# C Program for Egg Dropping Puzzle - DP-11?

This is a famous puzzle problem. Suppose there is a building with n floors, if we have m eggs, then how can we find the minimum number of drops needed to find a floor from which it is safe to drop an egg without breaking it.

There some important points to remember −

• When an egg does not break from a given floor, then it will not break for any lower floor also.
• If an egg breaks from a given floor, then it will break for all upper floors.
• When an egg breaks, it must be discarded, otherwise we can use it again.

Input - The number of eggs and the maximum floor. Say the number of eggs are 4 and the maximum floor is 10.

Output - Minimum number of trials  4.

## Algorithm

### eggTrialCount(eggs, floors)

Input − Number of eggs, maximum floor.

Output − Get minimum number of trials.

Begin
define matrix of size [eggs+1, floors+1]
for i:= 1 to eggs, do
minTrial[i, 1] := 1
minTrial[i, 0] := 0
done
for j := 1 to floors, do
minTrial[1, j] := j
done
for i := 2 to eggs, do
for j := 2 to floors, do
minTrial[i, j] := ∞
for k := 1 to j, do
res := 1 + max of minTrial[i-1, k-1] and minTrial[i, j-k]
if res < minTrial[i, j], then minTrial[i,j] := res
done
done
done
return minTrial[eggs, floors]
End

## Example

Live Demo

#include<stdio.h>
#define MAX_VAL 9999
int max(int a, int b) {
return (a > b)? a: b;
}
int eggTrialCount(int eggs, int floors) { //minimum trials for worst case
int minTrial[eggs+1][floors+1]; //to store minimum trials for i-th egg
and jth floor
int res, i, j, k;
for (i = 1; i <= eggs; i++) { //one trial to check from first floor, and
no trial for 0th floor
minTrial[i] = 1;
minTrial[i] = 0;
}
for (j = 1; j <= floors; j++) //when egg is 1, we need 1 trials for
each floor
minTrial[j] = j;
for (i = 2; i <= eggs; i++){ //for 2 or more than 2 eggs
for (j = 2; j <= floors; j++) { //for second or more than second
floor
minTrial[i][j] = MAX_VAL;
for (k = 1; k <= j; k++) {
res = 1 + max(minTrial[i-1][k-1], minTrial[i][j-k]);
if (res < minTrial[i][j])
minTrial[i][j] = res;
}
}
}
return minTrial[eggs][floors]; //number of trials for asked egg and
floor
}
int main () {
int egg, maxFloor;
printf("Enter number of eggs: ");
scanf("%d", &egg);
printf("Enter max Floor: ");
scanf("%d", &maxFloor);
printf("Minimum number of trials: %d", eggTrialCount(egg, maxFloor));
}

## Output

Enter number of eggs: 4
Enter max Floor: 10
Minimum number of trials: 4