<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:
        coins = sorted(coins)
        n = len(coins)
        # coins = coins[::-1]
        ans = 0
        if  amount == 0 :
            return 0 

        an = 100001


        def bfs (i=0):
            qu = deque()
            for i in coins:
                qu.append((i,1))
            su = 0
            co = 0
            vi = set()
            while qu:
                nn = len(qu)
                # print(qu)
                for i in range(nn):
                    cur , cnt = qu.popleft()
                    if cur == amount:
                        return cnt

                    for i in range(n):
                        if cur + coins[i] <= amount  and (cur,coins[i]) not in vi:
                            qu.append(((cur + coins[i]),cnt + 1))
                            vi.add((cur,coins[i]))
                
                            
                        
            return -1

        return bfs()
        

        
                    
                   
                        






















        def chk (i , s = 0, co = 1):
            nonlocal an
            if i >= n:
                return 
            s = coins[i] + s
            if  s > amount:
                print(co)
                return  

            if s == amount:
                an = min(an,co)
                return 
            for x in range(i,n):
               chk(i + x ,s , co+1)
                
        for i in range(n):
            chk(i)
            
        if an == 100001:
            return -1
        return an
            

    




        
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