<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 {
private:
    int count_ways(int idx, int target, vector<vector<int>>& dp, vector<int>& coins){
        if(idx == 0){
            if(target % coins[idx] == 0){
                return 1;
            }
            else{
                return 0;
            }
        }
        if(dp[idx][target] != -1){
            return dp[idx][target];
        }
        int not_take = count_ways(idx-1, target, dp, coins);
        int take = 0;
        if(coins[idx] <= target){
            take = count_ways(idx, target - coins[idx], dp, coins);
        }
        return dp[idx][target] = take + not_take;
    }
public:
    int change(int amount, vector<int>& coins) {
        int n = coins.size();
        vector<vector<long>> dp(n, vector<long> (amount + 1, 0));
        for(int i = 0;i <= amount;i++){
            if(i % coins[0] == 0){
                dp[0][i] = 1;
            }
        }
        for(int i = 1;i < n;i++){
            for(int j = 0;j <= amount;j++){
                long not_take = dp[i-1][j];
                long take = 0;
                if(coins[i] <= j){
                    take = dp[i][j - coins[i]];
                }
                dp[i][j] = take + not_take;
            }
        }
        return dp[n - 1][amount];
    }
};
30865