<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 number of combinations that make up that amount</em>. If that amount of money cannot be made up by any combination of the coins, return <code>0</code>.</p>

<p>You may assume that you have an infinite number of each kind of coin.</p>

<p>The answer is <strong>guaranteed</strong> to fit into a signed <strong>32-bit</strong> integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> amount = 5, coins = [1,2,5]
<strong>Output:</strong> 4
<strong>Explanation:</strong> there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> amount = 3, coins = [2]
<strong>Output:</strong> 0
<strong>Explanation:</strong> the amount of 3 cannot be made up just with coins of 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> amount = 10, coins = [10]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= coins.length &lt;= 300</code></li>
	<li><code>1 &lt;= coins[i] &lt;= 5000</code></li>
	<li>All the values of <code>coins</code> are <strong>unique</strong>.</li>
	<li><code>0 &lt;= amount &lt;= 5000</code></li>
</ul>

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        vector<int> dp(amount+1,0);
        dp[0]=1;
        // for(int j=0;j<coins.size();j++){
        //         dp[coins[j]]=1;
        // }
        int n=coins.size();
        int t[n+1][amount+1];

        for(int i=0;i<=n;i++){
            for(int j=0;j<=amount;j++){
                if(i==0){
                    t[i][j]=0;
                }
                if(j==0){
                    t[i][j]=1;
                }
            }
        }

        for(int i=1;i<=n;i++){
            for(int j=1;j<=amount;j++){
                if(coins[i-1]<=j){
                    t[i][j]=t[i][j-coins[i-1]]+t[i-1][j];
                }
                else{
                    t[i][j]=t[i-1][j];
                }

            }
        }
        return t[n][amount];
        // for(int i=0;i<=amount;i++){
        //     int sum=0;
        //     for(auto coin:coins){
        //         if(coin<=i){
        //             dp[i]+=(dp[i-coin]);
        //         }
        //     }
        //     //dp[i]+=sum;
        //     cout<<dp[i]<<" "<<i<<endl;
        // }

    }
};
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