<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[302][5002];
    int solvememo(int amount,vector<int>&arr,int idx){
        if(idx==arr.size()){
            if(amount==0) return 1;
            return 0;
        }
        if(dp[idx][amount]!=-1) return dp[idx][amount];
        int take = 0;
        int nottake=solvememo(amount,arr,idx+1);
        if(arr[idx]<=amount)
        take=solvememo(amount-arr[idx],arr,idx);
        return dp[idx][amount]=take+nottake;
    }
    int solvetabulation(vector<int>&arr,int amount){
       int n= arr.size();
        int dp[n+1][amount+1];
        for(int i=0;i<=n;i++){
            for(int j=0;j<=amount;j++){
                dp[i][j]=0;
                if(j==0) dp[i][j]=1;
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=amount;j++){
                if(arr[i-1]<=j){
                    dp[i][j]=dp[i-1][j]+dp[i][j-arr[i-1]];
                }
                else{
                    dp[i][j]=dp[i-1][j];
                }
            }
        }
        return dp[n][amount];
    }
    int solvetabulationspaceoptimisation(vector<int>&arr,int amount){
       int n= arr.size();
        vector<int> pre(amount+1,0);
        pre[0]=1;
        for(int i=1;i<=n;i++){
            vector<int> curr(amount+1,0);
            curr[0]=1;
            for(int j=1;j<=amount;j++){
                if(arr[i-1]<=j){
                    curr[j]=pre[j]+curr[j-arr[i-1]];
                }
                else{
                    curr[j]=pre[j];
                }
            }
            pre=curr;
        }
        return pre[amount];
    }
    int change(int amount, vector<int>& arr) {
        // memset(dp,-1,sizeof(dp));
        // solvememo(amount,arr,0);
        //solvetabulation(arr,amount);
        return solvetabulationspaceoptimisation(arr,amount);
    }
};
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