# Eggs dropping puzzle | Set 2

• Difficulty Level : Hard
• Last Updated : 02 Jun, 2021

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:
Explanation:
The first trial was on the 4th floor. Two cases arise after this:

1. If the 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 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:

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 ``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 > 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

 ``

Output:

`8`

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

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