<p>You are given an integer array <code>coins</code> representing coins of different denominations and an integer <code>amount</code> representing a total amount of money.</p>

<p>Return <em>the fewest number of coins that you need to make up that amount</em>. If that amount of money cannot be made up by any combination of the coins, return <code>-1</code>.</p>

<p>You may assume that you have an infinite number of each kind of coin.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> coins = [1,2,5], amount = 11
<strong>Output:</strong> 3
<strong>Explanation:</strong> 11 = 5 + 5 + 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> coins = [2], amount = 3
<strong>Output:</strong> -1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> coins = [1], amount = 0
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= coins.length &lt;= 12</code></li>
	<li><code>1 &lt;= coins[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>0 &lt;= amount &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        def tab():
            dp = [inf] * (amount + 1)
            dp[0] = 0
            for i in range(1, amount + 1):
                for coin in coins:
                    if coin == i:
                        dp[i] = 1
                        break
                    elif coin > i:
                        continue
                    else:
                        dp[i] = min(dp[i], 1 + dp[i-coin])
            return -1 if dp[amount] == inf else dp[amount]
        
        def memo(index, remain):
            if remain < 0:
                return inf
            if remain == 0:
                return 0
            if index == len(coins):
                return inf
            if remain in cache:
                return cache[remain]
            min_coin = inf
            for i in range(index, len(coins)):
                take = 1 + memo(index, remain - coins[i])
                skip = memo(index + 1, remain)
                min_coin = min(min_coin, take, skip)
            #return min_coin
            cache[remain] = min_coin
            return cache[remain]

        #return tab()
        cache = {}
        #min_coin = memo(0, amount)
        #return -1 if min_coin == inf else min_coin

        def dfs(rem):
            if rem < 0:
                return -1
            if rem == 0:
                return 0
            if rem in cache:
                return cache[rem]
            min_cost = float('inf')
            for coin in coins:
                res = dfs(rem - coin)
                if res != -1:
                    min_cost = min(min_cost, res + 1)
            cache[rem] = min_cost
            return cache[rem] if cache[rem] != float('inf') else -1

        return dfs(amount)
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