<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 mod = 1e9 + 7;
    long long dp[301][5001];
    long long rec(int level,int left,vector<int>& c){
        if(left < 0) return 0;
        
        if(level == c.size()){
            if(left == 0) return 1;
            else return 0;
        }

        if(dp[level][left] != -1) return dp[level][left];

        long long ans = 0;
        

        ans = ans + rec(level,left-c[level],c);
        ans = ans%mod;

        ans = ans + rec(level+1,left,c);
        ans = ans%mod;

        return dp[level][left] = ans;
    }
    int change(int target, vector<int>& c) {
        long long dp[2][target+2];

        long long mod = 1e17;
        memset(dp,0,sizeof(dp));
        int n = c.size();
        // dp[i][j] = dp[i][j-c[i]] + dp[i+1][j]
        for(int j = 0;j <= target;j++){
            if(j == 0){
                dp[(n)%2][j] = 1;
            }
            else{
                dp[(n)%2][j] = 0;
            }
        }
        for(int i = n-1;i >= 0;i--){
            for(int j = 0;j <= target;j++){
                dp[(i%2)][j] = 0;
            }

            for(int j = 0;j <= target;j++){
                long long x = dp[((i+1)%2)][j];
                x = x%mod;
                if(j - c[i] >= 0){
                    x = x + dp[(i%2)][j-c[i]];
                    x = x%mod;
                }
                dp[(i%2)][j] = x%mod;
            }
        }

        long long p = dp[0][target]%mod;

        return p;
    }
};
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