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43203 / 60 = 720 chocolates ( 3 dollars left and you cannot buy any more chocolate for 3 dollar )
720 / 5 = 144 chocolates ( we can buy 144 chocolates from 720 wrappers )
144 / 5 = 28 chocolates ( we can buy 28 chocolates from 144 wrappers, 4 wrappers left )
(28 + 4 ) / 5 = 6 ( 28 wrappers + 4 wrappers left from 144, we can buy 6 chocolates )
(6 + 2) / 5 = 1 ( 6 wrappers + 2 wrappers left from 32 wrappers, we can buy 1 chocolate )
This sums to 720 + 144 + 28 + 6 + 1 = 899.
this kinda worked
import java.io.;
import java.util.;
import java.text.;
import java.math.;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for(int a0 = 0; a0 < t; a0++){
int n = in.nextInt();
int c = in.nextInt();
int m = in.nextInt();
int totalChoc = n / c;
int totalWrappers = totalChoc;
int freeChocs = 0;
while (m <= totalWrappers) {
int wrapperTogive=totalWrappers-(totalWrappers%m) ;
freeChocs = totalWrappers / m;
totalWrappers = (totalWrappers - wrapperTogive) + freeChocs;
totalChoc=totalChoc + freeChocs;
}
System.out.println(totalChoc);
}
}
I did something similar. As several others have noted, it's an inefficient approach, but it was really quick to write and handled all the test cases so I considered it fit for purpose.
int chocolateFeast(int n, int c, int m) {
int i,sum,x,j;
sum=n/c;
x=sum;
if(sum==0)
return 0;
while(x>=m)
{
j=x%m;
i=x/m;
sum=sum+i;
x=i+j;
}
return sum;
}
thanks. I had a terrible time with this problem. maybe it was the noise in my environment, maybe it was something to do with the computer i'm using... I couldn't get it exactly right untill I read this. Thanks for the code.
Chocolate Feast
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Same problem with my code also.
thaks I miss that case
check out this code .. https://www.hackerrank.com/challenges/chocolate-feast/submissions/code/20066920
this kinda worked import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;
public class Solution {
}
I guess you are doing way to much unnecessary work. Heree is my while loop
This approach is clean and easy to follow, which is great.
However, it results in O(N) time (input of 10000 1 2 results in the loop running 9999 times.)
If you switch the loop to the following you end up with O( lg(N)) (or 14 in the worst case, since you are using integer math).
Hi Jason I kind of did the same thing what you did but mine didn't pass the test cases how whats the difference
import time
start_time = time.time()
def chocolateFeast1(he_has,cost,wrapper):
start_time = time.time() def chocolateFeast(cash, cost, wrapper_cost):
wrappers = wrappers % wrapper_cost instead of wrappers -= new_chocolate * wrapper_cost
was better understandable for me.
wrappers %= wrapper_cost would be even shorter
thanks.anyways, i got it.
Your code is flawless. Quick, simple and completely readable! Congratulations!
see this
def chocolateFeast(n, c, m):
your code will result in performance issues when the count is too high. Thats the reason suggested to do with / and % operators
I did something similar. As several others have noted, it's an inefficient approach, but it was really quick to write and handled all the test cases so I considered it fit for purpose.
int chocolateFeast(int n, int c, int m) { int i,sum,x,j; sum=n/c; x=sum; if(sum==0) return 0; while(x>=m) { j=x%m; i=x/m; sum=sum+i; x=i+j; } return sum; }
there is problem with ur code. 899 is correct.
any idea if 3 dollars left are carry to next trip?
thanks. I had a terrible time with this problem. maybe it was the noise in my environment, maybe it was something to do with the computer i'm using... I couldn't get it exactly right untill I read this. Thanks for the code.
Why does this time out?
try this out with time comlexity less than O(n) :)
can you explain why my code is not giving output for testcase 8 only.
def chocolateFeast(n, c, m): inichoc=n//c total=inichoc leftwrap=0 i=0 while inichoc+leftwrap>=m: addchoc=math.floor((inichoc+leftwrap)/m) leftwrap=(inichoc-i)%m total+=addchoc inichoc=addchoc i+=1 return total
What is the complexity of this? I have the same solution but I am not sure what my complexity is, I find it a bit difficult. :)
try this
def chocolateFeast(n, c, m):
GOT IT MAN
THANKS