<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, memo: dict = {}) -> int:
        # if floors are 0 or 1 only n moves required
        if n == 0 or n == 1:
            return n

        # if only one egg is there
        if k == 1:
            return n
        
        key = str(k) + ',' + str(n)

        if key in memo:
            return memo[key]
    
        min_answer,low,high = float('inf'),0,n # intialize low and high
        
        while low<=high:

            mid = (low + high) // 2

            # /*representing both the choices with memo
            # First one, if the egg will break, no. of eggs will decreased and we have to
            # down from that floor.
            # Second one, if the egg will not break, no. of eggs will not decreased and we
            # have to go above form that floor.*/

            left,right = self.superEggDrop(k-1,mid-1,memo), self.superEggDrop(k,n-mid,memo)

            temp = 1 + max(left,right)
            # since we need more temp values in worst case so we need to go above
            if left < right:
                low = mid + 1
            else:
                high = mid - 1      # move to downward

            min_answer = min(temp,min_answer)
        memo[key] = min_answer
        return memo[key]
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