= n). If the egg breaks, you can no longer use it. However, if the egg does not break, you may reuse it in future moves.

Return the minimum number of moves that you need to determine with certainty what the value of f is.

Example 1:

Input: k = 1, n = 2
Output: 2
Explanation: 
Drop the egg from floor 1. If it breaks, we know that f = 0.
Otherwise, drop the egg from floor 2. If it breaks, we know that f = 1.
If it does not break, then we know f = 2.
Hence, we need at minimum 2 moves to determine with certainty what the value of f is.

Example 2:

Input: k = 2, n = 6
Output: 3

Example 3:

Input: k = 3, n = 14
Output: 4

Constraints:

• 1 <= k <= 100
• 1 <= n <= 104

动态规划解法

掉落鸡蛋是一个非常经典的问题。
有些人可能会想出想法O(KN^2)
在那里dp[k][n] = 1 + max(dp[k - 1][i - 1], dp[k][n - i])，因为我在1 ... n。
然而，这个想法是非常暴力的，因为你要检查所有的可能性。

所以我以不同的方式考虑这个问题：
dp[m][k]意味着，给定k鸡蛋和m移动，
我们可以检查的最大楼层数是多少。

dp 方程是：
dp[m][k] = dp[m - 1][k - 1] + dp[m - 1][k] + 1，
这意味着我们移动到一个楼层，

1. 如果鸡蛋破了，那么我们可以检查dp[m - 1][k - 1]楼层。
2. 如果鸡蛋没有破，那么我们可以检查dp[m - 1][k]地板。

dp[m][k] 是组合的数量，它呈指数增长 n

复杂
对于时间， 空间复杂度O(nk)，时间复杂度O(klogn)，

class Solution 
    public int superEggDrop(int k, int n) 
        int[][] dp = new int[n + 1][k + 1];
        int m = 0;
        while (dp[m][k] < n) 
            ++m;
            for (int i = 1; i <= k; i++) 
                dp[m][i] = dp[m - 1][i - 1] + dp[m - 1][i] + 1;
            
        
        
        return m;
    

优化空间复杂度O(k)

class Solution 
    public int superEggDrop(int k, int n) 
        int[] dp = new int[k + 1];
        int m = 0;
        for (; dp[k] < n; ++m) 
            for (int i = k; i > 0; --i) 
                dp[i] += dp[i - 1] + 1;
            
        
        return m;
    

参考

https://leetcode.com/problems/super-egg-drop/discuss/158974/C%2B%2BJavaPython-2D-and-1D-DP-O(KlogN)