## Epic Systems Interview Question

System Administrators```
public class Denomination {
public static void main(String[] args) {
// TODO Auto-generated method stub
double amountPaid;
double itemCost, amountReturned;
int index = 0;
double denominations[] = {10, 5, 1, .25, .1, .05, .01};
itemCost = .5;
amountPaid = 5;
amountReturned = amountPaid - itemCost;
System.out.println("Return a total of $" + amountReturned + ".");
while(amountReturned > 0) {
int num_denominations = (int) (amountReturned/denominations[index]);
amountReturned = amountReturned - (num_denominations*denominations[index]);
System.out.println("Number of $" + denominations[index] + " to be returned: " + num_denominations);
index++;
}
}
}
```

This code may appear correct to the eye, but the while loop, under some conditions, turn into an infinite loop causing an out of bounds exception. When working with decimals you really need to be cautious about precision. If you don't believe me, try running this program using .52 as the cost instead of .5.

Good work. Only correction is -

DecimalFormat twoDForm = new DecimalFormat("#.##");

amountReturned =

Double.valueOf(twoDForm.format(amountReturned - (num_denominations*denominations[index])));

//take input cost as float cost_f

//multiply by 100 to and cast to int

cost_i = (int) (cost_f*100) //100 since 1c (1/100 of $1) is least

//repeatedly divide by domination *100 to get respective values

tens = cost_i/(10*100); modulo_ten = cost_i%(10*100);

fives = modulo_ten/(5*100); modulo_five = modulo_ten%500;

//and so on.. till one_cents

```
/*
At a geeks goodies store.. every thing is either one dollar or less. and they accept only 10 dollars or less than 10 as denominations.So make a function that takes in the cost and payment as input and that out puts the change.
The denominations are namely 1c,5c,10c,25c,$1,$5,$10.
*/
#include<iostream>
using namespace std;
float coins[]={5,2,1,0.25,0.10,0.05,0.01};
#define TOTALCOINS 7
void coin_change(float cost,float payment)
{
if (payment<0.01 || cost<0.01){cout<<"Ivalid Input";}
else if (payment>10){cout<<"We only accept denominations upto 10 dollars \n";}
else if (cost>payment){cout<<"Payment is less than the cost. Please give $"<<cost<<endl;}
else if (cost==payment){cout<<"Payment complete. Thank you for visit.\n";}
else
{
int num_of_coins[TOTALCOINS]={0};
float to_return=payment-cost;
int index=0,c=0;
while (to_return>0 && index<TOTALCOINS)
{
c=(int)(to_return/coins[index]);
num_of_coins[index]=c;
to_return=to_return-(c*coins[index]);
index++;
}
for(int i=0;i<TOTALCOINS;i++)
{
cout<<"Number of $"<<coins[i]<<" coins= "<<num_of_coins[i]<<endl;
}
}
}
int main()
{
float cost,payment;
cout <<"Enter the cost\n";
cin>>cost;
cout<<"Enter the payment\n";
cin>>payment;
coin_change(cost,payment);
}
```

Python working code

```
"""
12:50
@Python 2.7
At a geeks goodies store.. every thing is either one dollar or less. and they accept only 10 dollars or less than 10 as denominations.So make a function that takes in the cost and payment as input and that out puts the change.
The denominations are namely 1c,5c,10c,25c,$1,$5,$10
- rohith on May 11, 2010 Report Duplicate | Flag
"""
class CoinChange(object):
def __init__(self, payment, cost):
if payment is None or cost is None:
print 'invalid inputs'
raise SystemExit
self._payment = payment * 100
self._cost = cost * 100
self._coins = [1000, 500, 100, 25, 10, 5, 1]
def getChange(self):
total = self._payment - self._cost
pos = 0
while total > 0:
# print total, '/', self._coins[pos]
tmp = int(total // self._coins[pos])
# print 'tmp = ', tmp
total = total - tmp * self._coins[pos]
print str(tmp) + ' X $' + str(self._coins[pos] / 100.0)
pos += 1
if __name__ == '__main__':
cc = CoinChange(10, 0.15)
cc.getChange()
```

1. Calculate the Payment-Cost, say X

- cuppanomics May 11, 20102. Start dividing X with largest denomination and pay the remainder number of denominations. X = X - (Remainder no. * denomination)

3. Repeat step 2 until X == 0

Say cost is 75 cents

Payment done is 10$

X = 9.25

X % 10 = 0 ( 0 * 10)

X % 5 = 1 (1 * 5) X = 9.25 - 5 = 4.25

X % 1 = 4 (4 * 1) X = 4.25 - 4.00 = .25

X % 25 = 1 (1 * .25) X = .25 - .25 = 0

X == 0