Skip to main content

Given a set of coin denominators, find the minimum number of coins to give a certain amount of change.

File: Denominator.java
-------------------------------------------------------------
package org.developersbrain.solutions;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.Properties;

public class Denominator {
String currency;
Properties prop;
InputStream inS;
Denominator(){
currency = "INR";//Default Currency
prop     = null;
}

Denominator(String currency) throws IOException{
this.currency = currency;
prop = new Properties();
inS  = new FileInputStream("DenominatorSpec.properties");
prop.load(inS);
}

String returnDenominations(int amount){
String strDenom="";
String strAmount=""+amount;
int    denom = 0,denomOther=0,tmpDenom=0;
String denomValue;
int d=0;
int m=1;
for(int i=0,j=1;i<strAmount.length();i++){
denom = amount%10;
denomValue=prop.getProperty(currency+"."+m);
if(denomValue != null){
if(denom>0 && prop.getProperty(currency+"."+(m*10))!=null){
strDenom=strDenom+("\n"+m+" * "+denom);
}
tmpDenom=denom;
m=m*10;
}else{
denomOther=denomOther+(denom*j);
j=j*10;
}
amount=amount/10;
}
if(tmpDenom>0 && !strDenom.equals("\n"+(m/10)+" * "+(tmpDenom+denomOther))){
if(denomOther>0){
denomOther=denomOther*10;
}
strDenom= strDenom+"\n"+(m/10)+" * "+(tmpDenom+denomOther);
}

return strDenom+"\n"+currency;
}



}

File;MainClass.java
-------------------------
package org.developersbrain.solutions;

import java.io.IOException;

public class MainClass {
public static void main(String[] args) throws IOException{
Denominator dn=new Denominator("JPY");
System.out.println(dn.returnDenominations(12345));
}
}

Properties File: DenominatorSpec.properties
------------------------------------------------------------------

INR.1=1
INR.2=2
INR.5=5
INR.10=10
INR.20=20
INR.50=50
INR.100=100
INR.500=500
INR.1000=1000
JPY.1=1
JPY.5=5
JPY.10=10
JPY.50=50
JPY.100=100
JPY.500=500
JPY.1000=1000
JPY.5000=5000
JPY.10000=10000


Program Output:
---------------------------------------------------------------------------
1       * 5
10       * 4
100       * 3
1000       * 2
10000       * 1
JPY


Comments

Post a Comment

Popular posts from this blog

Java Interface

Problem Statement A Java interface can only contain method signatures and fields. Interface can be used to achieve polymorphism. In this problem you will practice your knowledge on interfaces. You are given an interface   AdvancedArithmetic   which contains a method signature   public abstract int divisorSum(int n) . You need to write a class called MyCalculator which implements the interface. divisorSum   function just takes an integer as input and return the sum of all its divisors. For example divisors of 6 are 1,2,3 and 6, so   divisorSum   should return 12. Value of n will be at most 1000. Read the partially completed code in the editor and complete it. You just need to write the MyCalculator class only.   Your class shouldn't be public. Sample Input 6 Sample Output I implemented: AdvancedArithmetic 12 Explanation Divisors of 6 are 1,2,3 and 6. 1+2+3+6=12. import java.util.*; interface AdvancedArithmetic{   p...
Post a Comment
Read more

change directory (cd) function for an abstract file system ( Java Implementation )

Write a function that provides change directory (cd) function for an abstract file system. Notes: Root path is '/'. Path separator is '/'. Parent directory is addressable as "..". Directory names consist only of English alphabet letters (A-Z and a-z). For example, new Path("/a/b/c/d").cd("../x").getPath() should return "/a/b/c/x". Note: The evaluation environment uses '\' as the path separator. public class Path {     private String path;     public Path(String path) {         this.path = path;     }     public String getPath() {         return path;     }     public Path cd(String newPath) {         //throw new UnsupportedOperationException("Waiting to be implemented."); String[] newP=newPath.split("/");     String[] oldP=path.split("/");     int lnCount=0;     for(String str:newP){     if(st...
6 comments
Read more

Calculate the number of elements of an array that are not divisors of each element.

Task description You are given a non-empty zero-indexed array A consisting of N integers. For each number A[i] such that 0 ≤ i < N, we want to count the number of elements of the array that are not the divisors of A[i]. We say that these elements are non-divisors. For example, consider integer N = 5 and array A such that: A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 3 A[4] = 6 For the following elements: A[0] = 3, the non-divisors are: 2, 6, A[1] = 1, the non-divisors are: 3, 2, 3, 6, A[2] = 2, the non-divisors are: 3, 3, 6, A[3] = 3, the non-divisors are: 2, 6, A[6] = 6, there aren't any non-divisors. Write a function: class Solution { public int[] solution(int[] A); } that, given a non-empty zero-indexed array A consisting of N integers, returns a sequence of integers representing the amount of non-divisors. The sequence should be returned as: a structure Results (in C), or a vector of integers (in C++), or a record Results (in Pascal), or...
Post a Comment
Read more