```
class Solution {
public int superEggDrop(int m, int n) {
m += 1;
n += 1;
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
dp[i][0] = 0;
dp[i][1] = 1;
}
for (int i = 0; i < n; i++) {
dp[0][i] = 0;
dp[1][i] = i;
}
for (int i = 2; i < m; i++) {
for (int j = 2; j < n; j++) {
int low = 1, high = j;
int ans = Integer.MAX_VALUE;
while (low <= high) {
int mid = (low + high) / 2;
int broken = dp[i - 1][mid - 1]; // egg breaks
int notBroken = dp[i][j - mid]; // egg doesn't break
int temp = 1 + Math.max(broken, notBroken);
if (broken < notBroken) {
low = mid + 1;
} else {
high = mid - 1;
}
ans = Math.min(ans, temp);
}
dp[i][j] = ans;
}
}
return dp[m - 1][n - 1];
}
}
```
explain this code
视频信息
答案文本
视频字幕
The Super Egg Drop problem is a classic dynamic programming challenge. Given m eggs and n floors, we need to find the minimum number of drops required to determine the critical floor where eggs start breaking. This requires an optimal strategy to handle the worst-case scenario.
The solution uses dynamic programming with a two-dimensional table. dp[i][j] represents the minimum number of drops needed with i eggs and j floors. We establish base cases: zero floors need zero drops, one floor needs one drop, and with one egg we must check floors sequentially. The recurrence relation considers dropping from each possible floor k and takes the minimum over all choices.
The code optimizes the solution using binary search. Instead of checking all possible floors k from 1 to j, it uses the fact that the broken case increases with k while the not broken case decreases. Binary search finds the optimal k where these two values balance, reducing time complexity from O of m n squared to O of m n log n.
The code implementation starts by adjusting m and n by adding 1 for one-based indexing. The main nested loops iterate through eggs from 2 to m minus 1 and floors from 2 to n minus 1. For each cell, binary search finds the optimal floor by comparing broken and not broken cases, updating the search bounds accordingly, and storing the minimum result in the dp table.
To summarize what we have learned: The Super Egg Drop problem demonstrates the power of dynamic programming combined with binary search optimization. The solution efficiently finds the minimum number of drops needed by building a table of subproblems and using binary search to find optimal drop floors. This technique reduces time complexity significantly and provides a robust framework for solving similar optimization challenges.