Formatted question description: https://leetcode.ca/all/887.html

887. Super Egg Drop (Hard)

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

TLE Solution

Assume we choose to throw the egg at floor i:

  • If the egg breaks, we continue throwing between floors [1, i - 1], with one less egg available
  • If the egg doesn’t break, we continue throwing between floors [i + 1, N], with the same number of eggs. In this way, for whichever floors region [m, n] (1 <= m <= n <= N), we can regard floor m - 1 is safe while floor n + 1 is not safe. So the 2nd case above is analogous to throwing between floors [1, N - i].

Denote f(K, N) as the result. The 1st case corresponds to f(K - 1, i - 1), and the 2nd case f(K, N - i). We have to pick the max of them to ensure certainty. Denote g(K, N, i) as this max.

Thus, f(K, N) = 1 + min{ g(K, N, i) | 1 <= i <= N }. This is the equation of DP.

In the worse case we need to visit all the combinations of k and n (k in [1, K], n in [1, N]), thus time complexity is O(KN).

// OJ: https://leetcode.com/problems/super-egg-drop/

// Time: O(KN)
// Space: O(KN)
struct pair_hash {
    template <class T1, class T2>
    std::size_t operator () (const std::pair<T1,T2> &p) const {
        return std::hash<T1>{}(p.first * 10000 + p.second);
    }
};
class Solution {
private:
    unordered_map<pair<int, int>, int, pair_hash> m;
public:
    int superEggDrop(int K, int N) {
        if (!K || !N) return 0;
        if (K == 1) return N;
        auto p = make_pair(K, N);
        if (m.find(p) != m.end()) return m[p];
        int val = INT_MAX;
        int prev = INT_MAX;
        for (int i = (N + 1) / 2; i >= 1; --i) {
            int v = max(superEggDrop(K - 1, i - 1), superEggDrop(K, N - i));
            if (v > prev) break;
            prev = val;
            val = min(val, v);
            
        }
        return m[p] = 1 + val;
    }
};

Solution 1.

Denote f(K, S) as the max number of floors that is solvable given K eggs and S steps.

After I throw an egg:

  • If the egg is broken, I should continue throw the eggs within lower floors. The max number of lower floors I can handle is f(K - 1, S - 1).
  • If the egg is not broken, I should continue throw the eggs within upper floors. The max number of upper floors I can handle is f(K, S - 1).

So the max total number of floors I can handle is 1 plus the result of the above two cases, i.e. f(K, S) = 1 + f(K - 1, S - 1) + f(K, S - 1).

// OJ: https://leetcode.com/problems/super-egg-drop/

// Time: O(SK) where S is the result.
// Space: O(K)
// Ref: https://leetcode.com/problems/super-egg-drop/discuss/159508/easy-to-understand
class Solution {
public:
    int superEggDrop(int K, int N) {
        int step = 0;
        vector<int> dp(K + 1);
        for (; dp[K] < N; ++step) {
            for (int k = K; k > 0; --k) {
                dp[k] += 1 + dp[k - 1];
            }
        }
        return step;
    }
};

Java

class Solution {
    public int superEggDrop(int K, int N) {
        int[][] dp = new int[N + 1][K + 1];
        for (int i = 0; i <= N; i++) {
            for (int j = 0; j <= K; j++)
                dp[i][j] = i;
        }
        dp[1][0] = 0;
        for (int i = 1; i <= K; i++)
            dp[1][i] = 1;
        for (int i = 0; i <= N; i++) {
            dp[i][0] = 0;
            dp[i][1] = i;
        }
        for (int i = 2; i <= N; i++) {
            for (int j = 2; j <= K; j++) {
                int low = 1, high = i;
                while (low < high) {
                    int mid = (high - low + 1) / 2 + low;
                    int remain1 = dp[mid - 1][j - 1];
                    int remain2 = dp[i - mid][j];
                    if (remain1 > remain2)
                        high = mid - 1;
                    else
                        low = mid;
                }
                dp[i][j] = Math.max(dp[low - 1][j - 1], dp[i - low][j]) + 1;
            }
        }
        return dp[N][K];
    }
}

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