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In this approach, we use dynamic programming (DP) to solve the problem efficiently. The idea is to set up a DP table where dp[k][n] represents the minimum number of moves required to find the critical floor with k eggs and n floors.
We can use binary search to determine which floor to drop the egg from in each step, which helps to optimize the number of trials needed. The recurrence relation can be formulated as:
dp[k][n] = 1 + min(max(dp[k-1][x-1], dp[k][n-x])) for all 1 <= x <= n.
This relation means that for each floor x, you either continue with the remaining floors if the egg doesn't break, or with one fewer egg if it does break. The goal is to find the minimal worst-case scenario.
Time Complexity: O(k * n * log n) because each fill operation of the DP table involves a binary search over the number of floors.
Space Complexity: O(k * n) because of the DP table holding computed results.
1class Solution:
2 def superEggDrop(self, k: int, n: int) -> int:
3 dp = [[0] *
In Python, the 2D list dp is used to represent the DP table. The method starts by initializing base cases and iteratively computes the minimum trials needed using a binary search approach for every possible egg-floor combination. This prevents the naive way of checking each floor sequentially. The test block calls the superEggDrop method with 3 eggs and 14 floors.
Here, we use a mathematical approach combined with dynamic programming. Instead of trying to calculate the minimum number of moves directly, we use a theoretical approach that considers the problem of breaking down moves into binary forms.
We change the problem into one of maximizing the number of floors we can test with a given number of drops. We can calculate m moves where the result is T[k][m] = T[k-1][m-1] + T[k][m-1] + 1. If T[k][m] ≥ n, we found the minimum m.
Time Complexity: O(k * n), because we directly shift through combinations.
Space Complexity: O(n) since we're using array-based exploratory floor encoding.
public class Solution {
public int SuperEggDrop(int k, int n) {
int[] dp = new int[n + 1];
for (int i = 0; i <= n; i++) dp[i] = i;
for (int i = 2; i <= k; i++) {
int[] dpNew = new int[n + 1];
int x = 1;
for (int j = 1; j <= n; j++) {
while (x < j && Math.Max(dp[x - 1], dpNew[j - x]) > Math.Max(dp[x], dpNew[j - x - 1])) {
x++;
}
dpNew[j] = 1 + Math.Max(dp[x - 1], dpNew[j - x]);
}
dp = dpNew;
}
return dp[n];
}
public static void Main(string[] args) {
Solution sol = new Solution();
Console.WriteLine(sol.SuperEggDrop(2, 6)); // Output: 3
}
}C# echoes the consistent premise of all compiled-managed scenarios leveraged here, fluctuating around each optimal O(n) adoption of conditions, eliminating overhead and unnecessary exploratory redundancy exploring maximal trends.