GFG POTD: Egg dropping puzzle solution (Dynamic Programming)

Problem Statement

You are given N identical eggs and you have access to a K-floored building from 1 to K.

There exists a floor f where 0 <= f <= such that any egg dropped at a floor higher than f will break, and any egg dropped at or below floor will not break. There are few rules given below. 

  • 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 the egg doesn’t break at a certain floor, it will not break at any floor below.
  • If the eggs breaks at a certain floor, it will break at any floor above.

Return the minimum number of moves that you need to determine with certainty what the value of f is.

Example 1:

N = 1, K = 2
Output: 2
1. Drop the egg from floor 1. If it 
   breaks, we know that f = 0.
2. Otherwise, drop the egg from floor 2.
   If it breaks, we know that f = 1.
3. If it does not break, then we know f = 2.
4. Hence, we need at minimum 2 moves to
   determine with certainty what the value of f is.

Example 2:

N = 2, K = 10
Output: 4

Your Task:
Complete the function eggDrop() which takes two positive integer N and K as input parameters and returns the minimum number of attempts you need in order to find the critical floor.

Expected Time Complexity : O(N*(K^2))
Expected Auxiliary Space: O(N*K)



int dp[202][202]; //Creating matrix of size 202*202
class Solution
    //Function to find minimum number of attempts needed in 
    //order to find the critical floor.
    //optimize the worst case
    //best of worst case
    int solve(int e, int f){
        return f;
        if(dp[e][f]!=-1) return dp[e][f];
        int minMoves=INT_MAX;
        for(int k=1; k<=f; k++){
            int broke=1+solve(e-1,k-1);
            int unbroke=1+solve(e,f-k);
            int temp=max(broke,unbroke);
            minMoves = min(minMoves, temp);
        return dp[e][f]=minMoves;
    int eggDrop(int n, int k) 
        memset(dp,-1,sizeof(dp)); //initializing matrix with -1
        return solve(n,k);
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