<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 memo(int idx, int sum, int amount, vector<int>& coins, vector<vector<int>>& dp){
        if(idx == 0) return !((amount - sum) % coins[0]);
        if(dp[idx][sum] != -1) return dp[idx][sum];

        int take = 0, notTake = memo(idx - 1, sum, amount, coins, dp);
        if(sum + coins[idx] <= amount) 
            take = memo(idx, sum + coins[idx], amount, coins, dp);

        return dp[idx][sum] = take + notTake;
    }

    int change(int amount, vector<int>& coins) {
        ios::sync_with_stdio(0); cin.tie(0);
        int n = coins.size();
        // vector<vector<int>> dp(n, vector<int>(amount + 1, -1));
        // for(auto& vec: dp) vec[amount] = 1;
        // return memo(n - 1, 0, amount, coins, dp);

        vector<int> prev(amount + 1, 0), cur(amount + 1, 0);
        for(int i = 0; i <= amount; i++) 
            prev[i] = !((amount - i) % coins[0]);

        for(int idx = 1; idx < n; idx++){
            for(int sum = amount; sum >= 0; sum--){
                int take = 0, notTake = prev[sum];
                if(sum + coins[idx] <= amount) 
                    take = cur[sum + coins[idx]];

                cur[sum] = take + notTake;
            }
            prev = cur;
        }

        return prev[0];
    }
};
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