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
function superEggDrop(k: number, n: number): number {
    const f: number[][] = new Array(n + 1).fill(0).map(() => new Array(k + 1).fill(0));
    const dfs = (i: number, j: number): number => {
        if (i < 1) {
            return 0;
        }
        if (j === 1) {
            return i;
        }
        if (f[i][j]) {
            return f[i][j];
        }
        let l = 1;
        let r = i;
        while (l < r) {
            const mid = (l + r + 1) >> 1;
            const a = dfs(mid - 1, j - 1);
            const b = dfs(i - mid, j);
            if (a <= b) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return (f[i][j] = Math.max(dfs(l - 1, j - 1), dfs(i - l, j)) + 1);
    };
    return dfs(n, k);
}
