Eggs dropping puzzle | Set 2

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: 4 Explanation: The first trial was on the 4th 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.
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 PuzzleApproach: 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.
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 kBelow is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
// 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 floorsint 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 codeint main(){    int n = 2, k = 36;     cout << eggDrop(n, k);     return 0;}
// Java implementation of the approachclass GFG{     // Function to return the minimum number// of trials needed in the worst case// with n eggs and k floorsstatic 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 codepublic static void main(String args[]){    int n = 2, k = 36;     System.out.println( eggDrop(n, k));}} // This code is contributed by Arnab Kundu
# Python implementation of the approach # Function to return the minimum number# of trials needed in the worst case# with n eggs and k floorsdef 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 codeif __name__ == '__main__':    n = 2; k = 36;     print(eggDrop(n, k)); # This code is contributed by PrinciRaj1992
// C# implementation of the approachusing System; class GFG{     // Function to return the minimum number// of trials needed in the worst case// with n eggs and k floorsstatic 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 codepublic static void Main(String []args){    int n = 2, k = 36;     Console.WriteLine(eggDrop(n, k));}} // This code is contributed by PrinciRaj1992
<script>// Javascript implementation of the approach // Function to return the minimum number// of trials needed in the worst case// with n eggs and k floorsfunction 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 codelet n = 2, k = 36;document.write(eggDrop(n, k));</script>
8
 
Time Complexity: O(NlogK) Space Complexity: O(N * K) 
andrew1234
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