<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:
    //Method - 01 (Memoization)
    int helper(int amount, int idx, vector<int>& coins, vector<vector<int>>& dp){
        if(amount == 0)
            return 1;
        if(idx < 0)
            return 0;
        if(amount < 0)
            return 0;
        if(dp[idx][amount] != -1)
            return dp[idx][amount];
        int notTake = helper(amount, idx-1, coins, dp);
        int take = helper(amount-coins[idx], idx, coins, dp);
        return dp[idx][amount] = notTake + take;
    }
    int change(int amount, vector<int>& coins) {
        int n = coins.size();

        //Method - 02 (Tabulation)

        // vector<vector<int>> dp(n, vector<int>(amount+1, 0)); 
        //dp[i][j] => no of ways of getting amount j up to index i
        // for(int i=0; i<n; i++)
        //     dp[i][0] = 1;
        // for(int i=0; i<n; i++){
        //     for(int j=1; j<=amount; j++){
        //         if(j-coins[i] >= 0)
        //             dp[i][j] = dp[i][j-coins[i]];
        //         if(i-1 >= 0)
        //             dp[i][j] += dp[i-1][j];
        //     }
        // }
        // return dp[n-1][amount];

        // Method-03(Space Optimization)
        vector<int> dp(amount+1, 0); 
        dp[0] = 1;
        for(int i=1; i<=amount; i++){
            if(i-coins[0] >= 0)
                dp[i] = dp[i-coins[0]];
        }
        for(int i=1; i<n; i++){
            vector<int> temp(amount+1, 0);
            temp[0] = 1;
            for(int j=1; j<=amount; j++){
                if(j-coins[i] >= 0)
                    temp[j] = temp[j-coins[i]];
                temp[j] += dp[j];
            }
            dp = temp;
        }
        return dp[amount];
    }
};
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