dp[j] represents the number of ways to make amount j. We process coins one at a time (outer loop) to avoid counting permutations as different combinations.

1. Create dp array of size amount + 1, initialized to 0.
2. dp[0] = 1 (one way to make amount 0: use no coins).
3. For each coin:
   • For each amount from coin to amount:
     • dp[j] += dp[j - coin].
4. Return dp[amount].

• Time Complexity: O(amount * len(coins)).
• Space Complexity: O(amount).

def change(amount, coins):
    dp = [0] * (amount + 1)
    dp[0] = 1
    
    for coin in coins:
        for j in range(coin, amount + 1):
            dp[j] += dp[j - coin]
    
    return dp[amount]