<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 solve(self, coins, t, i):
        # Base case: if we are at the last coin
        if i == len(coins) - 1:
            if t % coins[i] == 0:
                self.dp[i][t] = t // coins[i]
                return self.dp[i][t]
            self.dp[i][t] = float('inf')  # Impossible case, return inf
            return self.dp[i][t]
        
        # Return the already computed value if available
        if self.dp[i][t] != -1:
            return self.dp[i][t]
        
        # Case where we do not take the current coin
        not_take = self.solve(coins, t, i + 1)
        
        # Case where we take the current coin (if possible)
        take = float('inf')
        if coins[i] <= t:
            take = 1 + self.solve(coins, t - coins[i], i)
        
        # Store the result in dp table and return
        self.dp[i][t] = min(take, not_take)
        return self.dp[i][t]

    def coinChange(self, coins: list[int], amount: int) -> int:
        # Initialize dp table with -1 (for memoization)
        self.dp = [[-1 for _ in range(amount + 1)] for _ in range(len(coins))]
        
        # Solve the problem starting from the first coin
        ans = self.solve(coins, amount, 0)
        
        # If no valid solution exists, return -1
        if ans >= float('inf'):
            return -1
        
        return ans
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