Introduction

In this blog post, we will explore how to solve the LeetCode problem "518" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money. Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0. You may assume that you have an infinite number of each kind of coin. The answer is guaranteed to fit into a signed 32-bit integer. Example 1: Input: amount = 5, coins = [1,2,5] Output: 4 Explanation: there are four ways to make up the amount: 5=5 5=2+2+1 5=2+1+1+1 5=1+1+1+1+1 Example 2: Input: amount = 3, coins = [2] Output: 0 Explanation: the amount of 3 cannot be made up just with coins of 2. Example 3: Input: amount = 10, coins = [10] Output: 1 Constraints: 1 <= coins.length <= 300 1 <= coins[i] <= 5000 All the values of coins are unique. 0 <= amount <= 5000

Explanation

• Dynamic Programming: Use dynamic programming to build a table dp where dp[i] represents the number of combinations to make up the amount i.
  • Iterative Approach: Iterate through each coin denomination and update the dp table. For each coin, iterate through amounts greater than or equal to the coin value, adding the number of combinations that can be formed by using the coin to the existing combinations.
  • Base Case: Initialize dp[0] to 1, as there is one way to make up the amount 0 (using no coins).

• Time Complexity: O(amount * number of coins), Space Complexity: O(amount)

Code

    def change(amount: int, coins: list[int]) -> int:
    """
    Calculates the number of combinations to make up the given amount using the given coins.

    Args:
        amount: The total amount of money.
        coins: A list of coin denominations.

    Returns:
        The number of combinations that make up the amount.
    """

    dp = [0] * (amount + 1)
    dp[0] = 1

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

    return dp[amount]