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I was trying to do this problem, where given coins of certain denomination, I want to find the maximum number of coins to make change.

Example Say, I'm given coins of value 3 and 5, and I want to make change for 15, the solution would be {3,3,3,3,3} (Thanks JoSSte for pointing out) Similarly, say, given coins of value 3 and 5, and I want to make change for 7,I need to display "No solution possible"

I was able to do this for Finding Minimum number of coins as follows

import java.util.ArrayList;
import java.util.Arrays;


public class Minimum
{
    static int[] options = {5,3};

    public static void main(String[] args)
    {
        ArrayList<Integer> result = new ArrayList<Integer>();

        result = fun(15);
        if(result.size() == 999)
             System.out.println("Not possible to make change with this denomination");
        else 
        {
            for(int i = 0;i<result.size();i++)
                System.out.print(result.get(i));
        }
    }

    static ArrayList<Integer> fun(int n)
    {       
        if(n == 0)
        {
            ArrayList<Integer> totalret = new ArrayList<Integer>();
            return totalret;
        }

        if(n < 0)   
        {
            ArrayList<Integer> totalret  = new ArrayList<Integer>(Arrays.asList(new Integer[999]));
            return totalret;
        }

        ArrayList<Integer> totalret  = new ArrayList<Integer>(Arrays.asList(new Integer[999]));
        for(int i = 0;i<options.length;i++)
        {
            ArrayList<Integer> reci = fun(n-options[i]);

            ArrayList<Integer> reti = new ArrayList<Integer>();
            reti.addAll(reci);
            reti.add(options[i]);

            if(reti.size() < totalret.size())
                totalret = reti;
        }
        return totalret;
    }
}

Notice that I have a check called if(n<0) where combinations that do not add up to the sum are removed from the options by setting their size to a very large number that cannot be the minimum.

However, how can I modify the above to find the maximum number of coins

tubby
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    Wouldn't change for 15 be {3,3,3,3,3}? – JoSSte Sep 21 '15 at 04:52
  • @JoSSte, yes sorry. It's {3,3,3,3,3}. Editing it so as to not confuse people. – tubby Sep 21 '15 at 04:59
  • I would probably look at factoring. For each value find the biggest number divisible by the smallest coin denomination, and look at the remainder. If the remainder + the smallest coin denomination = a bigger denomination, you have a quick result. From there on it gets more complex, but at least it's a start – JoSSte Sep 21 '15 at 05:04

2 Answers2

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For your solution the you have to check the condition for n=1,2. if n=1,2 you can return ans as 999.

Your function must be as follow

 static ArrayList<Integer> fun(int n)
   {       
     if(n == 0)
     {
            ArrayList<Integer> totalret = new ArrayList<Integer>();
            return totalret;
        }

        if(n < 0 || n==1 || n==2)   
        {
            ArrayList<Integer> totalret  = new ArrayList<Integer>(Arrays.asList(new Integer[999]));
            return totalret;
        }

        ArrayList<Integer> totalret  = new ArrayList<Integer>(Arrays.asList(new Integer[999]));
        for(int i = 0;i<options.length;i++)
        {
            ArrayList<Integer> reci = fun(n-options[i]);

            ArrayList<Integer> reti = new ArrayList<Integer>();
            reti.addAll(reci);
            reti.add(options[i]);

            if(reti.size() < totalret.size())
                totalret = reti;
        }
        return totalret;
    }

Hope this should work..

Anuj Shah
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Here is coin change dp code in c++.

int coin[] = {3, 5}; //initialize array
int make;
int dp[2][100];

int call(int i, int amount)
{
     if(i >= 2) { // All coins have been taken
          if(amount == make) return 1;
          return 0;
     }

     if(dp[i][amount] != -1) return dp[i][amount];
     int ret1 = 0, ret2 = 0;
     // try to take coin i
     if(amount + coin[i] <= make) ret1 = call(i, amount + coin[i]); 

     ret2 = call(i+1, amount); // don't take coin i
     return dp[i][amount] = ret1 | ret2; // is possible or not?
}

int main()
{
    make = 7;
    memset(dp, -1, sizeof(dp));
    if(call(0, 0) == 0) cout << "not possible" << endl;
    else cout << "possible" << endl;

    return 0;
}
Rabiul Awal
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