# 887. Super Egg Drop

## Description

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 <= 100
• 1 <= n <= 104

## Solutions

• class Solution {
private int[][] f;

public int superEggDrop(int k, int n) {
f = new int[n + 1][k + 1];
return dfs(n, k);
}

private int dfs(int i, int j) {
if (i < 1) {
return 0;
}
if (j == 1) {
return i;
}
if (f[i][j] != 0) {
return f[i][j];
}
int l = 1, r = i;
while (l < r) {
int mid = (l + r + 1) >> 1;
int a = dfs(mid - 1, j - 1);
int b = dfs(i - mid, j);
if (a <= b) {
l = mid;
} else {
r = mid - 1;
}
}
return f[i][j] = Math.max(dfs(l - 1, j - 1), dfs(i - l, j)) + 1;
}
}

• class Solution {
public:
int superEggDrop(int k, int n) {
int f[n + 1][k + 1];
memset(f, 0, sizeof(f));
function<int(int, int)> dfs = [&](int i, int j) -> int {
if (i < 1) {
return 0;
}
if (j == 1) {
return i;
}
if (f[i][j]) {
return f[i][j];
}
int l = 1, r = i;
while (l < r) {
int mid = (l + r + 1) >> 1;
int a = dfs(mid - 1, j - 1);
int b = dfs(i - mid, j);
if (a <= b) {
l = mid;
} else {
r = mid - 1;
}
}
return f[i][j] = max(dfs(l - 1, j - 1), dfs(i - l, j)) + 1;
};
return dfs(n, k);
}
};

• class Solution:
def superEggDrop(self, k: int, n: int) -> int:
@cache
def dfs(i: int, j: int) -> int:
if i < 1:
return 0
if j == 1:
return i
l, r = 1, i
while l < r:
mid = (l + r + 1) >> 1
a = dfs(mid - 1, j - 1)
b = dfs(i - mid, j)
if a <= b:
l = mid
else:
r = mid - 1
return max(dfs(l - 1, j - 1), dfs(i - l, j)) + 1

return dfs(n, k)


• func superEggDrop(k int, n int) int {
f := make([][]int, n+1)
for i := range f {
f[i] = make([]int, k+1)
}
var dfs func(i, j int) int
dfs = func(i, j int) int {
if i < 1 {
return 0
}
if j == 1 {
return i
}
if f[i][j] != 0 {
return f[i][j]
}
l, r := 1, i
for l < r {
mid := (l + r + 1) >> 1
a, b := dfs(mid-1, j-1), dfs(i-mid, j)
if a <= b {
l = mid
} else {
r = mid - 1
}
}
f[i][j] = max(dfs(l-1, j-1), dfs(i-l, j)) + 1
return f[i][j]
}
return dfs(n, k)
}

• function superEggDrop(k: number, n: number): number {
const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(k + 1).fill(0));
const dfs = (i: number, j: number): number => {
if (i < 1) {
return 0;
}
if (j === 1) {
return i;
}
if (f[i][j]) {
return f[i][j];
}
let l = 1;
let r = i;
while (l < r) {
const mid = (l + r + 1) >> 1;
const a = dfs(mid - 1, j - 1);
const b = dfs(i - mid, j);
if (a <= b) {
l = mid;
} else {
r = mid - 1;
}
}
return (f[i][j] = Math.max(dfs(l - 1, j - 1), dfs(i - l, j)) + 1);
};
return dfs(n, k);
}