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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.
1#include <iostream>
2#include <vector>
3#include <climits>
4using namespace std;
5
6int superEggDrop(int k, int n) {
7 vector<vector<int>> dp(k + 1, vector<int>(n + 1, INT_MAX));
8 for (int i = 0; i <= k; ++i) {
9 dp[i][0] = 0;
10 if (i > 0) dp[i][1] = 1;
11 }
12 for (int j = 0; j <= n; ++j) {
13 dp[1][j] = j;
14 }
for (int i = 2; i <= k; ++i) {
for (int j = 2; j <= n; ++j) {
int low = 1, high = j, mid;
while (low <= high) {
mid = (low + high) / 2;
int broken = dp[i - 1][mid - 1];
int not_broken = dp[i][j - mid];
int cost = 1 + max(broken, not_broken);
dp[i][j] = min(dp[i][j], cost);
if (broken > not_broken) {
high = mid - 1;
} else {
low = mid + 1;
}
}
}
}
return dp[k][n];
}
int main() {
cout << superEggDrop(3, 14) << endl; // Output: 4
return 0;
}This C++ solution uses a vector to create and store the DP table with k + 1 eggs and n + 1 floors. The vector is initialized for all base cases.
For each k eggs and n floors, it calculates the minimum number of drops using binary search to optimize the solution and prepare the DP table with the proposed relationship. The main function tests the function for 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.
In this C solution, we use a similar 1D DP concept with two arrays that take shifts for operations. It reduces the grains of move possibilities (instead of examining choice permutations and trays them shallow iterations). For each egg beyond the first, the new entries find a balance between fewer and extra floors, inheriting an O(n) step management budget.