<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) {

        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>pq;

        pq.push({0,amount});

        unordered_set<int>st;

        st.insert(amount);

        while(!pq.empty()){

            int coin=pq.top().first;

            int amt=pq.top().second;

            // cout<<coin<<" "<<amt<<endl;

            pq.pop();

            if(amt==0)return coin;

            for(int j:coins){

                if(amt-j==0)return coin+1;

                else if(amt-j>0 && st.find(amt-j)==st.end()){

                    pq.push({coin+1,amt-j});

                    st.insert(amt-j);

                }

            }

        }

        return -1;

    }

};

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