<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 solve(vector<int>& coins,int i,int amount,vector<vector<int>>&dp){


if(amount==0) return 0;
if(i==coins.size() || amount<0) return 1e9;
if(dp[i][amount]!=-1) return dp[i][amount];

int ans= min(solve(coins,i+1,amount-coins[i],dp)+1,min(solve(coins,i,amount-coins[i],dp)+1,solve(coins,i+1,amount,dp)));


return dp[i][amount]=ans;
}

    int coinChange(vector<int>& coins, int amount) {
int n=coins.size();
vector<vector<int>>dp(n+2,vector<int>(amount+2,-1));
int c=solve(coins,0,amount,dp);
if(c>=1e9) return -1;
else return c;




    }
};
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