<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:
    vector<vector<int>> dp; // Memoization table

    int helper(int k, int n) {
        // Base cases
        if (n == 0 || n == 1) return n;
        if (k == 1) return n;

        // Check if the result is already in the memo table
        if (dp[k][n] != -1) return dp[k][n];

        int l = 1;
        int r = n;
        int ans = INT_MAX;

        // Binary search to minimize the worst-case scenario
        while (l <= r) {
            int mid = l + (r - l) / 2;
            int notBreak = helper(k, n - mid);     // Egg didn't break
            int breakEgg = helper(k - 1, mid-1);     // Egg broke

            // We need the worst-case number of trials
            if (breakEgg < notBreak) {
                l = mid + 1;
            } else {
                r = mid - 1;
            }

            ans = min(ans, max(notBreak, breakEgg) + 1); // +1 because we are counting the current trial
        }

        // Store the result in the memo table and return it
        return dp[k][n] = ans;
    }

    int superEggDrop(int k, int n) {
        // Initialize dp table with -1 (unvisited states)
        dp.clear();
        dp = vector<vector<int>>(k + 5, vector<int>(n + 5, -1));
        return helper(k, n);
    }
};

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