LeetCode 1884: Egg Drop With 2 Eggs and N Floors Solution

Master LeetCode problem 1884 (Egg Drop With 2 Eggs and N Floors), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1884. Egg Drop With 2 Eggs and N Floors

Medium

Math Dynamic Programming

Problem Explanation

Explanation

To solve this problem efficiently, we can use a dynamic programming approach. We will define a 2D array dp where dp[i][j] represents the minimum number of moves needed to determine the value of f with i floors and j eggs.

At each floor x, we have two choices: either the egg breaks or it doesn't. If the egg breaks, we need to continue with one less egg and search in the lower floors. If the egg doesn't break, we can continue with the same number of eggs and search in the upper floors. We choose the worst-case scenario of these two choices to get the minimum number of moves.

We can start by initializing the base cases where if there is only 1 floor or 0 floors, the number of moves required is equal to the number of floors. If there is only 1 egg, the number of moves required will be equal to the number of floors.

Then, we can iterate over each floor and each remaining egg to fill in the dp array. The final answer will be dp[n][2].

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

Solutions

Solution Code

class Solution {
    public int twoEggDrop(int n) {
        int[][] dp = new int[n + 1][3];
        
        for (int i = 0; i <= n; i++) {
            dp[i][1] = i;
            dp[i][2] = i;
        }
        
        for (int i = 2; i <= n; i++) {
            for (int j = 2; j <= 2; j++) {
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = 1; k <= i; k++) {
                    dp[i][j] = Math.min(dp[i][j], 1 + Math.max(dp[k - 1][j - 1], dp[i - k][j]));
                }
            }
        }
        
        return dp[n][2];
    }
}

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