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Case No. 5, input string is "43203 60 5", my output is "892", right answer is "899"
I just want to make sure whether there is a mistake at your end. Kindly check, I've checked plenty of time, If not Kindly explain the case? Much appreciaited.
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.
There is no doubt about the veracity of the answers. Do not falter. Try again. I too faced the same issue. Try to code it exactly as the problem has been described.
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 chocs = n/c; //chocolates bought with Money
int wrapper = chocs; //wrappers from chocolate
int ex_chocs = wrapper/m; //chocolates from wrappers
while(wrapper>=m){
ex_chocs = ex_chocs+(wrapper/m + wrapper%m)/m;
wrapper = wrapper/m;
}
System.out.println(chocs+ex_chocs);
}
}
Python also allows that
for _ in range(int(input())):n,c,m=map(int,input().split());print(n//c+(n//c-1)//(m-1))
But for some people is no more oneline because it cant used in another line and produce no return value
First, your total (initial) number of chocolates, and wrappers, is equal to what you can buy with cash, which is n/c. Now you have to iteratively give your wrappers to the shop keeper, in exchange for chocolates, costing you m wrappers (wrappers / m). During this iterative process you add the chocolates to your total count. You subtract the number of wrappers you have handed in and also add back the number of wrappers you receive from the chocolates you are getting back. Here's a java 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 total = n/c;
int wrappers = total;
while(wrappers >= m) {
total += (wrappers / m);
int leftover = wrappers % m;
wrappers = (wrappers / m) + leftover;
}
System.out.println(total);
}
}
static int cc;
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int t,n,c,m,r;
t = in.nextInt();
while(t-->0){
n = in.nextInt();
c = in.nextInt();
m = in.nextInt();
r=n/c;
cc=r;
while(r>=m){
cc=cc+r/m;
r=r%m+r/m;
}
System.out.println(cc);
}
}
I am (currently) just on hackerranks to have "exercises" to learn java/programming with and look at neat solutions and tips after getting the solution my own way and see if i can improve it.
I loved the neatness of the above solution and thought wow that must run so much faster than my solution, so i submitted them one after another a few times only to see my solution outperform the above.
Why?
Is the remainder operator in the above solution really so expensive and my jumble of code faster just because it doesn't use it?
import java.io.;
import java.util.;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = Integer.parseInt(sc.nextLine()); //t = number of test cases
int[][] tc = readInput(sc, t); //tc[t][0] = money. tc[t][1] = price. tc[t][2] = wrappers per free bar
for (int i = 0; i<t; i++){ //loop for all test cases
int choc = calcChoc(tc,i); //work out how much choc can be bought
System.out.println(choc); //print result for the test case
}
}
//calculate how much choc he can buy with m $ at p price with w wrappers needed for a free bar
public static int calcChoc(int[][] tc,int i){
int m = tc[i][0]; //money he has
int p = tc[i][1]; //price of choc
int w = tc[i][2]; //wrappers per free bar
int bars = m/p; //how many bars he can buy initially
int wrappers = bars; //each bar is a wrapper from initial purpose
//loop to turn in all wrappers while it is possible to do so
while (w<=wrappers){
int barsFromTurnIn = wrappers/w; //bars from turning in current wrappers.
bars = bars + barsFromTurnIn; //new bar count
wrappers = wrappers - (barsFromTurnIn * (w-1)); //wrapper count reduced by amount of wrappers turned in -1 wrapper per bar recieved from turn in.
if (w==1){ //break out of infinite loop when you get 1 bar for 1 wrapper!
System.out.print("Infinite Bars, exiting infinite loop at bars = ");
break;
}
}
return bars;
}
//read input for each test case and make 2d array of the info
public static int[][] readInput(Scanner sc, int t){
int[][] input = new int[t][3];
for (int i = 0; i<t; i++){
String[] inputLine = sc.nextLine().split(" ");
input[i][0] = Integer.parseInt(inputLine[0]);
input[i][1] = Integer.parseInt(inputLine[1]);
input[i][2] = Integer.parseInt(inputLine[2]);
}
return input;
}
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 s = n/c;
int wrapper = n/c;
while(wrapper-m>=0){
wrapper-=(m-1);
s++;
}
System.out.println(s);
}
}
The trick resides in while you are having a free chocolate, you have another more changeable wrapper. So, while you have wrappers, if you can change it for another choco (wrapper-m>=0), decrease your current wrappers as much as you can deliver for a new choco minus 1. That minus 1 means that you will have another more wrapper coming from your new free choco =)
void main() {
int numChocs = n / c;
if (numChocs >= m) {
numChocs = numChocs + test(numChocs, m);
}
System.out.println(numChocs);
}
static int test(int wraps, final int m) {
int remainder = wraps % m;
int numChocs = wraps / m;
if (numChocs + remainder >= m) {
numChocs = numChocs + test(numChocs + remainder, m);
}
return numChocs;
}
Bobby uses money on the first iteration only. So, after 'n //= c' operator there are only two numbers in use: n - number of bars/wrappers Bobby has at the beginning, m - price of the promotional bar in wrappers.
We can get rid of iterations when understanding that the chockolate bar "costs" not m wrappers, but m-1, because Bobby has one additional wrapper for every m of them.
So the promotional bars number can be calculated without 'for/while' operator, simply dividing wrapper number by m-1.
But Bobby can't spend all his wrappers. At least one wrapper remains. Thus the final formula is
result = n [bars count bought for money] + (n-1)[wrappers count except one wrapper] // (m-1)
// Complete the chocolateFeast function below.
static int chocolateFeast(int n, int c, int m)
{
int count=0;
count=n/c;
int chocolate=count;
while(chocolate>=m)
{
chocolate=chocolate-m+1;
count++;
}
return count;
}
Chocolate Feast
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Only 1 output worng among all cases?
Case No. 5, input string is "43203 60 5", my output is "892", right answer is "899" I just want to make sure whether there is a mistake at your end. Kindly check, I've checked plenty of time, If not Kindly explain the case? Much appreciaited.
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
I had the same problem, but I resolved using another approach. Tip while :)
There is no doubt about the veracity of the answers. Do not falter. Try again. I too faced the same issue. Try to code it exactly as the problem has been described.
I tried again :-)
import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;
public class Solution {
}
@pankaj113, You can simplify/reduce your code by moving this line
inside while loop, like so:
@cesarjom Elegant and straight-forward. Thanks for sharing.
just th tweak i needed
can be further simplified
check out this .. https://www.hackerrank.com/challenges/chocolate-feast/submissions/code/20066920
Against my better judgement, wanting a one line answer I got stuck here :D Hahahah
Yeah, I cannot see a way to make this a one-liner, can you?
Dude! This was amazing. Do you think you can break it to be able to understand?
Here's your one-liner
It is the same solution. And it is 100 times less readable for the sake of two lines of code.
P.S. Most of languages, other than Python allow converting programs into one line simply by deleting the newline characters. But nobody does that.
Python also allows that
for _ in range(int(input())):n,c,m=map(int,input().split());print(n//c+(n//c-1)//(m-1))But for some people is no more oneline because it cant used in another line and produce no return valuecheck my link for solution..its in python http://ideone.com/Al2fEX
Following code works for me:
wrappers = n/c;
count = wrappers;
while (wrappers >= m) {
wrappers = wrappers - m;
count = count + 1;
wrappers = wrappers + 1;
}
System.out.println (count);
I couldn't figure out the algo. Pls help.
First, your total (initial) number of chocolates, and wrappers, is equal to what you can buy with cash, which is n/c. Now you have to iteratively give your wrappers to the shop keeper, in exchange for chocolates, costing you m wrappers (wrappers / m). During this iterative process you add the chocolates to your total count. You subtract the number of wrappers you have handed in and also add back the number of wrappers you receive from the chocolates you are getting back. Here's a java solution.
That's exactly what I did.
That's basically what I did, except in Python.
A better version:
Similar here without iterating every bar/wrapper swap
Same as mine!
i dont get it, why sum plus one on wrappers at the end of the loop... ? that was my mistake...
every time you subtract m wrappers you get one chocolate so one more wrapper so you add 1
You are just right brother
static int cc; public static void main(String[] args) {
}
I am (currently) just on hackerranks to have "exercises" to learn java/programming with and look at neat solutions and tips after getting the solution my own way and see if i can improve it.
I loved the neatness of the above solution and thought wow that must run so much faster than my solution, so i submitted them one after another a few times only to see my solution outperform the above. Why? Is the remainder operator in the above solution really so expensive and my jumble of code faster just because it doesn't use it?
import java.io.; import java.util.;
public class Solution {
}
Could you explain your code?
+1 Used the exact same logic with c++
+1 Used the exact same logic with c++
For checking the visualisation of your code... TRY...
pythontutor.com
You can check the test cases and exact bug in your code. :-)
It works for me:
The trick resides in while you are having a free chocolate, you have another more changeable wrapper. So, while you have wrappers, if you can change it for another choco (wrapper-m>=0), decrease your current wrappers as much as you can deliver for a new choco minus 1. That minus 1 means that you will have another more wrapper coming from your new free choco =)
Why not ?
Why not ?
This Code Passes all the test cases and quite clear with variable names. https://www.hackerrank.com/challenges/chocolate-feast/submissions/code/21403363
This solution works:
include
include
include
include
include
include
include
int main() { int t,i; scanf("%d",&t); for(i=1;i<=t;i++) { int n,c,m,ans,x,sum=0; scanf("%d %d %d",&n,&c,&m); ans=n/c; x=ans; while(x>=m) { sum=sum+(x/m); x=(x/m)+(x%m);
}
we need to decrement number of wrappers as we use them
try this -
simple solution in c++
thanks broo
can you please explain how this works ?
So, how to solve the problem in two operators?
Bobby uses money on the first iteration only. So, after 'n //= c' operator there are only two numbers in use: n - number of bars/wrappers Bobby has at the beginning, m - price of the promotional bar in wrappers.
We can get rid of iterations when understanding that the chockolate bar "costs" not m wrappers, but m-1, because Bobby has one additional wrapper for every m of them. So the promotional bars number can be calculated without 'for/while' operator, simply dividing wrapper number by m-1.
But Bobby can't spend all his wrappers. At least one wrapper remains. Thus the final formula is
result = n [bars count bought for money] + (n-1)[wrappers count except one wrapper] // (m-1)
try out this
can you please explain this ?
right ans is 898
simpliest solution for any language solved for all test cases https://www.hackerrank.com/challenges/chocolate-feast/forum/comments/484679
me too. only test case5
You can check this out
https://www.hackerrank.com/challenges/chocolate-feast/forum/comments/561503
public class Solution {