<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:
    using node = pair<int, int>;

    int coinChange(vector<int>& coins, int amount) {
        // TLE
        // We can do a BFS
        // Regarding amount as the first node, coins are the edges
        // We make a node to track {current coin count, amount left}
        // In BFS, the first pop() elements from the queue are the "shortest" path.
        queue<node> q;
        set<int> amountSeen;
        int coinNum = coins.size();
        sort(coins.begin(), coins.end());
        q.push({amount, 0});
        amountSeen.insert(amount);
        int minCount = amount + 1;
        while(!q.empty()){
            int currAmount = q.front().first;
            int currCoinCount = q.front().second;
            q.pop();
            
            // cout<<currAmount<<endl;
            if (currAmount == 0)    return currCoinCount;
            // Traverse from 
            for (int i = coinNum - 1; i >= 0; i --) {
                int nextAmount = currAmount - coins[i];
                if (nextAmount >= 0)
                    if (!amountSeen.count(nextAmount)){
                        // ensure first-visit
                        q.push({nextAmount, currCoinCount+1});
                        amountSeen.insert(nextAmount);
                    }
            }
        }
        return -1;
    }
};
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