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Coin Change

In this article, we are going to see how to solve the coin change problem? Which can be solved using dynamic programming concept.
Submitted by Radib Kar, on March 22, 2019

Problem statement:

Given a value N, find the number of ways to make change for N cents, if we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins. The order of coins doesn't matter.

Example:

    Input:
    N = 4 
    S = {1, 2, 3} //infinite number of 1 cent, 2 cent, 3 cent coins

    Output:
    4
    {1,1,1,1} //four 1 cent coins
    {1,1,2} //two 1 cent coins, one 2 cent coins
    {2,2} //two 2 cent coins 
    {1,3} //one 1 cent and one 3 cent coins
    Thus total four ways (Order doesn’t matter)

Solution

Let's think of the solution. Let’s do it by hand first. An intuitive idea can be two checks whether the amount can be handled by the same valued coins.

Like 4 cent can be managed by four 1 cent coins.

Same time can be managed by two 2 cent coins.

For the rest, we need to take different valued coins.

Now we can think of memorization. We can simply store the sub-amounts that can have been managed still.

Like two 1 cent coin manage 2 cents. Thus 2 cent is our sub amount.

Now we pick two more 1 cent coins that will sum up to 4.

Else we can pick only one 2cent coin also summing up to 4.

This memorized approach helps us to solve the problem.

Let's revise the algorithm

Pre-requisite: Coins array, amount

Algorithm:

    1.  Create a DP table of size amount
        Table[amount+1]={0};
    2.  Base case table[0]=1
    3.  For each coins[i] from the coins array
            For j= 1: amount //j be the sub-amount
                IF j>=coins[i] 
                    Table[j]=table[j]+table[j-coins[i]]
                END IF
            END For
        END For
    4.  Return table[amount]
    Table[amount] refers to all possible way to make the amount

Example with explanation:

    We have pre added coin 0-cent as coin[0]=0
    For Coin index:  1 //coins[1]=1

    Sub amount:1

    Table status:
    1 | 1 | 0 | 0 | 0


    Sub amount:2
    Table status:
    1 | 1 | 1 | 0 | 0

    Sub amount:3

    1 | 1 | 1 | 1 | 0
    Sub amount:4
    1 | 1 | 1 | 1 | 1 

This actually shows that by coin[1] all the sub-amounts can be managed only by one way, so if we continue this way for other coins we will get all the possible ways.


C++ implementation:

#include <bits/stdc++.h>
using namespace std;

void findway(vector<int> a,int n,int amount){
	int table[amount+1]; //DP table
	memset(table,0,sizeof(table));
	table[0]=1;
	//j be the sub-amounts
	for(int i=1;i<=n;i++){
		for(int j=1;j<=amount;j++){
		if(j>=a[i])
			table[j]+=table[j-a[i]];
		}
	}
	cout<<table[amount]<<endl; //final result
}

int main()
{
	int n,item,amount;

	cout<<"Enter the number of coins\n";
	scanf("%d",&n);
	
	cout<<"Enter value of coins\n";
	vector<int> a; //coins array
	//we pre add 0-cent coin as coins[0] 
	//for sake of 1 indexing
	a.push_back(0); 
	
	for(int j=0;j<n;j++){
		scanf("%d",&item);
		a.push_back(item);
	}
	
	cout<<"Enter total amount\n";
	cin>>amount;
	cout<<"Number of ways to sum the amount is: ";
	findway(a,n,amount); 
	
	return 0;
}

Output

Enter the number of coins
3
Enter value of coins
1 2 3
Enter total amount
4
Number of ways to sum the amount is: 4





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