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.
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 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
// 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 |
Python3
# 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#
// 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 |
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 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> |
Output
8
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 floorsint 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 codeint 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 floorsdef 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 xn = 2k = 36print(eggDrop(n, k)) |
Javascript
// 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(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 caselet n = 2, k = 36;console.log(eggDrop(n, k)); |
Output
8
Time Complexity: O(N * K)
Auxiliary Space: O(N)
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