Coin change problem in python

The coin change problem is a classic algorithmic problem that involves finding the minimum number of coins needed to make a specific amount of money. Given a set of coin denominations and an amount, the goal is to determine the fewest number of coins required to make up that amount.

In Python, you can solve the coin change problem using dynamic programming. The dynamic programming approach involves building a table where each cell represents the minimum number of coins needed for a particular amount.

The algorithm starts by initializing a list, often referred to as “dp” (short for dynamic programming), to store the minimum number of coins needed for each amount. Initially, all cells in the table are set to infinity, except for dp[0], which is set to 0 since no coins are needed to make an amount of 0.

Then, the algorithm iterates through each coin denomination and checks if using that coin would result in a smaller number of coins needed for the current amount. If so, the algorithm updates the corresponding cell in the table with the new minimum value.

By the end of the iteration, dp[amount] will contain the minimum number of coins needed to make the desired amount. If dp[amount] remains infinity, it means that it is not possible to make that amount using the given coin denominations.

The algorithm efficiently solves the coin change problem by leveraging the principle of optimality and building solutions for smaller subproblems. It eliminates redundant calculations by gradually building on previously computed solutions.

The coin change problem has various applications, including making change at a cash register, optimizing vending machine transactions, and solving optimization problems in finance or resource allocation.

By implementing the coin change problem in Python, you can efficiently determine the minimum number of coins required to make a given amount, providing a versatile solution for dealing with currency-related challenges.

Program:

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

for coin in coins:
for i in range(coin, amount + 1):
dp[i] = min(dp[i], dp[i - coin] + 1)

return dp[amount] if dp[amount] != float('inf') else -1` </pre>

<pre> `coins = [1, 2, 5]
amount = 6
min_coins = coin_change(coins, amount)
print(min_coins)

Output: 3