<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 {
public:
    int dfs(int r, int e) {
        static int dp[10002][101] = {};
        if (r == 1 || e == 1) return r - 1;
        if (dp[r][e]) return dp[r][e];

        int &x = dp[r][e];
        x = INT_MAX;
        for (int m = 1; (m << 1) <= r; ++m)
            x = min(x, max(dfs(m, e-1), dfs(r-m, e)));
        return ++x;
    }

    int superEggDrop(int k, int n) {
        // dp bsearch
        // if egg lost, less bsearches before must linear search
        // thus need to find optimal place to split a range to result in least moves
        // use dp for this
        // state: range size, eggs remaining
        // dp[r][e] = 1 + min<k>(max(dp[m][e-1], dp[r-m][e])) for m in [1, r/2]
        // dp[1][e] = 0
        // dp[r][1] = r-1
        return dfs(n+1, k);
    }
};
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