Python Program for Coin Change

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.

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# Dynamic Programming Python implementation of Coin 
# Change problem
def count(S, m, n):
    # We need n+1 rows as the table is constructed 
    # in bottom up manner using the base case 0 value
    # case (n = 0)
    table = [[0 for x in range(m)] for x in range(n+1)]
  
    # Fill the entries for 0 value case (n = 0)
    for i in range(m):
        table[0][i] = 1
  
    # Fill rest of the table entries in bottom up manner
    for i in range(1, n+1):
        for j in range(m):
  
            # Count of solutions including S[j]
            x = table[i - S[j]][j] if i-S[j] >= 0 else 0
  
            # Count of solutions excluding S[j]
            y = table[i][j-1] if j >= 1 else 0
  
            # total count
            table[i][j] = x + y
  
    return table[n][m-1]
  
# Driver program to test above function
arr = [1, 2, 3]
m = len(arr)
n = 4
print(count(arr, m, n))
  
# This code is contributed by Bhavya Jain

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# Dynamic Programming Python implementation of Coin 
# Change problem
def count(S, m, 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)
    # Initialize all table values as 0
    table = [0 for k in range(n+1)]
  
    # 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 i in range(0,m):
        for j in range(S[i],n+1):
            table[j] += table[j-S[i]]
  
    return table[n]
  
# Driver program to test above function
arr = [1, 2, 3]
m = len(arr)
n = 4
x = count(arr, m, n)
print (x)
  
# This code is contributed by Afzal Ansari

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Please refer complete article on Dynamic Programming | Set 7 (Coin Change) for more details!




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