### Question

Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesnâ€™t matter.

For example, for N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}. So output should be 4. For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}. So the output should be 5.

### Analysis

I did this question before, it’s not a DFS search question, because question only asks the total number of ways.

So what method do we use? Remember the 4 types of DP?

Input cannot sort

Find minimum/maximum result

Check the feasibility

Count all possible solutions

So, this is a DP question!

### Solution

The solutions in in two parts:

- Solutions that do not contain mth coin (or Sm).
- Solutions that contain at least one Sm.

Using m to denote the types of coin used, and n denote the total value, the equation is:

count( S, m, n ) = count( S, m - 1, n ) + count( S, m, n-S[m-1] )

**Solution one is using recursion**. It’s not good because of a lot of repeated calculation. But the code is extremely easy to write:

```
int count( int S[], int m, int n ) {
// If n is 0 then there is 1 solution (do not include any coin)
if (n == 0)
return 1;
// If n is less than 0 then no solution exists
if (n < 0)
return 0;
// If there are no coins and n is greater than 0, then no solution exist
if (m <=0 && n >= 1)
return 0;
// count is sum of solutions (i) including S[m-1] (ii) excluding S[m-1]
return count( S, m - 1, n ) + count( S, m, n-S[m-1] );
}
```

**Solution two is DP**, it’s a little hard to write actually. Do it carefully.

### Code

C++ code not written by me:

```
int count( int S[], int m, int n ) {
int i, j, x, y;
// We need n+1 rows as the table is consturcted in bottom up manner using
// the base case 0 value case (n = 0)
int table[n+1][m];
// Fill the enteries for 0 value case (n = 0)
for (i=0; i<m; i++)
table[0][i] = 1;
// Fill rest of the table enteries in bottom up manner
for (i = 1; i < n+1; i++)
{
for (j = 0; j < m; j++)
{
// Count of solutions including S[j]
x = (i-S[j] >= 0)? table[i - S[j]][j]: 0;
// Count of solutions excluding S[j]
y = (j >= 1)? table[i][j-1]: 0;
// total count
table[i][j] = x + y;
}
}
return table[n][m-1];
}
```