<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;
        }
        vector<int> prev(amount+1, 0);
        prev[0] = 0;
        int n = coins.size();
        for(int j = 1;j<=amount;j++){
            if(coins[0]<=j && j%coins[0]==0){
                prev[j] = j/coins[0];
            }else{
                prev[j] = 1e9;
            }
        }
        for(int i = 1;i<n;i++){
            if(coins[i]>amount){
                continue;
            }
            vector<int> curr(amount+1, 0);
            for(int j=0;j<=amount;j++){
                int notTake = prev[j];
                int take = 1e9;
                if(coins[i]<=j){
                    take = 1+ curr[j-coins[i]];
                }
                curr[j] = min(notTake, take);
            }
            prev = curr;
        }
        if(prev[amount]==1e9){
            return -1;
        }
        return prev[amount];
    }
};
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