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.

• Optimal Substructure
  To count total number solutions, we can divide all set solutions in 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.

• Overlapping Subproblems

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

#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;
}

It should be noted that the above function computes the same subproblems again and again. See the following recursion tree for S = {1, 2, 3} and n = 5.
The function C({1}, 3) is called two times. If we draw the complete tree, then we can see that there are many subproblems being called more than once.

C() --> count()
                              C({1,2,3}, 5)                     
                           /                \
                         /                   \              
             C({1,2,3}, 2)                 C({1,2}, 5)
            /     \                        /         \
           /        \                     /           \
C({1,2,3}, -1)  C({1,2}, 2)        C({1,2}, 3)    C({1}, 5)
               /     \            /    \            /     \
             /        \          /      \          /       \
    C({1,2},0)  C({1},2)   C({1,2},1) C({1},3)    C({1}, 4)  C({}, 5)
                   / \      / \       / \        /     \    
                  /   \    /   \     /   \      /       \ 
                .      .  .     .   .     .   C({1}, 3) C({}, 4)
                                               /  \
                                              /    \  
                                             .      .

Since same suproblems are called again, this problem has Overlapping Subprolems property. So the Coin Change problem has both properties of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array table[][] in bottom up manner.

Dynamic Programming Solution

#include<stdio.h>
 
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];
}
 
// Driver program to test above function
int main()
{
    int arr[] = {1, 2, 3};
    int m = sizeof(arr)/sizeof(arr[0]);
    int n = 4;
    printf(" %d ", count(arr, m, n));
    return 0;
}

Output:

4

Time Complexity: O(mn)

Following is a simplified version of method 2. The auxiliary space required here is O(n) only.

int count( int S[], int m, int n )
{
    // table[i] will be storing the number of solutions for
    // value i. We need n+1 rows as the table is consturcted
    // in bottom up manner using the base case (n = 0)
    int table[n+1];
 
    // Initialize all table values as 0
    memset(table, 0, sizeof(table));
 
    // Base case (If given value is 0)
    table[0] = 1;
 
    // Pick all coins one by one and update the table[] values
    // after the index greater than or equal to the value of the
    // picked coin
    for(int i=0; i<m; i++)
        for(int j=S[i]; j<=n; j++)
            table[j] += table[j-S[i]];
 
    return table[n];
}