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

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π‘ Problem Formulation: The task is to determine the minimum number of coins that you need to make up a given amount of money, assuming you have an unlimited supply of coins of given denominations. For instance, if the input is `11` cents, and the coin denominations are `[1, 2, 5]`, the desired output is `3` because the optimal combination is one 5-cent coin and three 2-cent coins.

## Method 1: Recursive Approach

This method explores the use of a recursive solution to calculate the minimum number of coins. The function recursively reduces the change amount by deducting the largest coin value it can accommodate. Due to its recursive nature, it provides a clear and straightforward conceptual understanding of the problem, at the expense of efficiency for larger amounts.

Here’s an example:

```def coinChange(coins, amount):
if amount == 0:
return 0
res = float('inf')
for coin in coins:
if amount >= coin:
sub_res = coinChange(coins, amount - coin)
if sub_res != -1:
res = min(res, sub_res + 1)
return res if res != float('inf') else -1

print(coinChange([1, 2, 5], 11))```

The output of the code is `3`.

The code defines a function `coinChange()` which takes an array of coin denominations and the target amount. It uses recursion to find the minimum number of coins needed. If the amount becomes zero, it returns zero, signaling no more coins are required. For all other cases, it tries to reduce the problem size and calls itself with the remaining amount until the base case is reached.

## Method 2: Dynamic Programming – Top Down

Using a top-down dynamic programming approach involves storing the results of costly recursive calculations in a memoization table, avoiding repeated work. This optimization can drastically reduce computation time, especially when dealing with larger amounts of change.

Here’s an example:

```def coinChange(coins, amount, memo):
if amount in memo: return memo[amount]
if amount == 0: return 0
if amount = 0 and res < min_coins:
min_coins = 1 + res
memo[amount] = -1 if min_coins == float('inf') else min_coins
return memo[amount]

memo = {}
print(coinChange([1, 2, 5], 11, memo))```

The output is `3`.

The function `coinChange()` includes an additional parameter `memo` which is a dictionary used to store the results of subproblems. It significantly speeds up the process by returning stored results when the same amount is encountered again.

## Method 3: Dynamic Programming – Bottom Up

A bottom-up approach starts with solving the smallest subproblems first and builds up the solutions to larger subproblems. This method employs a tabulation technique where the element at index i of the array represents the minimum number of coins needed to make change for amount i.

Here’s an example:

```def coinChange(coins, amount):
dp = [float('inf')] * (amount + 1)
dp[0] = 0

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

print(coinChange([1, 2, 5], 11))```

The output is `3`.

This snippet introduces a bottom-up dynamic programming solution, where `dp` is a list initialized with infinity and a base case of 0 coins for the 0 amount. The algorithm iteratively computes the minimum coins needed for all values from 1 up to the target amount.

## Method 4: Iterative Greedy Algorithm

The greedy algorithm for making change picks the largest coin denomination that is smaller than the current amount of change required and subtracts it until the remaining amount is zero. This method may not always give an optimal solution for arbitrary coin systems but is efficient for canonical coin systems.

Here’s an example:

```def coinChange(coins, amount):
coins.sort(reverse=True)
count = 0
for coin in coins:
if amount >= coin:
count += amount // coin
amount %= coin
return count if amount == 0 else -1

print(coinChange([1, 2, 5], 11))```

The output is `3`.

In this code, the coins list is sorted in descending order so that we can start with the largest coin denomination. The function then iterates over the coins, using division and modulo operations to determine how many of each coin should be used, adjusting the amount as it goes.

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

In Python, list comprehensions offer a concise way to create lists. This one-liner provides the same greedy approach but with a more Pythonic style, compacting the iterative process into a single expression.

Here’s an example:

```def coinChange(coins, amount):
return sum((lambda x: (lambda y: (y, amount % x[0]))(amount // x[0]))(x) for x in sorted([(c, amount // c) for c in coins if c <= amount], reverse=True))

print(coinChange([1, 2, 5], 11))```

The output is `3`.

This method uses a list comprehension within a lambda function to determine the number of each type of coin needed. This example is compact and showcases the power of Python’s list comprehensions and lambda functions, although it sacrifices some readability.

## Summary/Discussion

Method 1: Recursive Approach. Simple and elegant; best for understanding the algorithm. Inefficient for large amounts.
Method 2: Dynamic Programming – Top Down. More efficient than pure recursion. Still not as efficient as bottom-up for some cases.
Method 3: Dynamic Programming – Bottom Up. Highly efficient; best for large problem sizes. Might be less intuitive than the recursive approach.
Method 4: Greedy Algorithm. Efficient for specific coin systems. Not guaranteed to work for all coin sets.
Method 5: Greedy with List Comprehension. Compact and stylish, but at the cost of readability. Best used when brevity is valued and the coin system is canonical.