How to find all combinations of coins when given s

Question

I found a piece of code that I was writing for interview prep few months ago.

According to the comment I had, it was trying to solve this problem:

Given some dollar value in cents (e.g. 200 = 2 dollars, 1000 = 10 dollars), find all the combinations of coins that make up the dollar value. There are only pennies, nickels, dimes, and quarters allowed. (quarter = 25 cents, dime = 10 cents, nickel = 5 cents, penny = 1 cent)

For example, if 100 was given, the answer should be:

4 quarter(s) 0 dime(s) 0 nickel(s) 0 pennies  
3 quarter(s) 1 dime(s) 0 nickel(s) 15 pennies  
etc.

This can be solved in both iterative and recursive ways, I believe. My recursive solution is quite buggy, and I was wondering how other people would solve this problem. The difficult part of this problem was making it as efficient as possible.

Answer 1

The sub problem is a typical Dynamic Programming problem.

/* Q: Given some dollar value in cents (e.g. 200 = 2 dollars, 1000 = 10 dollars),
      find the number of combinations of coins that make up the dollar value.
      There are only penny, nickel, dime, and quarter.
      (quarter = 25 cents, dime = 10 cents, nickel = 5 cents, penny = 1 cent) */
/* A:
Reference: http://andrew.neitsch.ca/publications/m496pres1.nb.pdf
f(n, k): number of ways of making change for n cents, using only the first
         k+1 types of coins.

          +- 0,                        n < 0 || k < 0
f(n, k) = |- 1,                        n == 0
          +- f(n, k-1) + f(n-C[k], k), else
 */

#include <iostream>
#include <vector>
using namespace std;

int C[] = {1, 5, 10, 25};

// Recursive: very slow, O(2^n)
int f(int n, int k)
{
    if (n < 0 || k < 0)
        return 0;

    if (n == 0)
        return 1;

    return f(n, k-1) + f(n-C[k], k); 
}

// Non-recursive: fast, but still O(nk)
int f_NonRec(int n, int k)
{
    vector<vector<int> > table(n+1, vector<int>(k+1, 1));

    for (int i = 0; i <= n; ++i)
    {
        for (int j = 0; j <= k; ++j)
        {
            if (i < 0 || j < 0) // Impossible, for illustration purpose
            {
                table[i][j] = 0;
            }
            else if (i == 0 || j == 0) // Very Important
            {
                table[i][j] = 1;
            }
            else
            {
                // The recursion. Be careful with the vector boundary
                table[i][j] = table[i][j-1] + 
                    (i < C[j] ? 0 : table[i-C[j]][j]);
            }
        }
    }

    return table[n][k];
}

int main()
{
    cout << f(100, 3) << ", " << f_NonRec(100, 3) << endl;
    cout << f(200, 3) << ", " << f_NonRec(200, 3) << endl;
    cout << f(1000, 3) << ", " << f_NonRec(1000, 3) << endl;

    return 0;
}

Answer 2

# short and sweet with O(n) table memory    

#include <iostream>
#include <vector>

int count( std::vector<int> s, int n )
{
  std::vector<int> table(n+1,0);

  table[0] = 1;
  for ( auto& k : s )
    for(int j=k; j<=n; ++j)
      table[j] += table[j-k];

  return table[n];
}

int main()
{
  std::cout <<  count({25, 10, 5, 1}, 100) << std::endl;
  return 0;
}

Answer 3

Straightforward java solution:

public static void main(String[] args) 
{    
    int[] denoms = {4,2,3,1};
    int[] vals = new int[denoms.length];
    int target = 6;
    printCombinations(0, denoms, target, vals);
}


public static void printCombinations(int index, int[] denom,int target, int[] vals)
{
  if(target==0)
  {
    System.out.println(Arrays.toString(vals));
    return;
  }
  if(index == denom.length) return;   
  int currDenom = denom[index];
  for(int i = 0; i*currDenom <= target;i++)
  {
    vals[index] = i;
    printCombinations(index+1, denom, target - i*currDenom, vals);
    vals[index] = 0;
  }
}

Answer 4

This is a simple recursive algorithm that takes a bill, then takes a smaller bill recursively until it reaches the sum, it then takes another bill of same denomination, and recurses again. See sample output below for illustration.

var bills = new int[] { 100, 50, 20, 10, 5, 1 };

void PrintAllWaysToMakeChange(int sumSoFar, int minBill, string changeSoFar)
{
    for (int i = minBill; i < bills.Length; i++)
    {
        var change = changeSoFar;
        var sum = sumSoFar;

        while (sum > 0)
        {
            if (!string.IsNullOrEmpty(change)) change += " + ";
            change += bills[i];

            sum -= bills[i]; 
            if (sum > 0)
            {
                PrintAllWaysToMakeChange(sum, i + 1, change);
            }
        }

        if (sum == 0)
        {
            Console.WriteLine(change);
        }
    }
}

PrintAllWaysToMakeChange(15, 0, "");

Prints the following:

10 + 5
10 + 1 + 1 + 1 + 1 + 1
5 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
5 + 5 + 1 + 1 + 1 + 1 + 1
5 + 5 + 5
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

Answer 5

PHP Code to breakdown an amount into denominations:

//Define the denominations    
private $denominations = array(1000, 500, 200, 100, 50, 40, 20, 10, 5, 1);
/**
 * S# countDenomination() function
 * 
 * @author Edwin Mugendi <[e-mail]>
 * 
 * Count denomination
 * 
 * @param float $original_amount Original amount
 * 
 * @return array with denomination and count
 */
public function countDenomination($original_amount) {
    $amount = $original_amount;
    $denomination_count_array = array();
    foreach ($this->denominations as $single_denomination) {

        $count = floor($amount / $single_denomination);

        $denomination_count_array[$single_denomination] = $count;

        $amount = fmod($amount, $single_denomination);
    }//E# foreach statement

    var_dump($denomination_count_array);
    return $denomination_count_array;
    //return $denomination_count_array;
}

//E# countDenomination() function

Answer 6

The code is using Java to solve this problem and it also works... This method may not be a good idea because of too many loops, but it's really a straight forward way.

public class RepresentCents {

    public static int sum(int n) {

        int count = 0;
        for (int i = 0; i <= n / 25; i++) {
            for (int j = 0; j <= n / 10; j++) {
                for (int k = 0; k <= n / 5; k++) {
                    for (int l = 0; l <= n; l++) {
                        int v = i * 25 + j * 10 + k * 5 + l;
                        if (v == n) {
                            count++;
                        } else if (v > n) {
                            break;
                        }
                    }
                }
            }
        }
        return count;
    }

    public static void main(String[] args) {
        System.out.println(sum(100));
    }
}