<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) {
        unordered_map<int, int> dp;
        dp[0] = 0;
        sort(begin(coins), end(coins));
        for (int i = 1; i < amount + 1; i++) {
            for(int coin : coins) {
                if (i - coin < 0) break;
                if (dp.find(i-coin) != dp.end()) {
                    if (dp.find(i) != dp.end()) {
                        dp[i] = min(dp[i], dp[i-coin] + 1);
                    } else {
                        dp[i] = dp[i-coin] + 1;
                    }
                }
            }
        }
        if (dp.find(amount) != dp.end()) {
            return dp[amount];
        } else {
            return -1;
        }
    }
};
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