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Eggs dropping puzzle | Set 2
  • Difficulty Level : Expert
  • Last Updated : 06 Feb, 2020

Given N eggs and K floors, the task is to find the minimum number of trials needed in the worst case to find the floor below which all floors are safe. A floor is safe if dropping an egg from it does not break the egg. Please refer n eggs and k floors for more insight.

Examples:

Input: N = 2, K = 10
Output: 4
Explanation:
The first trial is from 4th floor. Two cases arise after this:

  1. If egg breaks, we have one egg left so we need three more trials.
  2. If the egg does not break, the next try is from 7th floor. Again two cases arise.

    Notice that if 4th floor is chosen as first floor, 7th as the next floor
    and 9 as last floor, the number of trials can never exceed 4.

Input: N = 2, K = 100
Output: 14

Prerequisites: Egg Dropping Puzzle



Approach: Consider this problem in a different way:
Let dp[x][n] is the maximum number of floors that can be checked with given n eggs and x moves.

Then the equation is:

    dp[x][n] = dp[x – 1][n – 1] + dp[x – 1][n] + 1
    which means we take 1 move to a floor,

  1. If the egg breaks, then we can check dp[x – 1][n – 1] floors.
  2. If the egg doesn’t break, then we can check dp[x – 1][n] + 1 floors.

Since we need to cover k floors, dp[x][n] >= k.

dp[x][n] is similar to the number of combinations and it increases exponentially to k

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum number
// of trials needed in the worst case
// with n eggs and k floors
int eggDrop(int n, int k)
{
    vector<vector<int> > dp(k + 1, vector<int>(n + 1, 0));
  
    int x = 0;
  
    // Fill all the entries in table using
    // optimal substructure property
    while (dp[x][n] < k) {
  
        x++;
        for (int i = 1; i <= n; i++)
            dp[x][i] = dp[x - 1][i - 1] + dp[x - 1][i] + 1;
    }
  
    // Return the minimum number of moves
    return x;
}
  
// Driver code
int main()
{
    int n = 2, k = 36;
  
    cout << eggDrop(n, k);
  
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
      
// Function to return the minimum number
// of trials needed in the worst case
// with n eggs and k floors
static int eggDrop(int n, int k)
{
    int dp[][] = new int [k + 1][n + 1];
  
    int x = 0;
  
    // Fill all the entries in table using
    // optimal substructure property
    while (dp[x][n] < k)
    {
  
        x++;
        for (int i = 1; i <= n; i++)
            dp[x][i] = dp[x - 1][i - 1] + 
                       dp[x - 1][i] + 1;
    }
  
    // Return the minimum number of moves
    return x;
}
  
// Driver code
public static void main(String args[])
{
    int n = 2, k = 36;
  
    System.out.println( eggDrop(n, k));
}
}
  
// This code is contributed by Arnab Kundu

Python3




# Python implementation of the approach
  
# Function to return the minimum number
# of trials needed in the worst case
# with n eggs and k floors
def eggDrop(n, k):
    dp = [[0 for i in range(n + 1)] for 
           j in range(k + 1)]
  
    x = 0;
  
    # Fill all the entries in table using
    # optimal substructure property
    while (dp[x][n] < k):
  
        x += 1;
        for i in range(1, n + 1):
            dp[x][i] = dp[x - 1][i - 1] + \
                        dp[x - 1][i] + 1;
      
    # Return the minimum number of moves
    return x;
  
# Driver code
if __name__ == '__main__':
    n = 2; k = 36;
  
    print(eggDrop(n, k));
  
# This code is contributed by PrinciRaj1992

C#




// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to return the minimum number
// of trials needed in the worst case
// with n eggs and k floors
static int eggDrop(int n, int k)
{
    int [,]dp = new int [k + 1, n + 1];
  
    int x = 0;
  
    // Fill all the entries in table using
    // optimal substructure property
    while (dp[x, n] < k)
    {
  
        x++;
        for (int i = 1; i <= n; i++)
            dp[x, i] = dp[x - 1, i - 1] + 
                    dp[x - 1, i] + 1;
    }
  
    // Return the minimum number of moves
    return x;
}
  
// Driver code
public static void Main(String []args)
{
    int n = 2, k = 36;
  
    Console.WriteLine(eggDrop(n, k));
}
}
  
// This code is contributed by PrinciRaj1992
Output:
8

Time Complexity: O(NlogK)
Space Complexity: O(N * K)

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