<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://now lets solved this using 2d dp tabulation
    int coinChange(vector<int>& coins, int amount) {
        int n=coins.size();
        //lets declare our dp vector of 2d
        vector<vector<int>>dp(2,vector<int>(amount+1,INT_MAX));  //initializing with INT_MAX
        //now for 0th col setting every value as 0
        for(int i=0;i<=1;i++)  //for 0th column
        {
            dp[i][0]=0;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=amount;j++)
            {
                if(j-coins[i-1]<0)    dp[1][j]=dp[0][j];   //leave
                else
                {
                    long long take=1LL+dp[1][j-coins[i-1]];
                    long long leave=dp[0][j];
                    dp[1][j]=min(take,leave);
                }
            }
            //now we have to copy the 1th row and paste it into 0th row
            for(int k=1;k<=amount;k++)  //k=1 because we will not make any changes in the 0th col connected cell
            {
                dp[0][k]=dp[1][k];
            }
        }
        int ans=dp[1][amount];  //as size are (2)*(amount+1)
        if(ans==INT_MAX)    return -1;  //no valid combinations are present
        return ans;
    }
};
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