<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 countways(int amount,vector<int>&coins,int n,vector<vector<int>>&dp){
        if(amount==0)return 1;
        if(n<0)return 0;
        if(dp[amount][n]!=-1)return dp[amount][n];
        int take=0;
        if(amount>=coins[n])take=countways(amount-coins[n],coins,n,dp);
        int not_take=countways(amount,coins,n-1,dp);
        return dp[amount][n]=take+not_take;
    }
    int change(int amount, vector<int>& coins) {
        vector<vector<int>>dp(amount+1,vector<int>(coins.size(),0));
        // return countways(amount,coins,coins.size()-1,dp);
        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++){
                    int take=0;
                    if(i-coins[j]>=0 && amount>=coins[j])take=dp[i-coins[j]][j];
                    int not_take=j-1>=0?dp[i][j-1]:0;
                    dp[i][j]=take+not_take;
            }
        }
        return dp[amount][coins.size()-1];
    }
};
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