<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:

    vector<int> dp ;
    void nbCoins(vector<int> coins ,int amount){
        if(amount == 0){
            return ;
        }
        int nb = amount+1;
        for( int i =0 ; i<coins.size() ; i++){
            int diff = amount - coins[i];
            if(diff<0){
            continue;
            }
            if(dp[diff] == -2){
                nbCoins(coins,diff);
            }
            if(dp[diff] == -1){
                continue;
            }else{
                nb = min(nb,1+dp[diff]);
            }       
    }if(nb == amount+1)
        dp[amount] = -1;
    else
        dp[amount] = nb;
    }
    int coinChange(vector<int>& coins, int amount) {
        for( int  i =  0 ;i<=amount;i++){
            dp.push_back(-2);
        }
        dp[0] = 0;
        nbCoins(coins,amount);
       return dp[amount]; 
    }
};
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