Algorithm
Problem Name: 887. Super Egg Drop
You are given k identical eggs and you have access to a building with n floors labeled from 1 to n.
You know that there exists a floor f where 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 unbroken egg and drop it from any floor x (where 1 <= x <= n). If the egg breaks, you can no longer use it. However, if the egg does not break, you may reuse it in future moves.
Return the minimum number of moves that you need to determine with certainty what the value of f is.
Example 1:
Input: k = 1, n = 2 Output: 2 Explanation: Drop the egg from floor 1. If it breaks, we know that f = 0. Otherwise, drop the egg from floor 2. If it breaks, we know that f = 1. If it does not break, then we know f = 2. Hence, we need at minimum 2 moves to determine with certainty what the value of f is.
Example 2:
Input: k = 2, n = 6 Output: 3
Example 3:
Input: k = 3, n = 14 Output: 4
Constraints:
1 <= k <= 1001 <= n <= 104
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const superEggDrop = function(K, N) {
let lo = 1,
hi = N,
mi
while (lo < hi) {
mi = ((lo + hi) / 2) >> 0
if (f(mi, K, N) < N) lo = mi + 1
else hi = mi
}
return lo
}
function f(x, K, N) {
let ans = 0,
r = 1,
i = 1
for (let i = 1; i <= K; ++i) {
r = ((r * (x - i + 1)) / i) >> 0
ans += r
if (ans >= N) break
}
return ans
}
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def superEggDrop(self, K, N):
drops = 0 # the number of eggs dropped
floors = [0 for _ in range(K + 1)] # floors[i] is the number of floors that can be checked with i eggs
while floors[K] < N: # until we can reach N floors with K eggs
for eggs in range(K, 0, -1):
floors[eggs] += 1 + floors[eggs - 1]
drops += 1
return drops
Input
Output
#3 Code Example with C# Programming
Code -
C# Programming
start coding...