<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) {
        // if i - coins[j] >= 0:
        // coins[j] can be used(exclude or include)
        
        // if i - coins[j] < 0
        // coins[j] cannot be used(exclude)
        vector<vector<int>>dp(amount + 1, vector<int>(coins.size(), 0));
        
        for(int i = 0; i < coins.size(); i++)
        {
            dp[0][i] = 1;
        }
        
        for(int i = 1; i <= amount; i++)
        {
            for(int j = 0; j < coins.size(); j++)
            {
                if (i - coins[j] >= 0)
                {
                    if (j > 0)
                    {
                        dp[i][j] = dp[i - coins[j]][j] + dp[i][j - 1];
                    }
                    else
                    {
                        dp[i][j] = dp[i - coins[j]][j];
                    }
                }
                else
                {
                    if (j > 0)
                    // cannot use coins[j] to make amount i
                    dp[i][j] = dp[i][j - 1];
                }
            }
        }
        return dp[amount][coins.size() - 1];
    }
};
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