LeetCode 887: Super Egg Drop Solution

Problem Explanation

Explanation:

We can solve this problem using dynamic programming. Let dp[i][j] represent the minimum number of moves needed to determine the value of f with i eggs and j floors.

For each move, we can either choose to drop an egg from a certain floor x:

If the egg breaks, we have i-1 eggs remaining and we need to search for x-1 floors below x.
If the egg does not break, we still have i eggs remaining and we need to search for j-x floors above x.

The optimal strategy is to minimize the worst-case scenario, so we try all possible floor choices from 1 to j and take the maximum of the two situations mentioned above for each choice.

The time complexity of this dynamic programming solution is O(k * n^2) and the space complexity is O(k * n).

Solution Code

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

Frequently Asked Questions

How to solve LeetCode 887 (Super Egg Drop)?

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What is the time complexity of LeetCode 887 (Super Egg Drop)?

The time complexity for LeetCode 887 (Super Egg Drop) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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887. Super Egg Drop