887. Super Egg Drop
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-1eggs remaining and we need to search forx-1floors belowx. - If the egg does not break, we still have
ieggs remaining and we need to search forj-xfloors abovex.
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).
:
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;
}
}Code Editor (Testing phase)
Improve Your Solution
Use the editor below to refine the provided solution. Select a programming language and try the following:
- Add import statement if required.
- Optimize the code for better time or space complexity.
- Add test cases to validate edge cases and common scenarios.
- Handle error conditions or invalid inputs gracefully.
- Experiment with alternative approaches to deepen your understanding.
Click "Run Code" to execute your solution and view the output. If errors occur, check the line numbers and debug accordingly. Resize the editor by dragging its bottom edge.