You are given K eggs, and you have access to a building with N floors from 1 to N.
Each egg is identical in function, and if an egg breaks, you cannot drop it again.
You know that there exists a floor F with 0 <= F <= N such that any egg dropped at a floor higher than F will break, and any egg dropped at or below floor F will not break.
Each move, you may take an egg (if you have an unbroken one) and drop it from any floor X (with 1 <= X <= N).
Your goal is to know with certainty what the value of F is.
What is the minimum number of moves that you need to know with certainty what F is, regardless of the initial value of F?
Example 1:
Input: K = 1, N = 2
Output: 2
Explanation:
Drop the egg from floor 1. If it breaks, we know with certainty that F = 0.
Otherwise, drop the egg from floor 2. If it breaks, we know with certainty that F = 1.
If it didn't break, then we know with certainty F = 2.
Hence, we needed 2 moves in the worst case to know what F is with certainty.
Example 2:
Input: K = 2, N = 6
Output: 3
Example 3:
Input: K = 3, N = 14
Output: 4
Note:
1 <= K <= 100
1 <= N <= 10000
class Solution {
// O(K*N*logN)
public int superEggDrop(int K, int N) {
// dp[i][j] -> min tests required with i eggs and j floors
int[][] dp = new int[K + 1][N + 1];
// Base cases
// We need 1 trial for 1 floor and 0 trials for 0 floors
for (int i = 1; i <= K; i++) {
dp[i][0] = 0;
dp[i][1] = 1;
}
// We always need j trials with 1 egg and j floors.
for (int j = 1; j <= N; j++)
dp[1][j] = j;
for (int i = 2; i <= K; i++) {
for (int j = 2; j <= N; j++) {
// O(N) Linear Search
// dp[i][j] = Integer.MAX_VALUE;
// checking the best we can get out of worst case for each floor
// for (int x = 1; x <= j; x++) {
// If the egg breaks then we need to consider i-1 eggs and x-1 floors
// If the egg does not breaks then we need to consider i eggs and j-x floors
// int res = 1 + Math.max(dp[i - 1][x - 1], dp[i][j - x]);
// dp[i][j] = Math.min(dp[i][j], res);
// }
// O(logN) Binary Search
// We don't need to go over all floors from 1 to j
// As for a fixed j, dp[i][j] goes up as j increases.
// This means dp[i-1][x-1] will increase and dp[i][j-x] will decrease as x goes from 1 to j.
// The optimal value of x will be the middle point where the two meet().
// So to get the optimal x value for dp[i][j], we can do a binary search for x from 1 to j.
int low = 1, high = j;
int result = Integer.MAX_VALUE;
while (low < high) {
int mid = low + (high - low) / 2;
int left = dp[i - 1][mid - 1];
int right = dp[i][j - mid];
result = Math.min(result, Math.max(left, right) + 1);
if (left == right)
break;
else if (left < right)
low = mid + 1;
else
high = mid;
}
dp[i][j] = result;
}
}
return dp[K][N];
}
}