<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 superE(int k, int n){
        int dp[14][10001]={{0}};
        
        for(int i=1;i<=n;i++){
            dp[1][i]=i;
        }
        
        for(int i=1;i<=k;i++){
            dp[i][1]=1;
        }
        
        
        for(int i=2;i<=n;i++){
            for(int j=2;j<=k;j++){
                int mi=1000000;
                int logofi=log2(i);
                
                for(int p=1;p<=i;){
                    
                    mi= min(mi, max(dp[j][i-p] , dp[j-1][p-1]));
                    
                    if(dp[j][i-p] > dp[j-1][p-1])p=p+logofi;
                    else if(dp[j][i-p] < dp[j-1][p-1]){
                        p--;
                        if(dp[j][i-p] > dp[j-1][p-1])break;
                    }
                    else{
                        break;
                    }
                }
                dp[j][i]=1+mi;
            }
        }
        return dp[k][n];
    }
    
    int superEggDrop(int k, int n) {
        if(k>13){
            return log2(n)+1;
        }
        //top down
        // int dp[14][10001]={{0}};
        // return superEg(k,n,dp);
        
        //bottom up
        return superE(k,n);
        
    }
};
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