Super Egg Drop
<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>
Math; Binary Search; Dynamic Programming
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)
}