<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 {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = (int) coins.size();
        vector<vector<int>> dp(n, vector<int> (amount + 1, -1));
        function<int(int, int)> f = [&] (int i, int amount) {
            if (amount < 0) return int(1e9);
            if (i < 0) {
                if (amount == 0) {
                    return 0;
                } else {
                    return int(1e9);
                }
            }
            if (dp[i][amount] != -1) return dp[i][amount];
            return dp[i][amount] = min(f(i - 1, amount), f(i, amount - coins[i]) + 1);
        };
        int ans = f(n - 1, amount);
        return ans == 1e9 ? -1 : ans;
    }
};
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