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Tested Python solution for LeetCode 322 with 15 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 322 # by problem number
lcpy gen -s coin_change # by problem name
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.
Input: coins = [1,2,5], amount = 11
Output: 3
Input: coins = [2], amount = 3
Output: -1
Input: coins = [1], amount = 0
Output: 0
1 <= coins.length <= 121 <= coins[i] <= 2^31 - 10 <= amount <= 10^4class Solution:
# Time: O(amount * len(coins))
# Space: O(amount)
def coin_change(self, coins: list[int], amount: int) -> int:
if amount == 0:
return 0
# Initialize dp array with amount + 1 (impossible value)
# Since max coins needed is amount (using all 1-cent coins)
# amount + 1 serves as "infinity" to indicate impossible cases
dp = [amount + 1] * (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 result: -1 if impossible, otherwise minimum coins needed
return dp[amount] if dp[amount] <= amount else -1
| Time | Space |
|---|---|
| O(amount * len(coins)) | O(amount) |