<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>

'''
You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

amount = money
coins  = type of coins

return minimum number of coins to make that amount.

if not found, return -1

1 <= coins.length <= 12
0 <= amount <= 104

1 2 3 4 5

pick 5 until target > t
pick 4 
'''
class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        # n = len(coins)
        # hm = {}
        # def dp(i,target):
        #     if i == 0:
        #         if target%coins[i]==0:
        #             hm[(i,target)]  = target//coins[i]
        #             return target//coins[i]
        #         else:
        #             hm[(i,target)]  = math.inf
        #             return math.inf

        #     if (i,target) in hm:
        #         return  hm[(i,target)]

        #     pick = math.inf
        #     not_pick = 0 + dp(i-1,target)
        #     if coins[i] <= target:
        #         pick = 1 + dp(i,target-coins[i])
        #     min_cost = min(not_pick,pick)
        #     hm[(i,target)] = min_cost
        #     return min_cost
        
        # res= dp(n-1,amount)
        # if res == math.inf:
        #     return -1
        # else: return res
        default_dict = defaultdict(lambda: math.inf)

        # Initialize the base cases
        for t in range(amount + 1):
            if t % coins[0] == 0:
                default_dict[(0, t)] = t // coins[0]  # Use integer division

        # Fill the DP table
        for i in range(1, len(coins)):
            for j in range(amount + 1):
                not_pick = default_dict[(i - 1, j)]  # Case where we don't pick the current coin
                pick = math.inf  # Case where we pick the current coin
                if coins[i] <= j:
                    pick = 1 + default_dict[(i, j - coins[i])]
                default_dict[(i, j)] = min(not_pick, pick)

        # Get the result from the dictionary
        result = default_dict[(len(coins) - 1, amount)]
        return result if result != math.inf else -1
    
            

        
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