<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 ans = INT_MAX;
    vector < int > memo;
    int ways(int amount, vector < int > &coins)
    {
        if(amount<0) return INT_MAX;
        if(amount==0)
        {
            return 0;
        }
        if(memo[amount]!=-1) return memo[amount];
        int min_coins = INT_MAX;
        for(int i = 0;i<coins.size();i++)
        {
            int result = ways(amount-coins[i],coins);
            if(result!=INT_MAX) min_coins = min(min_coins, result+1);
        }
        return memo[amount] = min_coins;
    }
    int coinChange(vector<int>& coins, int amount) {
        memo.resize(amount+1,-1);
        ans = ways( amount, coins);
        return ans==INT_MAX?-1:ans;
    }
};
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