Super Egg Drop Puzzle — Dynamic Programming

Problem Statement :

You are given K eggs, and you have access to a building with N floors from 1 to N.

Each egg is identical in function, and if an egg breaks, you cannot drop it again.

You know that there exists a floor F with 0 <= F <= N such that any egg dropped at a floor higher than F will break, and any egg dropped at or below floor F will not break.

Each move, you may take an egg (if you have an unbroken one) and drop it from any floor X (with 1 <= X <= N).

Your goal is to know with certainty what the value of F is.

What is the minimum number of moves that you need to know with certainty what F is, regardless of the initial value of F?

Example 1:

Input: K = 1, N = 2
Output: 2
Explanation: 
Drop the egg from floor 1.  If it breaks, we know with certainty that F = 0.
Otherwise, drop the egg from floor 2.  If it breaks, we know with certainty that F = 1.
If it didn't break, then we know with certainty F = 2.
Hence, we needed 2 moves in the worst case to know what F is with certainty.

Example 2:

Input: K = 2, N = 6
Output: 3

Example 3:

Input: K = 3, N = 14
Output: 4

Note:

1. 1 <= K <= 100
2. 1 <= N <= 10000

Solution :

class Solution {
    Integer dp[][]=new Integer[101][10001];
    //Declared wrapper class because no need to initialise values, will be null defaultly
    
    public int superEggDrop(int K, int N) {
      
        if(K==1)//when there is only one egg
            return N;
        
        if(N==0 || N==1)//when no. of floors are 0 or 1
            return N;
        
        if(dp[K][N]!=null)//checking if it is filled
            return dp[K][N];
        
        int i,l=1,h=N;
        int ans=Integer.MAX_VALUE;

        while(l<=h)
        {
            int mid=(l+h)/2;
            
            int down_temp=superEggDrop(K-1,mid-1);//If egg breaks go down
            
            int up_temp=superEggDrop(K,N-mid);//if egg doesn't break go up
            
            int temp=1+Math.max(down_temp,up_temp);
            //adding one because we have used 1 attempt and max of up and down because
            //we need worst case attempts from both
            
            if(down_temp<up_temp)//if down attempts are less we only require upper ones and vise versa
                l=mid+1;
            else 
                h=mid-1;
            
            ans=Math.min(temp,ans);//this is beacuse we need minimum of all worst case attempts 
        }
        return dp[K][N]=ans;
    }
}