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.

How to Solve

Consider this problem in a different way: dp[M][K] means that, given K eggs and M moves, what is the maximum number of floor that we can check.

The dp equation is:

dp[m][k] = dp[m - 1][k - 1] + dp[m - 1][k] + 1,

which means we take 1 move to a floor, if egg breaks, then we can check dp[m - 1][k - 1] floors. if egg doesn’t breaks, then we can check dp[m - 1][k] floors.

dp[m][k] is the number of combinations, and it increases exponentially to N

Test Cases

Input:

(int) k = 1
(int) n = 2

Output:

(int) 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.

Input:

(int) k = 2
(int) n = 6

Output:

(int) 3

Input:

(int) k = 3
(int) n = 14

Output:

(int) 4

Solution

C++ Java Python

class Solution {
public:
    int superEggDrop(int K, int N) {
        vector<vector<int>> dp(N + 1, vector<int>(K + 1, 0));
        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;
    }
};

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;
    }
}

class Solution:
    def superEggDrop(self, K: int, N: int) -> int:
        dp = [[0] * (K + 1) for i in range(N + 1)]
        for m in range(1, N + 1):
            for k in range(1, K + 1):
                dp[m][k] = dp[m - 1][k - 1] + dp[m - 1][k] + 1
            if dp[m][K] >= N: return m