<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<pair<long long,long long>> queue;
    int coinChange(vector<int>& c, int amount) {
        if(amount==0)
            return 0;
        sort(c.begin(),c.end());
        reverse(c.begin(),c.end());
        
        int left = 0;
        queue.push_back({0,0});

        int dp[10002] = {0};
        while(left<queue.size()){
            long long currAmount = queue[left].first;
            long long numCoins = queue[left].second;
            
            
            for(int i = 0;i<c.size();i++){
                if(currAmount + c[i] >amount){
                    continue;
                }
                if((currAmount + c[i])==amount)
                    return numCoins+1;
                
                if(dp[currAmount + c[i]]){
                    continue;
                }
                dp[currAmount + c[i]] = 1;
                queue.push_back({currAmount + c[i],numCoins + 1});
                
            }
            left++;
        }

        return -1;
    }

};
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