The Coin Change Problem


Problem Statement :


Given an amount and the denominations of coins available, determine how many ways change can be made for amount. There is a limitless supply of each coin type.

Example
n = 3
c = [8,3,1,2]
There are 3 ways to make change for n=3: {1,1,1}, {1,2}, and {3}.

Function Description

Complete the getWays function in the editor below.

getWays has the following parameter(s):

   int n: the amount to make change for
   int c[m]: the available coin denominations

Returns

   int: the number of ways to make change

Input Format

The first line contains two space-separated integers n and m, where:
n is the amount to change
m is the number of coin types
The second line contains m space-separated integers that describe the values of each coin type.

Constraints

   1 <= c[i] <= 50
   1 <= n <= 250
   1 <= m <= 50
   Each c[i] is guaranteed to be distinct.


Solution :



title-img 


                            Solution in C :

In C++ :




#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>

using namespace std;


int main() {
	vector<int> vec;
	int n;
	
	string str;
	getline(cin, str);
	
	const char * temp = str.c_str();	
	char * cstr = new char[10000];
	strcpy(cstr, temp);
	
	cstr = strtok(cstr, ", ");

	while(cstr!=NULL) {
		vec.push_back(atoi(cstr));
		cstr = strtok(NULL, ", ");
	}

	cin>>n;

	sort(vec.begin(), vec.end());
	int M = vec.size();
	vector< vector<int> > mat(M, vector<int>(n+1, 0));
	for(int i=0; i<M; i++)
		mat[i][0] = 1;
	for(int i=1; i<=n; i++)
		mat[0][i] = (i%vec[0]==0) ? 1 : 0;

	for(int i=1; i<M; i++) {
		for(int j=1; j<=n; j++) {
			mat[i][j]=mat[i-1][j];
			if(j>=vec[i])
				mat[i][j]+=mat[i][j-vec[i]];
		}
	}

	cout << mat[M-1][n] << endl;
	return 0;
}









In Java :




import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    
    public static int ways(int[] a, int n){
        int len = a.length;
        int[] dp = new int[n+1];
        dp[0] = 1;
        
        for(int i=0;i<len;i++)
            for(int j=a[i];j<=n;j++)
                dp[j]+=dp[j-a[i]];
        
        return dp[n];
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String line = sc.nextLine();
        int n = Integer.parseInt(sc.nextLine().replace(" ",""));
        line = line.replace(" ","");
        String[] temp = line.split(",");
        int[] coins = new int[temp.length];
        for(int i=0;i<temp.length;i++)
            coins[i] = Integer.parseInt(temp[i]);
        
        System.out.println(ways(coins, n));
    }
}









In c :





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() 
{
    char s[500];
    int a[50],k=0,i,j,n,l,x;
    gets(s);
    scanf("%d",&n);
    static int t[251];
    l=strlen(s);
    for(i=0;i<l;i++)
    {
        x=0;
        if(s[i]<='9'&&s[i]>='0')
            x=x*10+s[i]-'0';
        i++;
        if(i<n&&s[i]<='9'&&s[i]>='0'){
            x=x*10+s[i++]-'0';}
        i++;
            a[k++]=x;
    }
    t[0]=1;
    for(i=0;i<k;i++)
        for(j=a[i];j<=n;j++)
            t[j]+=t[j-a[i]];
    printf("%d",t[n]);
    return 0;
}









In Python3 :





cs = [int(i) for i in input().strip().split(',')]  # coin denominations
n = int(input())  # value for which we shall make change
ways = [0 for i in range(n + 1)]
ways[0] = 1

for i in range(len(cs)):
    for j in range(cs[i], n + 1):
        ways[j] += ways[j - cs[i]]

print(ways[n])








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