動機

複習dp

Problem

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 = 11Output: 3Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3Output: -1

Example 3:

Input: coins = [1], amount = 0Output: 0

Example 4:

Input: coins = [1], amount = 1Output: 1

Example 5:

Input: coins = [1], amount = 2Output: 2

 

Constraints:

  • 1 <= coins.length <= 12
  • 1 <= coins[i] <= 231 - 1
  • 0 <= amount <= 104

Sol

下面多了i就做了無用功，就變慢了

class Solution:
    @functools.lru_cache(None)
    def dp(self,cnt,i):
        if cnt == 0:
            return 0
        elif cnt < 0 or i < 0:
            return float('inf')
        else:
            ret = float('inf')
            for x in reversed(range(i)):
                ret = min(ret, self.dp(cnt-self.cs[x],x+1))
                
            return ret+1
    def coinChange(self, coins: List[int], amount: int) -> int:
        self.cs = coins
        self.cnt = amount
        self.ret = float('inf')
        self.cs.sort()
        ans = self.dp(amount,len(coins))
        
        return ans if ans != float('inf') else -1

只要總和做維度就好…

class Solution:
    @functools.lru_cache(None)
    def dp(self,cnt):
        if cnt == 0:
            return 0
        elif cnt < 0:
            return float('inf')
        else:
            ret = float('inf')
            for x in self.cs:
                ret = min(ret, self.dp(cnt-x))
                
            return ret+1
    def coinChange(self, coins: List[int], amount: int) -> int:
        self.cs = coins
        self.cnt = amount
        self.ret = float('inf')
        self.cs.sort(reverse=True)
        ans = self.dp(amount)
        
        return ans if ans != float('inf') else -1