5 Best Ways to Find the Number of Coins Needed for Change in Python

💡 Problem Formulation: Suppose you are building a vending machine software that needs to return change to customers in the least number of coins possible. The input is the total change amount and an array representing coin denominations, while the desired output is the minimum number of coins that make up that amount. For example, given an amount of 11 and coin denominations [1, 2, 5], the output should be 3 (indicating 5+5+1).

Method 1: Greedy Algorithm

Greedy algorithms work by always choosing the largest denomination coin available to reduce the remaining change amount. This method is simple and efficient for currencies like the US dollar where it can guarantee an optimal solution; however, it may not always result in the least number of coins for arbitrary coin systems.

Here’s an example:

def greedy_coin_change(coins, amount):
    coins.sort(reverse=True)
    count = 0
    for coin in coins:
        count += amount // coin
        amount %= coin
    return count

print(greedy_coin_change([1, 2, 5], 11))

Output: 3

The function greedy_coin_change() sorts the denominations in descending order and iteratively reduces the amount by the largest coin value while incrementing the count. This process is repeated until the amount is reduced to zero.

Method 2: Dynamic Programming – Bottom-Up Approach

The bottom-up dynamic programming approach builds up a solution by starting from the smallest subproblem and iteratively solving larger ones based on this. It does not suffer from the potential suboptimality of the greedy approach and is guaranteed to find the minimum number of coins required for change.

Here’s an example:

def dp_coin_change(coins, amount):
    dp = [0] + [float('inf')] * amount
    for coin in coins:
        for x in range(coin, amount + 1):
            dp[x] = min(dp[x], dp[x - coin] + 1)
    return dp[amount] if dp[amount] != float('inf') else -1

print(dp_coin_change([1, 2, 5], 11))

Output: 3

Here, dp_coin_change() initializes a list dp to store the minimum number of coins that make each amount from 0 to the target amount. It iteratively updates the list whenever a smaller number of coins is found for a specific amount.

Method 3: Dynamic Programming – Top-Down Approach

A top-down dynamic programming method uses recursion with memoization to solve the problem. It starts with the original problem and breaks it down into smaller subproblems, storing their results to avoid redundant calculations. It is as efficient as the bottom-up approach, yet can be more intuitive to understand due to its recursive nature.

Here’s an example:

def dp_coin_change_memo(coins, amount, memo):
    if amount < 0: return float('inf')
    if amount == 0: return 0
    if amount in memo: return memo[amount]
    memo[amount] = min(dp_coin_change_memo(coins, amount - coin, memo) + 1 for coin in coins)
    return memo[amount]

print(dp_coin_change_memo([1, 2, 5], 11, {}))

Output: 3

The function dp_coin_change_memo() uses a memoization dictionary to store previously computed results. This significantly reduces the number of calculations by ensuring that we only calculate the minimum coins for a specific amount once.

Method 4: Recursive Brute Force

The brute force recursive method exhaustively tries all possible combinations of coins to find the minimum number of coins needed for change. This method is simple to implement but is not time-efficient, especially for larger amounts as the number of recursive calls increases exponentially.

Here’s an example:

def brute_force_coins(coins, amount):
    if amount < 0: return float('inf')
    if amount == 0: return 0
    return min(brute_force_coins(coins, amount - coin) + 1 for coin in coins)

print(brute_force_coins([1, 2, 5], 11))

Output: 3

The function brute_force_coins() uses recursion to try every possible combination of coins. It’s the most straightforward approach but also the most computationally expensive.

Bonus One-Liner Method 5: Recursive One-Liner

Sometimes, for the sake of brevity or a challenge, a recursive solution can be condensed into a single line. This is not generally advisable for production code due to readability concerns, but it can be an interesting exercise to understand recursion.

Here’s an example:

coin_change_one_liner = lambda coins, amount: amount if amount == 0 else min([coin_change_one_liner(coins, amount - c) + 1 for c in coins if c <= amount], default=float('inf'))

print(coin_change_one_liner([1, 2, 5], 11))

Output: 3

This one-liner encapsulates the recursive brute force method in a lambda function. It uses a list comprehension to recursively calculate the number of coins and returns the minimum value using Python’s min() function, with a safeguard for when no coin combination is found.

Summary/Discussion

• Method 1: Greedy Algorithm. Fast and simple. May not always provide the optimal solution for arbitrary coin systems.
• Method 2: Dynamic Programming – Bottom-Up. Guaranteed to find the minimum number of coins. Requires additional space for tabulation.
• Method 3: Dynamic Programming – Top-Down. As optimal as bottom-up. More intuitive. Suffers from recursion overhead and stack size limitations in Python.
• Method 4: Recursive Brute Force. Simple but highly inefficient for larger input sizes due to its exponential time complexity.
• Bonus Method 5: Recursive One-Liner. Interesting as a coding exercise. Lack of readability makes it unsuitable for real-world applications.