<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 sz = size(coins);
        vector<vector<int>> dp(sz + 1, vector<int>(amount + 1, INT_MAX));
        function<int(int, int)> minCoins = [&](int i, int amount) {
            if(amount == 0)
                return 0;
            if(amount < 0 || i == coins.size())
                return 100000;
            if(dp[i][amount] != INT_MAX)
                return dp[i][amount];
            int minCount = minCoins(i, amount - coins[i]);
            if(minCount != 100000)
                minCount++;
            return dp[i][amount] = min(minCount, minCoins(i + 1, amount));
        };
        int minCount = minCoins(0, amount);
        return minCount != 100000 ? minCount : -1;
    }
};
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