<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 dp[101][10005];
    int solve(int k,int n){
        if(n==0 or n==1)return n;
        if(k==1)return n;
        if(dp[k][n]!=-1){
            return dp[k][n];
        }
        int answer=INT_MAX;
        int start=1;
        int end=n;
        while(start<=end){
            int middle=(start+end)/2;
            int left=solve(k-1,middle-1);
            int right=solve(k,n-middle);
            int temp=1+max(left,right);
            if(left<right){
                start=middle+1;
            }
            else{
                end=middle-1;
            }
            answer=min(answer,temp);
        }
    return dp[k][n]=answer;
    }
    int superEggDrop(int k, int n) {
        memset(dp,-1,sizeof(dp));
        return solve(k,n);
    }
};
11900