<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 fewest number of coins that you need to make up that amount</em>. If that amount of money cannot be made up by any combination of the coins, return <code>-1</code>.</p>

<p>You may assume that you have an infinite number of each kind of coin.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> coins = [1,2,5], amount = 11
<strong>Output:</strong> 3
<strong>Explanation:</strong> 11 = 5 + 5 + 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> coins = [2], amount = 3
<strong>Output:</strong> -1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> coins = [1], amount = 0
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= coins.length &lt;= 12</code></li>
	<li><code>1 &lt;= coins[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>0 &lt;= amount &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int f(int ind,int target, vector<int>& coins,vector<int> prev,vector<int> curr){
        int n=coins.size();
        
        for (int t = 0; t <= target; t++) {
            if (t % coins[0] == 0) {
                prev[t] = t / coins[0];
            }
        }
              
        for(int i=1;i<n;i++){
            for(int j=0;j<target+1;j++){
                
            int take=1e9;
            int nottake=prev[j];
            if(coins[i]<=j){
                take=1+curr[j-coins[i]];}
            curr[j]=min(take,nottake);
            

            }
            prev=curr;
            
        }
        return  prev[target];
        

    }
    int coinChange(vector<int>& coins, int amount) {
        if(amount==0){return 0;}
        int n=coins.size();
        vector<int> prev(amount+1,1e9);
        vector<int> curr(amount+1,1e9);
        
        
        int ans= f(n-1,amount,coins,prev,curr);
        if(ans==1e9){return -1;}
        return ans;
    }
};
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