<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>

vector<vector<int>> dp(101,vector<int> (10001,0));
int p=0;
class Solution {
public:
    void solve()
    {
        int k=100,n=10000;
        for(int i=0;i<=1;i++)
        {
            for(int j=0;j<=k;j++)
            {
                dp[j][i]=i;
            }
        }
        for(int j=0;j<=n;j++) dp[1][j]=j;

        for(int j=2;j<=n;j++)
        {
            for(int i=2;i<=k;i++)
            {
                int mn=INT_MAX;
                int l=1,u=j;
                while(l<=u)
                {
                    int mid=(l+u)/2;
                    int upper=dp[i][j-mid];
                    int lower=dp[i-1][mid-1];
                    int temp=1+max(upper,lower);
                    if(upper>lower)
                    {
                        l=mid+1;
                    }
                    else
                    {
                        u=mid-1;
                    }
                    mn=min(temp,mn);
                }
                dp[i][j]=mn;
            }
        }
    }
    int superEggDrop(int k, int n) {
        if(p) return dp[k][n];
        solve();
        p=1;
        // for(int i=0;i<=k;i++){for(int j=0;j<=n;j++){cout<<dp[i][j]<<" ";}cout<<endl;}
        return dp[k][n];
    }
};
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