<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) 
    {
        vector <int> prev(10001), now(10001);
        for(int am=0; am<=amount; am++)
            prev[am] = am%coins[0]==0 ? am/coins[0] : 1e9;
        for(int cur=1; cur<coins.size(); cur++)
        {
            for(int am=0; am<=amount; am++)
            {
                int doNotTakeCoin = 0 + prev[am];
                int takeCoin = INT_MAX;
                if(coins[cur] <= am) 
                    takeCoin = 1 + now[am-coins[cur]];
                now[am] = min(takeCoin, doNotTakeCoin);
            }
            prev = now;
        }
        return (prev[amount] >= 1e9) ? -1 : prev[amount];
    }
};
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