Coin Change

Problem Link

Approach

Find the fewest coins needed to make amount amount, or -1 if impossible.

Let dp[a] be the minimum coins for amount a. Initialize dp[0] = 0 and others to infinity.

For each amount from 1 to amount, try every coin:

• dp[a] = min(dp[a], 1 + dp[a - coin]) when a - coin >= 0

Bottom-up avoids recursion overhead. Unbounded knapsack style: each coin can be reused.

Time Complexity: O(amount × n)
Space Complexity: O(amount)

Code

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        dp = [float("inf")] * (amount + 1)
        dp[0] = 0

        for a in range(1, amount + 1):
            for coin in coins:
                if a >= coin:
                    dp[a] = min(dp[a], 1 + dp[a - coin])

        return dp[amount] if dp[amount] != float("inf") else -1