<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:
        # Create a memoization table
        memo = {}
        
        def dp(k, n):
            if (k, n) in memo:
                return memo[(k, n)]
            
            # Base cases
            if n == 0 or n == 1:
                return n
            if k == 1:
                return n
            
            # Binary search optimization
            left, right = 1, n
            while left + 1 < right:
                mid = (left + right) // 2
                broken = dp(k - 1, mid - 1)
                not_broken = dp(k, n - mid)
                if broken < not_broken:
                    left = mid
                else:
                    right = mid
            
            # Minimum moves considering worst case
            res = 1 + min(max(dp(k - 1, x - 1), dp(k, n - x)) for x in range(left, right + 1))
            
            # Store the result in memo
            memo[(k, n)] = res
            return res
        
        return dp(k, n)

# Example usage:
solution = Solution()
print(solution.superEggDrop(1, 2))  # Output: 2
print(solution.superEggDrop(2, 6))  # Output: 3
print(solution.superEggDrop(3, 14))  # Output: 4
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