Calculating Distinct Coin Sums with Python: Top Methods Explored

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πŸ’‘ Problem Formulation: Given a set of coins of different denominations and the number of coins available, what is the total number of distinct sum values that can be made? For instance, with coins of denominations [1, 2] and quantities [2, 2], we can make sums {1, 2, 3, 4, 5} – a total of 5 distinct sums.

Method 1: Dynamic Programming Approach

The dynamic programming method exploits the fact that the problem has an optimal substructure and overlapping subproblems, building up a solution iteratively using past results. We construct a boolean table indicating the possible sums. The space complexity of O(S*n) could be a concern, where S is the total sum and n the coin count.

Here’s an example:

def distinct_coin_sums(coins, quantities):
    sums = {0}
    for coin, quant in zip(coins, quantities):
        sums = {x + coin * i for x in sums for i in range(quant + 1)}
    return len(sums)

coins = [1, 2]
quantities = [2, 2]
print(distinct_coin_sums(coins, quantities))

Output:

5

This function iteratively updates the possible sums. For each coin, it expands the current set of sums by adding the coin’s value multiplied by all possible quantities. It then returns the total count of unique sums.

Method 2: Recursive Enumeration with Memoization

This method entails recursion to explore all combinations and memoization to cache results of explored branches, shrinking the overall problem space and improving efficiency. It works well with a smaller number of coins and quantities but could lead to stack overflow with large inputs due to deep recursion calls.

Here’s an example:

def count_sums(coins, quantities, current_sum, index, memo):
    if index == len(coins):
        return {current_sum}
    if (current_sum, index) in memo:
        return memo[(current_sum, index)]

    coin_sums = set()
    for i in range(quantities[index] + 1):
        coin_sums |= count_sums(coins, quantities, current_sum + coins[index] * i, index + 1, memo)

    memo[(current_sum, index)] = coin_sums
    return coin_sums

coins = [1, 2]
quantities = [2, 2]
print(len(count_sums(coins, quantities, 0, 0, {})))

Output:

5

This code snippet uses a recursive approach to sum over all possible coin combinations. The memoization technique significantly reduces the number of redundant calculations, leading to better performance on larger inputs.

Method 3: Optimized Dynamic Programming

This method improves the space complexity of the dynamic programming approach using a space-optimized table. Instead of keeping a row for every coin, we iterative update a single list, being more memory efficient especially for large numbers of coins.

Here’s an example:

def optimized_distinct_coin_sums(coins, quantities):
    sums = [0] * (sum(coins[i]*quantities[i] for i in range(len(coins)))+1)
    sums[0] = 1
    for i in range(len(coins)):
        for j in range(coins[i], len(sums)):
            for k in range(1, quantities[i]+1):
                sums[j] += sums[j-coins[i]*k] if j >= coins[i]*k else 0
    return sum(1 for s in sums if s)

coins = [1, 2]
quantities = [2, 2]
print(optimized_distinct_coin_sums(coins, quantities))

Output:

5

This function maintains a list of sums that are initially set to 0 with the exception of the first element. It then updates the list based on the coins and quantities we have, and counts distinct sum counts in the end.

Method 4: Bitset Operations

Bitset operations leverage bitwise logic to compute reachable sums. This can be extremely fast and parallel, utilizing bit-level parallelism intrinsic in modern processors. However, this method might not be intuitive for those unfamiliar with bitwise operations.

Here’s an example:

def bitset_distinct_coin_sums(coins, quantities):
    max_sum = sum(coins[i]*quantities[i] for i in range(len(coins)))
    bitmap = 1
    for coin, quant in zip(coins, quantities):
        bitmap |= bitmap << (coin * quant)
    return bin(bitmap).count('1') - 1

coins = [1, 2]
quantities = [2, 2]
print(bitset_distinct_coin_sums(coins, quantities))

Output:

5

The function initializes a bitset representation of sum combinations and employs left bitwise shifts combined with an ‘or’ operation to populate it based on the coins and their quantities. The final count of distinct sums is given by counting the ‘1’ bits in the bitset.

Bonus One-Liner Method 5: Using functools and itertools

This concise method uses Python’s functools and itertools modules to quickly calculate combinations of sums, handling the iterations internally. It’s elegant, but it’s not the most memory or time-efficient for larger inputs due to the generation of all distinct combinations.

Here’s an example:

import itertools
import functools

coins = [1, 2]
quantities = [2, 2]
print(len(set(functools.reduce(lambda x, y: x + y, (list(itertools.combinations_with_replacement(coins, i)) for i in range(sum(quantities) + 1))))))

Output:

5

The one-liner leverages a combination of itertools’ combinations_with_replacement and functools’ reduce function to condense the problem into a single line of code that calculates the distinct sums of all possible combinations.

Summary/Discussion

  • Method 1: Dynamic Programming Approach. Best for precision and a comprehensive resolution of the problem. It can, however, be space-intensive for large input sets.
  • Method 2: Recursive Enumeration with Memoization. It’s a straightforward conceptually and handles the problem combinatorially. Nonetheless, it suffers from potential stack overflows and can be inefficient for large quantities.
  • Method 3: Optimized Dynamic Programming. Good for larger datasets where space is a concern. It might be less straightforward to implement compared to basic dynamic programming.
  • Method 4: Bitset Operations. It’s efficient and concise, ideal for performance-critical applications. On the flip side, it’s less readable and could be conceptually challenging.
  • Bonus One-Liner Method 5: Using functools and itertools. It’s elegant and great for one-off calculations or smaller sets of coins and quantities. For large coin sets, it may become sluggish due to the handling of numerous combinations.