We use bottom-up DP. dp[i] represents the minimum number of coins needed to make amount i. We try each coin and take the minimum.

1. Create dp array of size amount + 1, initialized to infinity.
2. dp[0] = 0 (0 coins for amount 0).
3. For each amount from 1 to target:
   • For each coin: if coin <= amount, dp[amount] = min(dp[amount], dp[amount - coin] + 1).
4. Return dp[amount] if reachable, else -1.

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

def coin_change(coins, amount):
    dp = [float('inf')] * (amount + 1)
    dp[0] = 0
    
    for i in range(1, amount + 1):
        for coin in coins:
            if coin <= i:
                dp[i] = min(dp[i], dp[i - coin] + 1)
    
    return dp[amount] if dp[amount] != float('inf') else -1