<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 solve(int k, int n, vector<vector<int>>& dp) {
        if(k == 1) return n; // If one egg, try every floor
        if(n == 0 || n == 1) return n; // If no or one floor, return n

        if(dp[k][n] != -1) return dp[k][n]; // Correct memoization check

        int maxi = INT_MAX;
        int low = 1; // Start binary search at floor 1
        int high = n;
        int num = 0;

        while(low <= high) {
            int mid = low + (high - low) / 2;

            int breakEgg = solve(k-1, mid-1, dp);  // Egg breaks
            int noBreakEgg = solve(k, n-mid, dp);  // Egg doesn't break

            num = 1 + max(breakEgg, noBreakEgg);

            // Binary search decision: minimize the worst-case attempts
            if(breakEgg > noBreakEgg) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }

            maxi = min(num, maxi);
        }

        return dp[k][n] = maxi;
    }

    int superEggDrop(int k, int n) {
        vector<vector<int>> dp(k+1, vector<int>(n+1, -1)); // Initialize memoization table
        return solve(k, n, dp); // Start recursive solution
    }
};

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