<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 superEggDrop(int K, int N) {
        if (N == 0 || N == 1) return N;
        if (K == 1) return N;

        vector<int> prev(N + 1, 0);
        vector<int> cur(N + 1, 0);

        for (int i = 0; i <= N; i++) {
            prev[i] = i;
        }

        for (int k = 2; k <= K; k++) {
            for (int n = 1; n <= N; n++) {
                int l = 1, h = n, ans = N;
                while (l <= h) {
                    int mid = (l + h) / 2;
                    int broken = prev[mid - 1];
                    int not_broken = cur[n - mid];

                    int worst = 1 + max(broken, not_broken);
                    if (broken > not_broken) {
                        h = mid - 1;
                    } 
                    else {
                        l = mid + 1;
                    }

                    ans = min(ans, worst);
                }
                cur[n] = ans;
            }
            swap(prev, cur);
        }

        return prev[N];
    }
};
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