You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.
Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.
You may assume that you have an infinite number of each kind of coin.
Example 1:
Input: coins = [1,2,5], amount = 11 Output: 3 Explanation: 11 = 5 + 5 + 1
Example 2:
Input: coins = [2], amount = 3 Output: -1
Example 3:
Input: coins = [1], amount = 0 Output: 0
1 <= coins.length <= 121 <= coins[i] <= 231 - 10 <= amount <= 104
class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
dp = [float('inf')] * (amount + 1)
dp[0] = 0 # Base case: no coins needed to make amount 0
# Iterate over each amount
for j in range(1, amount + 1):
# Check each coin for this amount
for coin in coins:
if j >= coin:
dp[j] = min(dp[j], dp[j - coin] + 1)
return dp[amount] if dp[amount] != float('inf') else -1
O(n * m), where n is the number of different coins and m is the total amount.
A dynamic programming table (dp) of size m + 1 is used, where m is the amount. This array is used to store the minimum number of coins needed for each amount from 0 to m. Thus, the space complexity is O(m), as the primary storage cost is the dp array which has an entry for each amount up to m.