<p>You are given <code>k</code> identical eggs and you have access to a building with <code>n</code> floors labeled from <code>1</code> to <code>n</code>.</p>

<p>You know that there exists a floor <code>f</code> where <code>0 &lt;= f &lt;= n</code> such that any egg dropped at a floor <strong>higher</strong> than <code>f</code> will <strong>break</strong>, and any egg dropped <strong>at or below</strong> floor <code>f</code> will <strong>not break</strong>.</p>

<p>Each move, you may take an unbroken egg and drop it from any floor <code>x</code> (where <code>1 &lt;= x &lt;= n</code>). If the egg breaks, you can no longer use it. However, if the egg does not break, you may <strong>reuse</strong> it in future moves.</p>

<p>Return <em>the <strong>minimum number of moves</strong> that you need to determine <strong>with certainty</strong> what the value of </em><code>f</code> is.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> k = 1, n = 2
<strong>Output:</strong> 2
<strong>Explanation: </strong>
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> k = 2, n = 6
<strong>Output:</strong> 3
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> k = 3, n = 14
<strong>Output:</strong> 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= 100</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def superEggDrop(self, k: int, n: int) -> int:
        self.memo = [[-1 for _ in range(n+1)] for _ in range(k+1)]
        def dp(cur_k, cur_n):
            if cur_n == 0:
                return 0
            if cur_k == 1:
                return cur_n
            if self.memo[cur_k][cur_n] != -1:
                return self.memo[cur_k][cur_n]
            
            
            left, right = 1, cur_n
            res = cur_n
            while left <= right:
                mid = left + (right-left)//2
                t1 = dp(cur_k-1, mid-1)
                t2 = dp(cur_k, cur_n-mid)
                if t1 == t2:
                    self.memo[cur_k][cur_n] = t1 + 1
                    return 1 + t1
                elif t1 < t2:
                    res = min(res, t2 + 1)
                    left = mid + 1
                elif t1 > t2:
                    right = mid -1
                    res = min(res, t1+1)
            # for i in range(1, cur_n+1):
            #     res = min(res, 1 + max(dp(cur_k-1, i-1), dp(cur_k, cur_n-i)))
            self.memo[cur_k][cur_n] = res
            return res
        return dp(k, n)
39773