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.

C
#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.

[ad type=”banner”] 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

Java
/* Dynamic Programming Java implementation of Coin
Change problem */
import java.util.Arrays;

class CoinChange
{
static long countWays(int S[], int m, int n)
{
//Time complexity of this function: O(mn)
//Space Complexity of this function: O(n)

// table[i] will be storing the number of solutions
// for value i. We need n+1 rows as the table is
// constructed in bottom up manner using the base
// case (n = 0)
long[] table = new long[n+1];

// Initialize all table values as 0
Arrays.fill(table, 0); //O(n)

// 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];
}

// Driver Function to test above function
public static void main(String args[])
{
int arr[] = {1, 2, 3};
int m = arr.length;
int n = 4;
System.out.println(countWays(arr, m, n));
}
}

Output:

4

Time Complexity: O(mn)

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