<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 dp[5001][301];
    int help(int amt, int idx, vector<int>& nums) {
        if (amt == 0)
            return 1;
        if (idx >= nums.size())
            return 0;
        if (dp[amt][idx] != -1)
            return dp[amt][idx];
        int take = 0;
        if (amt - nums[idx] >= 0)
            take = help(amt - nums[idx], idx, nums);
        int notTake = help(amt, idx + 1, nums);

        return dp[amt][idx] = take + notTake;
    }
    int change(int amount, vector<int>& coins) {
        // memset(dp , -1 , sizeof (dp)) ;
        // return help(amount, 0, coins);
        int n = coins.size();
        vector<vector<int>> dp(amount + 2, vector<int>(n + 2, 0));

        for (int idx = 0; idx <= n; idx++)
            dp[0][idx] = 1;

        for (int amt = 0; amt <= amount; amt++)
            for (int idx = n - 1; idx >= 0; idx--) {
                int take = 0;
                if (amt - coins[idx] >= 0)
                    take = dp[amt - coins[idx]][idx];
                int notTake = dp[amt][idx + 1]; //(amt, idx + 1, nums);

                dp[amt][idx] = take + notTake;
            }
        return dp[amount][0];
    }
};
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