<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) {
        queue<pair<int,int>> q;
        vector<int> vis(amount+1,0);
        if(amount == 0){
            return 0;
        }
        q.push({amount,0});
        vis[amount] = 1;
        while(!q.empty()){
            int nums = q.front().first;
            int steps = q.front().second;
            q.pop();
            for(int i=0; i<coins.size(); i++){
                if(nums-coins[i] == 0){
                    return steps+1;
                }
                if(nums-coins[i] > 0 && !vis[nums-coins[i]] ){
                    q.push({nums-coins[i],steps+1});
                    vis[nums-coins[i]] = 1;
                }
            }
        }
        return -1;
    }
};
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