DP - Counting coin change

Question

The problem requires to count number of coin changes for a particular cost.

For example, if I have coin values of 50, 20, 10, 5, 1, I can form costs of:

5 => (5), (11111), which are 2 ways.

10 => (10), (5, 5), (5, 11111), (11111, 11111), which are 4 ways.

Here is my function. It is returning wrong results begging from cost of 10 (returns 9 ways while the actual number of ways is only 4)

int dp[10000];
int coins[] = { 50, 20, 10, 5, 1 };
int rec(int n)
{
  if (n == 0) return 1;
  if (dp[n] != -1) return dp[n];
  int cnt = 0;
  for (int i = 0; i < 5; i++)
    if (coins[i] <= n) cnt += rec(n - coins[i]);
  return dp[n] = cnt;
}

How can I fix this function to give the correct number of ways? Is this algorithm correct even? see the complete code and its output here

NOTE: my problem is not with dp array initialization. I am using memset to initialize it to -1 each time before calling rec.

Answer 1

(5, 1, 1, 1, 1, 1) and (1, 1, 1, 5, 1, 1) is different way in you algorithm, you should keep it decreasing.

int dp[10000][5];  // dp[20][2] means, if the biggest coin is coins[2],
                   // how much ways for 20 ?
int coins[] = { 1, 5, 10, 20, 50 }; // here
int rec(int n, int m)
{
  int cnt = 0;
  int i;
  if (n == 0) return 1;
  //if (m == 0) return 1;
  if (dp[n][m] != -1) return dp[n][m];
  for (i = 0; i <= m; i++)
    if (coins[i] <= n) cnt += rec(n - coins[i], i);
  return dp[n][m] = cnt;
}

int main()
{
        memset(dp, -1, sizeof(dp));
        printf("%d\n", rec(10, 4));
}