<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) {

        int n=coins.size();
        vector<vector<int>> dp(n,vector<int>(amount+1,0));

        for (int i = 0; i <= amount; i++) 
        {
            if (i % coins[0] == 0)
                dp[0][i] = i / coins[0];
            else
                dp[0][i] = 1e9; // Set it to a very large value if not possible

    }
        for(int i=1;i<n;i++)
        {
            for(int j=0;j<=amount;j++)
            {
                int nottake=dp[i-1][j];
                
                int take=1e9;
                if(coins[i]<=j)
                {
                    take=1+dp[i][j-coins[i]];
                }
                dp[i][j]=min(take,nottake);
            }
        }
        return dp[n-1][amount]>=1e9?-1:dp[n-1][amount];
        
    }
};
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