<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>

#include <vector>
#include <algorithm>
#include <climits>

class Solution {
public:
    int superEggDrop(int k, int n) {
        // dp[i][j] will be the minimum number of moves required for i floors and j eggs
        vector<vector<int>> dp(n + 1, vector<int>(k + 1, 0));
        
        // Base cases
        for (int j = 1; j <= k; ++j) {
            dp[0][j] = 0; // 0 floors require 0 moves
            dp[1][j] = 1; // 1 floor requires 1 move
        }
        
        for (int i = 1; i <= n; ++i) {
            dp[i][1] = i; // If we have only one egg, we need i moves for i floors (linear search)
        }
        
        // Fill the dp table
        for (int j = 2; j <= k; ++j) {
            for (int i = 2; i <= n; ++i) {
                int low = 1, high = i;
                while (low + 1 < high) {
                    int mid = low + (high - low) / 2;
                    if (dp[mid - 1][j - 1] < dp[i - mid][j]) {
                        low = mid;
                    } else {
                        high = mid;
                    }
                }
                dp[i][j] = 1 + min(max(dp[low - 1][j - 1], dp[i - low][j]),
                                  max(dp[high - 1][j - 1], dp[i - high][j]));
            }
        }
        
        return dp[n][k];
    }
};

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