<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) {
        if (amount == 0) {
            return 0;
        }

        // Sort coins in reverse order (largest to smallest)
        std::sort(coins.rbegin(), coins.rend());

        // Priority queue to store (count of coins used, remaining amount)
        std::priority_queue<std::pair<int, int>, std::vector<std::pair<int, int>>, std::greater<>> q;

        // Initialize the priority queue with the first set of states
        for (int coin : coins) {
            if (amount >= coin) {
                q.push({1, amount - coin});
            }
        }

        // Hash map to track the smallest count to reach a particular total
        std::unordered_map<int, int> seen;
        for (int i = 0; i <= amount; ++i) {
            seen[i] = INT_MAX;
        }

        // Process the priority queue
        while (!q.empty()) {
            auto [count, total] = q.top();
            q.pop();

            // If total is 0, we found the solution
            if (total == 0) {
                return count;
            }

            // For each coin, push the new state if it's more efficient
            for (int coin : coins) {
                int next_total = total - coin;
                if (next_total >= 0) {
                    // Proceed if it's more efficient
                    if (seen[next_total] > count) {
                        seen[next_total] = count;
                        q.push({count + 1, next_total});
                    }
                }
            }
        }

        // If no solution was found, return -1
        return -1;
    }
};
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