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Egg Dropping Puzzle

Sumit Rawal answered on October 19, 2022 Popularity 7/10 Helpfulness 1/10

answer Egg Dropping Puzzle

Egg Dropping Puzzle

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class Solution 
{
    //Function to find minimum number of attempts needed in 
    //order to find the critical floor.
    static int eggDrop(int n, int k) 
	{
	    // Your code here
	    int dp[][]=new int[n+1][k+1];
	    for(int i=1;i<=n;i++){//i==>eggs
	        for(int j=1;j<=k;j++){//j==>floors
	            if(i==1){//if num of eggs is 1 it will take floors time to find f
	                dp[i][j]=j;
	            }
	            else if(j==1){//if there is only one floor then only one move to find f
	                dp[i][j]=1;
	            }
	            else{
	                int min=Integer.MAX_VALUE;
	                for(int mj=j-1,pj=0;mj>=0;mj--,pj++){
	                    int val1=dp[i-1][pj];
	                    int val2=dp[i][mj];
	                    int max=Math.max(val1,val2);
	                    min=Math.min(min,max);
	                }
	                dp[i][j]=min+1;
	            }
	        }
	    }
	    
	    return dp[n][k];
	    
	    
	    
	    
	    /*if(n==0){
	        return 0;
	    }
	    if(n==1){
	        return k;
	    }
	    if(k==0){
	        return 0;
	    }
	    if(k==1){
	        return 1;
	    }
	    int min=Integer.MAX_VALUE;
	    for(int f=1;f<=k;f++)
	    {
	       int broken=eggDrop(n-1,f-1);
	       int NotBroken=eggDrop(n,k-f);
	       int max=Math.max(broken,NotBroken);
	       min=Math.min(min,max+1);
	    }
	    return min;*/
	}
}

Popularity 7/10 Helpfulness 1/10 Language whatever

Source: Grepper

Tags: egg puzzle whatever

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Contributed on Oct 19 2022

Sumit Rawal

0 Answers Avg Quality 2/10

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