Coin Change

 

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. Write clean and readable code:

    python
  6. 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
  7. 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
  8. print(coinChange(11, [1, 2, 5])) # Output: 3 print(coinChange(3, [2])) # Output: -1 print(coinChange(0, [1, 2, 5])) # Output: 0
  9. Optimize if necessary:

    • The solution is already optimized using dynamic programming and has a time complexity of O(amount * len(coins)).
  10. Handle error cases:

    • The code does not handle invalid input cases explicitly. However, it assumes that the input follows the problem constraints.
  11. 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)).
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