<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:
        if amount == 0:
            return 0
        n = len(coins)
        dp = [[[0,0]]*(amount + 1) for _ in range(n + 1)]
        coins.insert(0, 0)
        for i in range(1, n + 1):
            for j in range(amount + 1):
                if coins[i] > j:
                    dp[i][j] = dp[i-1][j]
                else:
                    number = None
                    val1 = dp[i - 1][j - coins[i]]
                    val2 = dp[i][j - coins[i]]
                    val3 = dp[i-1][j]
                    candidate = None
                    
                    if val1[0] > val2[0]:
                        number = dp[i - 1][j - coins[i]][1]
                        candidate = val1[0]
                    elif val1[0] < val2[0]:
                        number = dp[i][j - coins[i]][1]
                        candidate = val2[0]
                    else:
                        candidate = val1[0]
                        number = min(val1[1], val2[1])
                    
                    
                    if candidate + coins[i] < val3[0]:
                        candidate = val3[0]
                        number = val3[1]
                    elif candidate + coins[i] == val3[0]:
                        candidate += coins[i]
                        number = min(number + 1, val3[1]) 
                    else:
                        candidate += coins[i]
                        number += 1
                    dp[i][j] = [candidate, number]

        if dp[n][amount][0] < amount:
            return -1
        return dp[n][amount][1]
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