<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:
        dp=[[-1]*(n+1) for _ in range(k+1)]
        def solve(e,f,dp):
            if e==1:
                return f
            if f==0 or f==1:
                return f
            if dp[e][f]!=-1:
                return dp[e][f]
            res=float('inf')
            l=1
            h=f
            while(l<=h):
                k=(l+h)//2
                #checj if we already computed broken egg [e-1][k-1] case
                if dp[e-1][k-1]!=-1: 
                    low=dp[e-1][k-1]
                else:
                    low=solve(e-1,k-1,dp)
                if dp[e][f-k]!=-1:
                    high=dp[e][f-k]
                else:
                    high=solve(e,f-k,dp)
                temp=1+max(low,high)
                if high>low: #If high > low (i.e., if the number of floors to search above k is worse than below k), we should search the upper half of the current range (l = k + 1).
                    l=k+1
                else:
                    h=k-1
                res=min(res,temp)
            dp[e][f]=res
            return res
        return solve(k,n,dp)


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