<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 f(int ind, int amt, vector<int>& coins) {

        if (amt == 0)
            return 0;

        if (ind < 0)
            return INT_MAX - 1;

        // if (amt % coins[ind] == 0)
        //     return amt / coins[ind];
        int ans = INT_MAX - 1;
        if (coins[ind] <= amt) {
            ans = min(ans, 1 + f(ind, amt - coins[ind], coins));
        }
        ans = min(ans, f(ind - 1, amt, coins));

        return ans;
    }

    int coinChange(vector<int>& coins, int amt) {
        sort(coins.begin(), coins.end());
        int n = coins.size();
        vector<vector<int>> dp(n + 1, vector<int>(amt + 1, INT_MAX - 1));
        int off = 1;
        vector<int> prev(amt + 1, INT_MAX - 1);
        vector<int> curr(amt + 1);
        prev[0] = 0;
        for (int i = 0; i < n; i++) {
            fill(curr.begin() + 1, curr.end(), INT_MAX - 1);
            curr[0] = 0;
            for (int j = 1; j <= amt; j++) {
                if (coins[i] <= j) {
                    curr[j] = min(1 + curr[j - coins[i]], curr[j]);
                }
                curr[j] = min(curr[j], prev[j]);
            }
            prev = curr;
        }

        return curr[amt] == INT_MAX - 1 ? -1 : curr[amt];
    }
};
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