<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) {
        // return dp(coins, amount);
        return bfs(coins, amount);
    }
private:
    int dp(vector<int>& coins, int amount) {
        if (amount == 0) return 0;

        vector<int> dp(amount+1, amount+1);
        dp[0] = 0;

        for (int i = 1; i <= amount; ++i) {
            for (int coin : coins) {
                if (i >= coin) {
                    dp[i] = min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        if (dp[amount] == amount + 1) return -1;
        return dp[amount];
    }

    int bfs(vector<int>& coins, int amount) {
        if (amount == 0) return 0;
        
        // vector<bool> seen(amount + 1, false);
        unordered_set<int> seen;
        queue<int> q;
        int count = 0;
        
        q.push(0);
        
        while (!q.empty()) {
            int size = q.size();
            count++;
            
            for (int i = 0; i < size; i++) {
                int curr = q.front();
                q.pop();
                
                for (int coin : coins) {
                    int next = curr + coin;
                    if (next == amount) return count;
                    // if (next < amount && !seen[next]) {
                    if (next < amount && seen.count(next) == 0) {
                        seen.insert(next);
                        q.push(next);
                    }
                }
            }
        }
        return -1;
    }
};

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