# Coin Change

Clarify the problem:

- The problem requires us to find the minimum number of coins needed to make up a given amount.
- We are given a list of coin denominations and the amount we need to make up.
- We need to return the minimum number of coins required to make up the amount. If it is not possible to make up the amount using the given coins, we return -1.

Analyze the problem:

- Input: An integer amount and a list of coin denominations (positive integers).
- Output: An integer representing the minimum number of coins needed to make up the amount, or -1 if it is not possible.
- Constraints:
- The list of coin denominations is non-empty.
- The amount is a positive integer not exceeding 10,000.
- It is guaranteed that we have an infinite supply of each coin denomination.

Design an algorithm:

- We can solve this problem using dynamic programming.
- We create an array called
`dp`

of size`amount + 1`

and initialize all elements to a value greater than the maximum possible number of coins (amount + 1). - We set
`dp[0]`

to 0 since no coins are needed to make up the amount 0. - For each coin denomination, we iterate through the
`dp`

array from the current coin denomination to the target amount. - For each value, we update
`dp[i]`

by taking the minimum of the current value of`dp[i]`

and`dp[i - coin] + 1`

, where`coin`

is the current coin denomination. - Finally, we return
`dp[amount]`

if it is less than or equal to the target amount; otherwise, we return -1.

Explain your approach:

- We will implement a function called
`coinChange`

that takes an amount and a list of coin denominations as input and returns the minimum number of coins needed to make up the amount. - We initialize the
`dp`

array with size`amount + 1`

and set all elements to`amount + 1`

. - We set
`dp[0]`

to 0 since no coins are needed to make up the amount 0. - We iterate through the coin denominations and the
`dp`

array. - For each value, we update
`dp[i]`

by taking the minimum of the current value of`dp[i]`

and`dp[i - coin] + 1`

, where`coin`

is the current coin denomination. - Finally, we return
`dp[amount]`

if it is less than or equal to the target amount; otherwise, we return -1.

- We will implement a function called
Write clean and readable code:

python`def coinChange(amount, coins): dp = [amount + 1] * (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] <= amount else -1`

Test your code:

- We test the code with the following cases:
- Input: amount = 11, coins = [1, 2, 5]. The expected output is 3, as we can make up 11 using 3 coins of denomination 5.
- Input: amount = 3, coins = [2]. The expected output is -1, as it is not possible to make up 3 using only coins of denomination 2.
- Input: amount = 0, coins = [1, 2, 5]. The expected output is 0, as no coins are needed to make up 0.

python- We test the code with the following cases:
`print(coinChange(11, [1, 2, 5])) # Output: 3 print(coinChange(3, [2])) # Output: -1 print(coinChange(0, [1, 2, 5])) # Output: 0`

Optimize if necessary:

- The solution is already optimized using dynamic programming and has a time complexity of O(amount * len(coins)).

Handle error cases:

- The code does not handle invalid input cases explicitly. However, it assumes that the input follows the problem constraints.

Discuss complexity analysis:

- Let N be the length of the coins list and M be the target amount.
- The time complexity of the solution is O(N * M) since we iterate through the coins list for each value in the
`dp`

array. - The space complexity is O(M) to store the
`dp`

array. - The solution is efficient and performs well for moderate-sized inputs. However, if the target amount is very large, the space complexity may become a concern. In such cases, we can optimize the solution to use a rolling array to reduce the space complexity to O(min(N, M)).