<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 noofcoins(vector<int>coins,int a,vector<vector<int>>&dp,int i,int n)
{ vector<int>prev(a+1,0);
for(int i=0;i<=a;i++) {if(i%coins[0]==0) prev[i]=i/coins[0];
else prev[i]=1e9;
}

for(int i=1;i<n;i++){
 vector<int>curr(a+1,0);    
for(int j=0;j<=a;j++){
int pick=INT_MAX;
int npick=prev[j];
if(coins[i]<=j) pick=1+curr[j-coins[i]];
 curr[j]=min(pick,npick);
}
prev=curr;
}
return prev[a];
}
    int coinChange(vector<int>& coins, int amount) {
    int n=coins.size();
    vector<vector<int>>dp(n,vector<int>(amount+1,0));
    int ans= noofcoins(coins,amount,dp,n-1,n);
    if(ans>=1e9) return -1;
    return ans;
    }
};
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