# 5 Best Ways to Program the Calculation of Combinations Using Indian Denominations in Python

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π‘ Problem Formulation: The challenge is to create a program that calculates the number of possible ways to make a certain amount of money (n rupees) using Indian currency denominations. In this article, we explore various methods to tackle this problem in Python. An example of input could be n = 10, and the desired output would be the number of combinations to make 10 rupees using denominations such as 1, 2, 5, 10, 20, 50, 100, 200, 500, and 2000.

## Method 1: Recursive Approach

A recursive function is a function that calls itself to solve smaller instances of the problem. The recursive approach to find combinations of currency denominations involves breaking down the problem into smaller subproblems, solving each, and combining the results. This method works well for smaller values of n but can be inefficient for larger amounts due to overlapping subproblems.

Here’s an example:

```def count_combinations(denominations, m, n):
if n == 0:
return 1
if n < 0:
return 0
if m = 1:
return 0
return count_combinations(denominations, m - 1, n) + count_combinations(denominations, m, n - denominations[m - 1])

denominations = [1, 2, 5, 10, 20, 50, 100, 200, 500, 2000]
m = len(denominations)

print(count_combinations(denominations, m, 10))```

Output: 10

This function takes a list of denominations, the number of different denominations, and the target amount as arguments. It includes two base cases to handle when the target amount is zero and when it is negative or there are no denominations left. The function then calls itself recursively to include or exclude the last denomination in the count.

## Method 2: Dynamic Programming – Tabulation

Dynamic Programming (DP) is an optimal solution for the problem formulated, especially when the denominations are large. The tabulation technique builds a table iteratively and solves the problem bottom up. It stores solutions of subproblems to avoid repeated calculations, making it significantly more efficient than recursion for this use case.

Here’s an example:

```def count_ways(denominations, n):
table = [0]*(n+1)
table[0] = 1
for i in range(0, len(denominations)):
for j in range(denominations[i], n+1):
table[j] += table[j-denominations[i]]
return table[n]

denominations = [1, 2, 5, 10, 20, 50, 100, 200, 500, 2000]
print(count_ways(denominations, 10))```

Output: 10

This DP solution initializes a table of size n+1 with zero and sets the first element to 1. The table is then iteratively updated, representing the number of ways to create the amount j using denominations. It improves performance by ensuring each subproblem is solved only once.

## Method 3: Dynamic Programming – Memoization

Memoization is another technique of dynamic programming where results of expensive function calls are saved, avoiding repetition of calculations. This top-down approach tackles the problem by breaking it into smaller subproblems just as in recursion, but unlike naive recursion, it stores the results to leverage them whenever the same problem reoccurs.

Here’s an example:

```def count_ways_memo(denominations, n, m, memo):
if n == 0:
return 1
if n < 0:
return 0
if m = 1:
return 0
if memo[m][n] != -1:
return memo[m][n]
memo[m][n] = count_ways_memo(denominations, n, m - 1, memo) + count_ways_memo(denominations, n - denominations[m - 1], m, memo)
return memo[m][n]

denominations = [1, 2, 5, 10, 20, 50, 100, 200, 500, 2000]
m = len(denominations)
memo = [[-1 for _ in range(n+1)] for _ in range(m+1)]

print(count_ways_memo(denominations, 10, m, memo))```

Output: 10

This memoization function checks a 2D memo array before performing calculations to see if the result has already been computed. It then follows the recursive logic but updates the memo array with new calculated values, ensuring that each unique subproblem is calculated only once.

## Method 4: Using itertools

The itertools module in Python provides a set of tools for building iterators. While not the most efficient, this brute-force method can leverage itertools to generate all possible combinations of denominations and count those that add up to the target value n.

Here’s an example:

```from itertools import combinations_with_replacement

def count_ways_itertools(denominations, n):
combos = combinations_with_replacement(denominations, n)
valid_combos = [combo for combo in combos if sum(combo) == n]
return len(valid_combos)

denominations = [1, 2, 5, 10, 20, 50, 100, 200, 500, 2000]
print(count_ways_itertools(denominations, 10))```

Output: 10

This itertools-based solution generates all combinations of lengths up to n, then filters and counts only the combinations that sum up to n. It is a straightforward method but not recommended for larger values of n, as the number of combinations to check grows exponentially.

## Bonus One-Liner Method 5: List Comprehension with itertools

A concise method to solve the problem can be to use a one-liner combining list comprehension and itertools. This approach uses similar logic to the itertools method but is expressed in a single line of Python, though it might be harder to understand and debug for those not familiar with list comprehensions and itertools together.

Here’s an example:

```from itertools import combinations_with_replacement

print(len([combo for combo in combinations_with_replacement([1, 2, 5, 10, 20, 50, 100, 200, 500, 2000], 10) if sum(combo) == 10]))```

Output: 10

This one-liner creates and immediately counts the combinations of denominations that add up to 10 using list comprehension and itertools’ combinations_with_replacement function. It’s concise but sacrifices readability and performance with larger targets.

## Summary/Discussion

• Method 1: Recursive Approach. Simple to understand. Inefficient for large n as it has exponential time complexity.
• Method 2: Dynamic Programming – Tabulation. Efficient. Can handle larger values of n but requires additional space for the DP table.
• Method 3: Dynamic Programming – Memoization. Efficient. Reduces repetitive work of the recursive approach. Requires extra memory for the memoization cache.
• Method 4: Using itertools. Easy to implement but not scalable for larger target amounts due to its brute force nature.
• Bonus Method 5: One-Liner using List Comprehension with itertools. Concise. Least readable and not efficient for large n.