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.
// C program for coin change problem. #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 constructed
// in bottom up manner using the base case 0
// value case (n = 0)
int table[n+1][m];
// Fill the entries for 0 value case (n = 0)
for (i=0; i<m; i++)
table[0][i] = 1;
// Fill rest of the table entries 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 constructed
// 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];
} |
Please refer complete article on Dynamic Programming | Set 7 (Coin Change) for more details!