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This method involves using a 2D array to keep track of number of ways to make up a certain amount using coins up to a particular denomination. The rows in the array represent the number of coins available (up to that index), and the columns represent each possible amount up to the target.
Define dp[i][j] as the number of ways to make amount j using the first i types of coins. Initialize dp[i][0] to 1 since there is one way to make amount 0 (choosing no coin). For each coin, iterate over all amounts and accumulate the number of ways by including the current coin.
Time complexity is O(n*m) where n is the amount and m is the number of coins. Space complexity is also O(n*m).
Time Complexity: O(n*m), Space Complexity: O(n*m) where `n` is the `amount` and `m` is the number of coins.
1function change(amount, coins) {
2    let dp = Array(coins.length + 1).fill(0).map(() => Array
The JavaScript solution uses a 2D array to store solutions to the subproblems and facilitates adding a specific coin along with previous coins to achieve the required total during iterations on all potential amounts.
This method optimizes space usage by employing a 1D array to store results. It iterates over each coin and updates the number of ways for each possible amount.
Define dp[j] as the number of ways to make amount j. Initially, set dp[0] to 1 (to denote the empty set that sums up to zero). Iterate through each coin, then iterate over possible amounts in nested for-loops from current coin's value to amount, updating the dp array by adding ways from previous calculations.
Time complexity remains O(n*m), but space complexity improves to O(n).
Time Complexity: O(n*m), Space Complexity: O(n) where `n` is the `amount` and `m` is the number of coins.
public class CoinChange {
    public int Change(int amount, int[] coins) {
        int[] dp = new int[amount + 1];
        dp[0] = 1;
        for (int i = 0; i < coins.Length; i++) {
            for (int j = coins[i]; j <= amount; j++) {
                dp[j] += dp[j - coins[i]];
            }
        }
        return dp[amount];
    }
    public static void Main(string[] args) {
        CoinChange cc = new CoinChange();
        int[] coins = { 1, 2, 5 };
        int amount = 5;
        Console.WriteLine("Number of combinations: " + cc.Change(amount, coins));
    }
}Using a single-dimensional `dp` array in C# ensures that past state solutions are preserved while reducing unnecessary overhead, directing changes and coin allocations smoothly through every iteration.