Egg Dropping Puzzle | DP-11
- Difficulty Level : Hard
- Last Updated : 21 Jul, 2022
The following is a description of the instance of this famous puzzle involving n=2 eggs and a building with k=36 floors.
Suppose that we wish to know which stories in a 36-storey building are safe to drop eggs from, and which will cause the eggs to break on landing. We make a few assumptions:
…..An egg that survives a fall can be used again.
…..A broken egg must be discarded.
…..The effect of a fall is the same for all eggs.
…..If an egg breaks when dropped, then it would break if dropped from a higher floor.
…..If an egg survives a fall then it would survive a shorter fall.
…..It is not ruled out that the first-floor windows break eggs, nor is it ruled out that the 36th-floor do not cause an egg to break.
If only one egg is available and we wish to be sure of obtaining the right result, the experiment can be carried out in only one way. Drop the egg from the first-floor window; if it survives, drop it from the second-floor window. Continue upward until it breaks. In the worst case, this method may require 36 droppings. Suppose 2 eggs are available. What is the least number of egg-droppings that is guaranteed to work in all cases?
The problem is not actually to find the critical floor, but merely to decide floors from which eggs should be dropped so that the total number of trials are minimized.
Source: Wiki for Dynamic Programming
Method 1: Recursion.
In this post, we will discuss a solution to a general problem with ‘n’ eggs and ‘k’ floors. The solution is to try dropping an egg from every floor(from 1 to k) and recursively calculate the minimum number of droppings needed in the worst case. The floor which gives the minimum value in the worst case is going to be part of the solution.
In the following solutions, we return the minimum number of trials in the worst case; these solutions can be easily modified to print floor numbers of every trial also.
Meaning of a worst-case scenario: Worst case scenario gives the user the surety of the threshold floor. For example- If we have ‘1’ egg and ‘k’ floors, we will start dropping the egg from the first floor till the egg breaks suppose on the ‘kth’ floor so the number of tries to give us surety is ‘k’.
1) Optimal Substructure:
When we drop an egg from a floor x, there can be two cases (1) The egg breaks (2) The egg doesn’t break.
- If the egg breaks after dropping from ‘xth’ floor, then we only need to check for floors lower than ‘x’ with remaining eggs as some floor should exist lower than ‘x’ in which egg would not break; so the problem reduces to x-1 floors and n-1 eggs.
- If the egg doesn’t break after dropping from the ‘xth’ floor, then we only need to check for floors higher than ‘x’; so the problem reduces to ‘k-x’ floors and n eggs.
Since we need to minimize the number of trials in worst case, we take the maximum of two cases. We consider the max of above two cases for every floor and choose the floor which yields minimum number of trials.
k ==> Number of floors
n ==> Number of Eggs
eggDrop(n, k) ==> Minimum number of trials needed to find the critical
floor in worst case.
eggDrop(n, k) = 1 + min{max(eggDrop(n – 1, x – 1), eggDrop(n, k – x)), where x is in {1, 2, …, k}}
Concept of worst case:
For example :
Let there be ‘2’ eggs and ‘2’ floors then-:
If we try throwing from ‘1st’ floor:
Number of tries in worst case= 1+max(0, 1)
0=>If the egg breaks from first floor then it is threshold floor (best case possibility).
1=>If the egg does not break from first floor we will now have ‘2’ eggs and 1 floor to test which will give answer as
‘1’.(worst case possibility)
We take the worst case possibility in account, so 1+max(0, 1)=2
If we try throwing from ‘2nd’ floor:
Number of tries in worst case= 1+max(1, 0)
1=>If the egg breaks from second floor then we will have 1 egg and 1 floor to find threshold floor.(Worst Case)
0=>If egg does not break from second floor then it is threshold floor.(Best Case)
We take worst case possibility for surety, so 1+max(1, 0)=2.
The final answer is min(1st, 2nd, 3rd….., kth floor)
So answer here is ‘2’.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;// A utility function to get// maximum of two integersint max(int a, int b){ return (a > b) ? a : b;}// Function to get minimum// number of trials needed in worst// case with n eggs and k floorsint eggDrop(int n, int k){ // If there are no floors, // then no trials needed. // OR if there is one floor, // one trial needed. if (k == 1 || k == 0) return k; // We need k trials for one // egg and k floors if (n == 1) return k; int min = INT_MAX, x, res; // Consider all droppings from // 1st floor to kth floor and // return the minimum of these // values plus 1. for (x = 1; x <= k; x++) { res = max( eggDrop(n - 1, x - 1), eggDrop(n, k - x)); if (res < min) min = res; } return min + 1;}// Driver program to test// to printDupsint main(){ int n = 2, k = 10; cout << "Minimum number of trials " "in worst case with " << n << " eggs and " << k << " floors is " << eggDrop(n, k) << endl; return 0;}// This code is contributed// by Akanksha Rai |
C
#include <limits.h>#include <stdio.h>// A utility function to get// maximum of two integersint max(int a, int b){ return (a > b) ? a : b;}/* Function to get minimum number of trials needed in worst case with n eggs and k floors */int eggDrop(int n, int k){ // If there are no floors, then no // trials needed. OR if there is // one floor, one trial needed. if (k == 1 || k == 0) return k; // We need k trials for one egg and // k floors if (n == 1) return k; int min = INT_MAX, x, res; // Consider all droppings from 1st // floor to kth floor and // return the minimum of these values // plus 1. for (x = 1; x <= k; x++) { res = max( eggDrop(n - 1, x - 1), eggDrop(n, k - x)); if (res < min) min = res; } return min + 1;}/* Driver program to test to printDups*/int main(){ int n = 2, k = 10; printf("nMinimum number of trials in " "worst case with %d eggs and " "%d floors is %d \n", n, k, eggDrop(n, k)); return 0;} |
Java
public class GFG { /* Function to get minimum number of trials needed in worst case with n eggs and k floors */ static int eggDrop(int n, int k) { // If there are no floors, then // no trials needed. OR if there // is one floor, one trial needed. if (k == 1 || k == 0) return k; // We need k trials for one egg // and k floors if (n == 1) return k; int min = Integer.MAX_VALUE; int x, res; // Consider all droppings from // 1st floor to kth floor and // return the minimum of these // values plus 1. for (x = 1; x <= k; x++) { res = Math.max(eggDrop(n - 1, x - 1), eggDrop(n, k - x)); if (res < min) min = res; } return min + 1; } // Driver code public static void main(String args[]) { int n = 2, k = 10; System.out.print("Minimum number of " + "trials in worst case with " + n + " eggs and " + k + " floors is " + eggDrop(n, k)); } // This code is contributed by Ryuga.} |
Python 3
import sys# Function to get minimum number of trials# needed in worst case with n eggs and k floorsdef eggDrop(n, k): # If there are no floors, then no trials # needed. OR if there is one floor, one # trial needed. if (k == 1 or k == 0): return k # We need k trials for one egg # and k floors if (n == 1): return k min = sys.maxsize # Consider all droppings from 1st # floor to kth floor and return # the minimum of these values plus 1. for x in range(1, k + 1): res = max(eggDrop(n - 1, x - 1), eggDrop(n, k - x)) if (res < min): min = res return min + 1# Driver Codeif __name__ == "__main__": n = 2 k = 10 print("Minimum number of trials in worst case with", n, "eggs and", k, "floors is", eggDrop(n, k))# This code is contributed by ita_c |
C#
using System;class GFG { /* Function to get minimum number of trials needed in worst case with n eggs and k floors */ static int eggDrop(int n, int k) { // If there are no floors, then // no trials needed. OR if there // is one floor, one trial needed. if (k == 1 || k == 0) return k; // We need k trials for one egg // and k floors if (n == 1) return k; int min = int.MaxValue; int x, res; // Consider all droppings from // 1st floor to kth floor and // return the minimum of these // values plus 1. for (x = 1; x <= k; x++) { res = Math.Max(eggDrop(n - 1, x - 1), eggDrop(n, k - x)); if (res < min) min = res; } return min + 1; } // Driver code static void Main() { int n = 2, k = 10; Console.Write("Minimum number of " + "trials in worst case with " + n + " eggs and " + k + " floors is " + eggDrop(n, k)); }}// This code is contributed by Sam007. |
Javascript
<script> /* Function to get minimum number of trials needed in worst case with n eggs and k floors */ function eggDrop(n,k) { // If there are no floors, then // no trials needed. OR if there // is one floor, one trial needed. if (k == 1 || k == 0) return k; // We need k trials for one egg // and k floors if (n == 1) return k; let min = Number.MAX_VALUE; let x, res; // Consider all droppings from // 1st floor to kth floor and // return the minimum of these // values plus 1. for (x = 1; x <= k; x++) { res = Math.max(eggDrop(n - 1, x - 1), eggDrop(n, k - x)); if (res < min) min = res; } return min + 1; } // Driver code let n = 2, k = 10; document.write("Minimum number of " + "trials in worst case with " + n + " eggs and " + k + " floors is " + eggDrop(n, k)); // This code is contributed by avanitrachhadiya2155</script> |
Minimum number of trials in worst case with 2 eggs and 10 floors is 4
Output:
Minimum number of trials in worst case with 2 eggs and 10 floors is 4
It should be noted that the above function computes the same subproblems again and again. See the following partial recursion tree, E(2, 2) is being evaluated twice. There will many repeated subproblems when you draw the complete recursion tree even for small values of n and k.
E(2, 4)
|
-------------------------------------
| | | |
| | | |
x=1/ x=2/ x=3/ x=4/
/ / .... ....
/ /
E(1, 0) E(2, 3) E(1, 1) E(2, 2)
/ /... /
x=1/ .....
/
E(1, 0) E(2, 2)
/
......
Partial recursion tree for 2 eggs and 4 floors.Complexity Analysis:
- Time Complexity: As there is a case of overlapping sub-problems the time complexity is exponential.
- Auxiliary Space :O(1). As there was no use of any data structure for storing values.
Since same subproblems are called again, this problem has Overlapping Subproblems property. So Egg Dropping Puzzle has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array eggFloor[][] in bottom up manner.
Method 2: Dynamic Programming.
In this approach, we work on the same idea as described above neglecting the case of calculating the answers to sub-problems again and again.. The approach will be to make a table which will store the results of sub-problems so that to solve a sub-problem, it would only require a look-up from the table which will take constant time, which earlier took exponential time.
Formally for filling DP[i][j] state where ‘i’ is the number of eggs and ‘j’ is the number of floors:
- We have to traverse for each floor ‘x’ from ‘1’ to ‘j’ and find minimum of:
(1 + max( DP[i-1][j-1], DP[i][j-x] )).
This simulation will make things clear:
i => Number of eggs
j => Number of floors
Look up find maximum
Lets fill the table for the following case:
Floors = ‘4’
Eggs = ‘2’
1 2 3 4
1 2 3 4 => 1
1 2 2 3 => 2
For ‘egg-1’ each case is the base case so the
number of attempts is equal to floor number.
For ‘egg-2’ it will take ‘1’ attempt for 1st
floor which is base case.
For floor-2 =>
Taking 1st floor 1 + max(0, DP[1][1])
Taking 2nd floor 1 + max(DP[1][1], 0)
DP[2][2] = min(1 + max(0, DP[1][1]), 1 + max(DP[1][1], 0))
For floor-3 =>
Taking 1st floor 1 + max(0, DP[2][2])
Taking 2nd floor 1 + max(DP[1][1], DP[2][1])
Taking 3rd floor 1 + max(0, DP[2][2])
DP[2][3]= min(‘all three floors’) = 2
For floor-4 =>
Taking 1st floor 1 + max(0, DP[2][3])
Taking 2nd floor 1 + max(DP[1][1], DP[2][2])
Taking 3rd floor 1 + max(DP[1][2], DP[2][1])
Taking 4th floor 1 + max(0, DP[2][3])
DP[2][4]= min(‘all four floors’) = 3
C++
// A Dynamic Programming based for// the Egg Dropping Puzzle#include <bits/stdc++.h>using namespace std;// A utility function to get// maximum of two integersint max(int a, int b){ return (a > b) ? a : b;}/* Function to get minimumnumber of trials needed in worstcase with n eggs and k floors */int eggDrop(int n, int k){ /* A 2D table where entry eggFloor[i][j] will represent minimum number of trials needed for i eggs and j floors. */ int eggFloor[n + 1][k + 1]; int res; int i, j, x; // We need one trial for one floor and 0 // trials for 0 floors for (i = 1; i <= n; i++) { eggFloor[i][1] = 1; eggFloor[i][0] = 0; } // We always need j trials for one egg // and j floors. for (j = 1; j <= k; j++) eggFloor[1][j] = j; // Fill rest of the entries in table using // optimal substructure property for (i = 2; i <= n; i++) { for (j = 2; j <= k; j++) { eggFloor[i][j] = INT_MAX; for (x = 1; x <= j; x++) { res = 1 + max( eggFloor[i - 1][x - 1], eggFloor[i][j - x]); if (res < eggFloor[i][j]) eggFloor[i][j] = res; } } } // eggFloor[n][k] holds the result return eggFloor[n][k];}/* Driver program to test to printDups*/int main(){ int n = 2, k = 36; cout << "\nMinimum number of trials " "in worst case with " << n<< " eggs and "<< k<< " floors is "<< eggDrop(n, k); return 0;}// this code is contributed by shivanisinghss2110 |
C
// A Dynamic Programming based for// the Egg Dropping Puzzle#include <limits.h>#include <stdio.h>// A utility function to get// maximum of two integersint max(int a, int b){ return (a > b) ? a : b;}/* Function to get minimumnumber of trials needed in worstcase with n eggs and k floors */int eggDrop(int n, int k){ /* A 2D table where entry eggFloor[i][j] will represent minimum number of trials needed for i eggs and j floors. */ int eggFloor[n + 1][k + 1]; int res; int i, j, x; // We need one trial for one floor and 0 // trials for 0 floors for (i = 1; i <= n; i++) { eggFloor[i][1] = 1; eggFloor[i][0] = 0; } // We always need j trials for one egg // and j floors. for (j = 1; j <= k; j++) eggFloor[1][j] = j; // Fill rest of the entries in table using // optimal substructure property for (i = 2; i <= n; i++) { for (j = 2; j <= k; j++) { eggFloor[i][j] = INT_MAX; for (x = 1; x <= j; x++) { res = 1 + max( eggFloor[i - 1][x - 1], eggFloor[i][j - x]); if (res < eggFloor[i][j]) eggFloor[i][j] = res; } } } // eggFloor[n][k] holds the result return eggFloor[n][k];}/* Driver program to test to printDups*/int main(){ int n = 2, k = 36; printf("\nMinimum number of trials " "in worst case with %d eggs and " "%d floors is %d \n", n, k, eggDrop(n, k)); return 0;} |
Java
// A Dynamic Programming based Java// Program for the Egg Dropping Puzzleclass EggDrop { // A utility function to get // maximum of two integers static int max(int a, int b) { return (a > b) ? a : b; } /* Function to get minimum number of trials needed in worst case with n eggs and k floors */ static int eggDrop(int n, int k) { /* A 2D table where entry eggFloor[i][j] will represent minimum number of trialsneeded for i eggs and j floors. */ int eggFloor[][] = new int[n + 1][k + 1]; int res; int i, j, x; // We need one trial for one floor and // 0 trials for 0 floors for (i = 1; i <= n; i++) { eggFloor[i][1] = 1; eggFloor[i][0] = 0; } // We always need j trials for one egg // and j floors. for (j = 1; j <= k; j++) eggFloor[1][j] = j; // Fill rest of the entries in table using // optimal substructure property for (i = 2; i <= n; i++) { for (j = 2; j <= k; j++) { eggFloor[i][j] = Integer.MAX_VALUE; for (x = 1; x <= j; x++) { res = 1 + max( eggFloor[i - 1][x - 1], eggFloor[i][j - x]); if (res < eggFloor[i][j]) eggFloor[i][j] = res; } } } // eggFloor[n][k] holds the result return eggFloor[n][k]; } /* Driver program to test to printDups*/ public static void main(String args[]) { int n = 2, k = 10; System.out.println("Minimum number of trials in worst" + " case with " + n + " eggs and " + k + " floors is " + eggDrop(n, k)); }}/*This code is contributed by Rajat Mishra*/ |
Python3
# A Dynamic Programming based Python Program for the Egg Dropping PuzzleINT_MAX = 32767# Function to get minimum number of trials needed in worst# case with n eggs and k floorsdef eggDrop(n, k): # A 2D table where entry eggFloor[i][j] will represent minimum # number of trials needed for i eggs and j floors. eggFloor = [[0 for x in range(k + 1)] for x in range(n + 1)] # We need one trial for one floor and0 trials for 0 floors for i in range(1, n + 1): eggFloor[i][1] = 1 eggFloor[i][0] = 0 # We always need j trials for one egg and j floors. for j in range(1, k + 1): eggFloor[1][j] = j # Fill rest of the entries in table using optimal substructure # property for i in range(2, n + 1): for j in range(2, k + 1): eggFloor[i][j] = INT_MAX for x in range(1, j + 1): res = 1 + max(eggFloor[i-1][x-1], eggFloor[i][j-x]) if res < eggFloor[i][j]: eggFloor[i][j] = res # eggFloor[n][k] holds the result return eggFloor[n][k]# Driver program to test to print printDupsn = 2k = 36print("Minimum number of trials in worst case with" + str(n) + "eggs and " + str(k) + " floors is " + str(eggDrop(n, k)))# This code is contributed by Bhavya Jain |
C#
// A Dynamic Programming based C# Program// for the Egg Dropping Puzzleusing System;class GFG { // A utility function to get maximum of // two integers static int max(int a, int b) { return (a > b) ? a : b; } /* Function to get minimum number of trials needed in worst case with n eggs and k floors */ static int eggDrop(int n, int k) { /* A 2D table where entry eggFloor[i][j] will represent minimum number of trials needed for i eggs and j floors. */ int[, ] eggFloor = new int[n + 1, k + 1]; int res; int i, j, x; // We need one trial for one floor and0 // trials for 0 floors for (i = 1; i <= n; i++) { eggFloor[i, 1] = 1; eggFloor[i, 0] = 0; } // We always need j trials for one egg // and j floors. for (j = 1; j <= k; j++) eggFloor[1, j] = j; // Fill rest of the entries in table // using optimal substructure property for (i = 2; i <= n; i++) { for (j = 2; j <= k; j++) { eggFloor[i, j] = int.MaxValue; for (x = 1; x <= j; x++) { res = 1 + max(eggFloor[i - 1, x - 1], eggFloor[i, j - x]); if (res < eggFloor[i, j]) eggFloor[i, j] = res; } } } // eggFloor[n][k] holds the result return eggFloor[n, k]; } // Driver function public static void Main() { int n = 2, k = 36; Console.WriteLine("Minimum number of trials " + "in worst case with " + n + " eggs and " + k + "floors is " + eggDrop(n, k)); }}// This code is contributed by Sam007. |
PHP
<?php// A Dynamic Programming based PHP// Program for the Egg Dropping Puzzle/* Function to get minimum number of trials needed in worst case with n eggs and k floors */function eggDrop($n, $k){ /* A 2D table where entry eggFloor[i][j] will represent minimum number of trials needed for i eggs and j floors. */ $eggFloor = array(array());; // We need one trial for one // floor and0 trials for 0 floors for ($i = 1; $i <=$n;$i++) { $eggFloor[$i][1] = 1; $eggFloor[$i][0] = 0; } // We always need j trials // for one egg and j floors. for ($j = 1; $j <= $k; $j++) $eggFloor[1][$j] = $j; // Fill rest of the entries in // table using optimal substructure // property for ($i = 2; $i <= $n; $i++) { for ($j = 2; $j <= $k; $j++) { $eggFloor[$i][$j] = 999999; for ($x = 1; $x <= $j; $x++) { $res = 1 + max($eggFloor[$i - 1][$x - 1], $eggFloor[$i][$j - $x]); if ($res < $eggFloor[$i][$j]) $eggFloor[$i][$j] = $res; } } } // eggFloor[n][k] holds the result return $eggFloor[$n][$k];} // Driver Code $n = 2; $k = 36; echo "Minimum number of trials in worst case with " .$n. " eggs and " .$k. " floors is " .eggDrop($n, $k) ; // This code is contributed by Sam007?> |
Javascript
<script>// A Dynamic Programming based Javascript// Program for the Egg Dropping Puzzle// A utility function to get // maximum of two integersfunction max(a,b){ return (a > b) ? a : b;}/* Function to get minimum number of trials needed in worst case with n eggs and k floors */function eggDrop(n,k){ /* A 2D table where entry eggFloor[i][j] will represent minimum number of trialsneeded for i eggs and j floors. */ let eggFloor = new Array(n + 1); for(let i=0;i<(n+1);i++) { eggFloor[i]=new Array(k+1); } let res; let i, j, x; // We need one trial for one floor and // 0 trials for 0 floors for (i = 1; i <= n; i++) { eggFloor[i][1] = 1; eggFloor[i][0] = 0; } // We always need j trials for one egg // and j floors. for (j = 1; j <= k; j++) eggFloor[1][j] = j; // Fill rest of the entries in table using // optimal substructure property for (i = 2; i <= n; i++) { for (j = 2; j <= k; j++) { eggFloor[i][j] = Number.MAX_VALUE; for (x = 1; x <= j; x++) { res = 1 + max( eggFloor[i - 1][x - 1], eggFloor[i][j - x]); if (res < eggFloor[i][j]) eggFloor[i][j] = res; } } } // eggFloor[n][k] holds the result return eggFloor[n][k];}/* Driver program to test to printDups*/let n = 2, k = 36;document.write("Minimum number of trials in worst" + " case with " + n + " eggs and " + k + " floors is " + eggDrop(n, k)); // This code is contributed by ab2127</script> |
Minimum number of trials in worst case with 2 eggs and 36 floors is 8
Complexity Analysis:
- Time Complexity: O(n*k^2).
Where ‘n’ is the number of eggs and ‘k’ is the number of floors, as we use a nested for loop ‘k^2’ times for each egg - Auxiliary Space: O(n*k).
As a 2-D array of size ‘n*k’ is used for storing elements.
Method 3: Dynamic Programming using memoization.
C++
#include <bits/stdc++.h>using namespace std;#define MAX 1000vector<vector<int>> memo(MAX, vector<int> (MAX, -1));int solveEggDrop(int n, int k) { if(memo[n][k] != -1) { return memo[n][k];} if (k == 1 || k == 0) return k; if (n == 1) return k; int min = INT_MAX, x, res; for (x = 1; x <= k; x++) { res = max( solveEggDrop(n - 1, x - 1), solveEggDrop(n, k - x)); if (res < min) min = res; } memo[n][k] = min+1; return min + 1; }int main() { int n = 2, k = 36; cout<<solveEggDrop(n, k); return 0;}// contributed by Shivam Agrawal(shivamagrawal3) |
C
#include <stdio.h>#include<limits.h>#include<string.h>#define MAX 1000int memo[MAX][MAX];int solveEggDrop(int n, int k) { if(memo[n][k] != -1) { return memo[n][k];} if (k == 1 || k == 0) return k; if (n == 1) return k; int min = INT_MAX, x, res; for (x = 1; x <= k; x++) { int a = solveEggDrop(n - 1, x - 1); int b = solveEggDrop(n, k - x); res = a>b?a:b; if (res < min) min = res; } memo[n][k] = min+1; return min + 1; }int main() { memset( memo, -1,MAX * MAX * sizeof( int ) ); int n = 2, k = 36; printf("%d",solveEggDrop(n, k)); return 0;}// This code is contributed by repakaeswaripriya. |
Java
import java.util.Arrays;class GFG { static final int MAX = 1000; static int[][] memo = new int[MAX][MAX]; static int solveEggDrop(int n, int k) { if (memo[n][k] != -1) { return memo[n][k]; } if (k == 1 || k == 0) return k; if (n == 1) return k; int min = Integer.MAX_VALUE, x, res; for (x = 1; x <= k; x++) { res = Math.max(solveEggDrop(n - 1, x - 1), solveEggDrop(n, k - x)); if (res < min) min = res; } memo[n][k] = min + 1; return min + 1; } public static void main(String[] args) { for (int i = 0; i < memo.length; i++) Arrays.fill(memo[i], -1); int n = 2, k = 36; System.out.print(solveEggDrop(n, k)); }}// This code IS contributed by umadevi9616 |
Python3
import sysMAX = 1000;memo = [[-1 for i in range(MAX)] for j in range(MAX)] ;def solveEggDrop(n, k): if (memo[n][k] != -1): return memo[n][k]; if (k == 1 or k == 0): return k; if (n == 1): return k; min = sys.maxsize; res = 0; for x in range(1,k+1): res = max(solveEggDrop(n - 1, x - 1), solveEggDrop(n, k - x)); if (res < min): min = res; memo[n][k] = min + 1; return min + 1;# Driver codeif __name__ == '__main__': n = 2; k = 36; print(solveEggDrop(n, k)); # This code is contributed by gauravrajput1 |
C#
using System;public class GFG { static readonly int MAX = 1000; static int[,] memo = new int[MAX,MAX]; static int solveEggDrop(int n, int k) { if (memo[n,k] != -1) { return memo[n,k]; } if (k == 1 || k == 0) return k; if (n == 1) return k; int min = int.MaxValue, x, res; for (x = 1; x <= k; x++) { res = Math.Max(solveEggDrop(n - 1, x - 1), solveEggDrop(n, k - x)); if (res < min) min = res; } memo[n,k] = min + 1; return min + 1; } public static void Main(String[] args) { for (int i = 0; i < memo.GetLength(0); i++) for(int j =0;j<memo.GetLength(1);j++) memo[i,j] = -1; int n = 2, k = 36; Console.Write(solveEggDrop(n, k)); }}// This code is contributed by gauravrajput1 |
Javascript
<script>var MAX = 1000; var memo = Array(MAX).fill().map(()=>Array(MAX).fill(-1)); function solveEggDrop(n , k) { if (memo[n][k] != -1) { return memo[n][k]; } if (k == 1 || k == 0) return k; if (n == 1) return k; var min = Number.MAX_VALUE, x, res; for (x = 1; x <= k; x++) { res = Math.max(solveEggDrop(n - 1, x - 1), solveEggDrop(n, k - x)); if (res < min) min = res; } memo[n][k] = min + 1; return min + 1; } var n = 2, k = 36; document.write(solveEggDrop(n, k));// This code is contributed by gauravrajput1</script> |
8
Another Approach:
The approach with O(n * k^2) has been discussed before, where dp[n][k] = 1 + max(dp[n – 1][i – 1], dp[n][k – i]) for i in 1…k. You checked all the possibilities in that approach.
Consider the problem in a different way:
dp[m][x] means that, given x eggs and m moves, what is the maximum number of floors that can be checked The dp equation is: dp[m][x] = 1 + dp[m - 1][x - 1] + dp[m - 1][x], which means we take 1 move to a floor. If egg breaks, then we can check dp[m - 1][x - 1] floors. If egg doesn't break, then we can check dp[m - 1][x] floors.
C++
// C++ program to find minimum number of trials in worst// case.#include <bits/stdc++.h>using namespace std;int minTrials(int n, int k){ // Initialize 2D of size (k+1) * (n+1). vector<vector<int> > dp(k + 1, vector<int>(n + 1, 0)); int m = 0; // Number of moves while (dp[m][n] < k) { m++; for (int x = 1; x <= n; x++) { dp[m][x] = 1 + dp[m - 1][x - 1] + dp[m - 1][x]; } } return m;}/* Driver code*/int main(){ cout << minTrials(2, 10); return 0;}// This code is contributed by Arihant Jain (arihantjain01) |
Output
4
Optimization to one-dimensional DP
The above solution can be optimized to one-dimensional DP as follows:
C++
// C++ program to find minimum number of trials in worst// case.#include <bits/stdc++.h>using namespace std;int minTrials(int n, int k){ // Initialize array of size (n+1) and m as moves. int dp[n + 1] = { 0 }, m; for (m = 0; dp[n] < k; m++) { for (int x = n; x > 0; x--) { dp[x] += 1 + dp[x - 1]; } } return m;}/* Driver code*/int main(){ cout << minTrials(2, 10); return 0;}// This code is contributed by Arihant Jain (arihantjain01) |
Output
4
Complexity Analysis:
- Time Complexity: O(n * log k)
- Auxiliary Space: O(n)
As an exercise, you may try modifying the above DP solution to print all intermediate floors (The floors used for minimum trial solution).
More Efficient Solution : Eggs dropping puzzle (Binomial Coefficient and Binary Search Solution)
Egg Dropping Puzzle with 2 Eggs and K Floors
2 Eggs and 100 Floor Puzzle
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References:
http://archive.ite.journal.informs.org/Vol4No1/Sniedovich/index.php
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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