<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 {
    vector<long> dp;
public:
    Solution() {
        dp = vector<long>(10001, -1);   
    }
public:
    int coinChange(vector<int>& coins, int amount) {
        //init(amount);
        long ans = numCoins(amount, coins);
        return (ans >= INT_MAX) ? -1 : ans;
    }

    long numCoins(int n, vector<int>& coins) {
        if (n == 0) {
            return 0;
        }

        if (dp[n - 1] != -1) {
            return dp[n - 1];
        }

        long ans = INT_MAX;
        for (int i = 0; i < coins.size(); ++i) {
            if ((n - coins[i]) >= 0) {
                ans = min(ans, numCoins(n - coins[i], coins) + 1);
            }
        }
        
        return dp[n - 1] = ans;
    }
};
30809