Eggs dropping puzzle | Set 2

19 June 2025

0

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 was on the 4^th floor. Two cases arise after this:

If the egg breaks, we have one egg left, so we need three more trials.

If the egg does not break, the next try is from the 7th floor. Again, two cases arise.

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:

If the egg breaks, then we can check dp[x – 1][n – 1] floors.
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

Javascript

<script>
// Javascript implementation of the approach
 
// Function to return the minimum number
// of trials needed in the worst case
// with n eggs and k floors
function eggDrop(n, k)
{
    let dp = new Array();
 
    for(let i = 0; i < k + 1; i++){
        dp.push(new Array(n + 1).fill(0))
    }
 
    let x = 0;
 
    // Fill all the entries in table using
    // optimal substructure property
    while (dp[x][n] < k) {
 
        x++;
        for (let 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
let n = 2, k = 36;
document.write(eggDrop(n, k));
</script>

Output

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

Efficient Approach : Space Optimization

To optimize the space complexity of the above approach, we can observe that each entry dp[x][i] only depends on the previous row dp[x-1][i-1] and dp[x-1][i]. Therefore, we can use a 1D array instead of a 2D array to store the values of the current row and update it iteratively.

Step-by-step approach:

Initialize a 1D vector dp of size n+1 with all elements set to 0.
Initialize a variable x to 0.
Enter a loop until the number of trials needed for n eggs is less than k.
Inside the loop, iterate from n down to 1.
- Update the dp array at index i with the sum of the values at dp[i-1], dp[i], and 1.
- Increment x.
- Repeat the loop until the number of trials needed for n eggs is greater than or equal to k.
Return the value of x.

Below is the implementation of the above approach:

C++

#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<int> dp(n + 1, 0);
 
    int x = 0;
 
    // Fill the entries in the array
    // using optimal substructure property
    while (dp[n] < k) {
 
        x++;
        for (int i = n; i >= 1; i--)
            dp[i] = dp[i - 1] + dp[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;
}

Python3

# 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] * (n + 1)
    x = 0
     
    # Fill the entries in the array
    # using optimal substructure property
    while dp[n] < k:
        x += 1
        for i in range(n, 0, -1):
            dp[i] = dp[i - 1] + dp[i] + 1
     
    # Return the minimum number of moves
    return x
 
n = 2
k = 36
print(eggDrop(n, k))

Javascript

// Function to return the minimum number
// of trials needed in the worst case
// with n eggs and k floors
function eggDrop(n, k) {
 
    let dp = new Array(n + 1).fill(0);
    let x = 0;
     
     
    // Fill the entries in the array
    // using optimal substructure property
    while (dp[n] < k) {
        x++;
        for (let i = n; i >= 1; i--)
            dp[i] = dp[i - 1] + dp[i] + 1;
    }
     
    // Return the minimum number of moves
    return x;
}
 
// Test case
let n = 2, k = 36;
console.log(eggDrop(n, k));

Output

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

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Eggs dropping puzzle | Set 2

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