Find the least number of coins required that can make any change from 1 to 99 cents

Question

Recently I challenged my co-worker to write an algorithm to solve this problem:

Find the least number of coins required that can make any change from

Answer 1

This might be a generic solution in C#

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace CoinProblem
{
    class Program
    {
        static void Main(string[] args)
        {
            var coins = new int[] { 1, 5, 10, 25 }; // sorted lowest to highest
            int upperBound = 99;
            int numCoinsRequired = upperBound / coins.Last();
            for (int i = coins.Length - 1; i > 0; --i)
            {
                numCoinsRequired += coins[i] / coins[i - 1] - (coins[i] % coins[i - 1] == 0 ? 1 : 0);
            }
            Console.WriteLine(numCoinsRequired);
            Console.ReadLine();
        }
    }
}

I haven't fully thought it through...it's too late at night. I thought the answer should be 9 in this case, but Gabe said it should be 10... which is what this yields. I guess it depends how you interpret the question... are we looking for the least number of coins that can produce any value <= X, or the least number of coins that can produce any value <= X using the least number of coins? For example... I'm pretty sure we can make any value with only 9 coins, without nickels, but then to produce 9... you need...oh... I see. You'd need 9 pennies, which you don't have, because that's not what we chose... in that case, I think this answer is right. Just a recursive implementation of Thomas' idea, but I don't know why he stopped there.. you don't need to brute force anything.

Edit: This be wrong.

Answer 2

Came across this one today, while studying https://www.coursera.org/course/bioinformatics

DPCHANGE(money, coins)
 MinNumCoins(0) ← 0
 for m ← 1 to money
        MinNumCoins(m) ← ∞
        for i ← 1 to |coins|
            if m ≥ coini
                if MinNumCoins(m - coini) + 1 < MinNumCoins(m)
                    MinNumCoins(m) ← MinNumCoins(m - coini) + 1
    output MinNumCoins(money)

Takes a comma-separated string of denominations available, and the target amount.

C# implementation:

    public static void DPCHANGE(int val, string denoms)
    {
        int[] idenoms = Array.ConvertAll(denoms.Split(','), int.Parse);
        Array.Sort(idenoms);
        int[] minNumCoins = new int[val + 1];

        minNumCoins[0] = 0;
        for (int m = 1; m <= val; m++)
        {
            minNumCoins[m] = Int32.MaxValue - 1;
            for (int i = 1; i <= idenoms.Count() - 1; i++)
            {
                if (m >= idenoms[i])
                {
                    if (minNumCoins[m - idenoms[i]] + 1 < minNumCoins[m])
                    {
                        minNumCoins[m] = minNumCoins[m - idenoms[i]] + 1;
                    }
                }
            }
        }
    }

Answer 3

Example Program:

#include<stdio.h> 

    #define LEN 9 // array length
    int main(){
        int coins[LEN]={0,0,0,0,0,0,0,0,0}; // coin count
        int cointypes[LEN]={1000,500,100,50,20,10,5,2,1}; // declare your coins and note here {ASC order}   
        int sum =0; //temp variable for sum
        int inc=0; // for loop
        int amount=0; // for total amount
        printf("Enter Amount :");
        scanf("%d",&amount);
        while(sum<amount){
            if((sum+cointypes[inc])<=amount){
                   sum = sum+  cointypes[inc];
                    //printf("%d[1] - %d\n",cointypes[inc],sum);
                    //switch case to count number of notes and coin
                   switch(cointypes[inc]){
                    case 1000:
                           coins[0]++;
                           break;
                    case 500:
                           coins[1]++;
                           break;
                    case 100:
                           coins[2]++;
                           break;
                    case 50:
                           coins[3]++;
                           break;               
                    case 20:
                           coins[4]++; 
                           break;
                    case 10:
                           coins[5]++;
                           break;
                    case 5:
                           coins[6]++;
                           break;
                    case 2:
                           coins[7]++;
                           break;
                    case 1:
                           coins[8]++;
                           break;
                       }
                }else{
                   inc++;
                }
            }
        printf("note for %d in\n note 1000 * %d\n note 500 * %d\n note 100 * %d\n note 50 * %d\n note 20 * %d\n note 10 * %d\n coin 5 * %d\n coin 2 * %d\n coin 1 * %d\n",amount,coins[0],coins[1],coins[2],coins[3],coins[4],coins[5],coins[6],coins[7],coins[8]); 

    }