Hello Everyone,

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 doesn’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.

**1) Optimal Substructure**

To count the total number of solutions, we can divide all set solutions into two sets.

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

Let count(S[], m, n) be the function to count the number of solutions, then it can be written as sum of count(S[], m-1, n) and count(S[], m, n-Sm).

Therefore, the problem has optimal substructure property as the problem can be solved using solutions to subproblems.

**2) Overlapping Subproblems**

Following is a simple recursive implementation of the Coin Change problem. The implementation simply follows the recursive structure mentioned above.

`// Recursive C program for`

`// coin change problem.`

`#include<stdio.h>`

`// Returns the count of ways we can`

`// sum S[0...m-1] coins to get sum n`

`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] );`

`}`

`// Driver program to test above function`

`int`

`main()`

`{`

` `

`int`

`i, j;`

` `

`int`

`arr[] = {1, 2, 3};`

` `

`int`

`m = `

`sizeof`

`(arr)/`

`sizeof`

`(arr[0]);`

` `

`printf`

`(`

`"%d "`

`, count(arr, m, 4));`

` `

`getchar`

`();`

` `

`return`

`0;`

`}`

Java

```
// Recursive java program for
// coin change problem.
import java.io.*;
class GFG {
// Returns the count of ways we can
// sum S[0...m-1] coins to get sum n
static 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] );
}
// Driver program to test above function
public static void main(String[] args)
{
int arr[] = {1, 2, 3};
int m = arr.length;
System.out.println( count(arr, m, 4));
}
}
```

**Output :**

4