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.

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:
  1. Input: amount = 11, coins = [1, 2, 5]. The expected output is 3, as we can make up 11 using 3 coins of denomination 5.
  2. 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.
  3. Input: amount = 0, coins = [1, 2, 5]. The expected output is 0, as no coins are needed to make up 0.

python

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)).