<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 f(int ind, int amount, vector<int>& coins, vector<vector<int>>& dp){
//     if(ind == 0){
//         if(amount % coins[0] == 0) return 1;
//         return 0;
//     }
//     if(dp[ind][amount] != -1) return dp[ind][amount];
//     int nottake = f(ind-1, amount, coins, dp);
//     int take = 0;
//     if(coins[ind] <= amount){
//         take = f(ind, amount-coins[ind], coins, dp);
//     }
//     return dp[ind][amount] = take + nottake;
// }
    int change(int amount, vector<int>& coins) {
        int n = coins.size();
        vector<int> prev(amount+1, 0);
        for(int i = 0 ; i <= amount; i++){
            if(i%coins[0] == 0) prev[i] = 1;
        }
        for(int i = 1; i < n; i++){
            vector<int> curr(amount+1, 0);
            for(int j = 0; j <= amount; j++){
                int nottake = prev[j];
                int take = 0;
                if(coins[i] <= j){
                    take = curr[j-coins[i]];
                }
                curr[j] = take + nottake;
            }
            prev = curr;
        }
        // int ans = dp[n-1][amount];
        return prev[amount];
    }

};
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