887. Super Egg Drop

You are given K eggs, and you have access to a building with N floors from 1 to N. 

Each egg is identical in function, and if an egg breaks, you cannot drop it again.

You know that there exists a floor F with 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 egg (if you have an unbroken one) and drop it from any floor X (with 1 <= X <= N). 

Your goal is to know with certainty what the value of F is.

What is the minimum number of moves that you need to know with certainty what F is, regardless of the initial value of F?

 

Example 1:

Input: K = 1, N = 2
Output: 2
Explanation: 
Drop the egg from floor 1.  If it breaks, we know with certainty that F = 0.
Otherwise, drop the egg from floor 2.  If it breaks, we know with certainty that F = 1.
If it didn't break, then we know with certainty F = 2.
Hence, we needed 2 moves in the worst case to know what F is with certainty.

Example 2:

Input: K = 2, N = 6
Output: 3

Example 3:

Input: K = 3, N = 14
Output: 4

 

Note:

1. 1 <= K <= 100
2. 1 <= N <= 10000

 

Approach #1: DP. [C++][TLE][O(K*N^2)]

class Solution {
public:
    int superEggDrop(int K, int N) {
        int c = 0;
        vector<vector<int>> dp(K+1, vector<int>(N+1, 0));
        for (int i = 1; i <= N; ++i) dp[1][i] = i;
        for (int i = 2; i <= K; ++i) {
            for (int j = 1; j <= N; ++j) {
                dp[i][j] = INT_MAX;
                for (int k = 1; k <= j; ++k) {
                    c = 1 + max(dp[i-1][k-1], dp[i][j-k]);
                    if (c < dp[i][j])
                        dp[i][j] = c;
                }
            }
        }
        return dp[K][N];
    }
};

　　

 

Approach #2: DP. [Java]

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 k = 1; k <= K; ++k) 
                dp[m][k] = dp[m-1][k-1] + dp[m-1][k] + 1;
        }
        
        return m;
    }
}

　　

Analysis:

Firstly, if we have K eggs and s steps to detect a buliding with Q(k, s) floors.

Secondly, we use 1 egg and 1 step to detect one floor,

if egg break, we can use (k-1) eggs and (s-1) to detect with Q(k-1, s-1),

if egg isn't broken, we can use k eggs and (s-1) step to detech with Q(k, s-1),

So, Q(k,s) = 1 + Q(k, s-1) + Q(k-1, s-1);

 

dp[i] is max floors we can use i eggs and s step to detect.

 

Reference:

https://leetcode.com/problems/super-egg-drop/discuss/159508/easy-to-understand