Program to find number of ways we can get n R.s using Indian denominations in Python

Suppose we have limited coins of denominations (?1, ?2, ?5 and ?10). We have to find in how many ways can you sum them up to a total of ?n? We have an array count of size 4, where count[0] indicates coins of ?1, count[1] indicates coins of ?2 and so on.

So, if the input is like n = 25 count = [7,3,2,2], then the output will be 9.

To solve this, we will follow these steps −

• denom := [1,2,5,10]
• A := an array of size (n + 1) and fill with 0
• B := a new list from A
• for i in range 0 to (minimum of count[0] and n), do
  • A[i] := 1
• for i in range 1 to 3, do
  • for j in range 0 to count[i], do
    • for k in range 0 to n + 1 - j *denom[i], do
      • B[k + j * denom[i]] := B[k + j * denom[i]] + A[k]
  • for j in range 0 to n, do
    • A[j] := B[j]
    • B[j] := 0
• return A[n]

Example

Let us see the following implementation to get better understanding −

denom = [1,2,5,10]
def solve(n, count):
   A = [0] * (n + 1)
   B = list(A)
   for i in range(min(count[0], n) + 1):
      A[i] = 1
   for i in range(1, 4):
      for j in range(0, count[i] + 1):
         for k in range(n + 1 - j *denom[i]):
            B[k + j * denom[i]] += A[k]
      for j in range(0, n + 1):
         A[j] = B[j]
         B[j] = 0
   return A[n]

n = 25
count = [7,3,2,2]
print(solve(n, count))

Input

25, [7,3,2,2]

Output

9