<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 fun(int ind, int rem, vector<int>& coins, vector<vector<int>>& mem){

        if(ind == 0){
            return rem == 0 ? 0 : INT_MAX-1;
        }

        if(mem[ind][rem] != -1)
            return mem[ind][rem];

        if(rem >= coins[ind-1]){
            return mem[ind][rem] =  min(1 + fun(ind, rem-coins[ind-1], coins, mem), fun(ind-1, rem, coins, mem));
        }
        else{
            return mem[ind][rem] = fun(ind-1, rem, coins, mem);
        }
    }
    int coinChange(vector<int>& coins, int amount) {
        
        int m = coins.size();
        vector<vector<int>> dp(m+1, vector<int>(amount+1, INT_MAX-1));

        for(int i=0; i<=m; i++)
            dp[i][0] = 0;
        
        for(int i=1; i<=m; i++){
            for(int j=1; j<=amount; j++){
                if(j >= coins[i-1]){
                    dp[i][j] = min(1+dp[i][j-coins[i-1]], dp[i-1][j]);
                }
                else{
                    dp[i][j] = dp[i-1][j];
                }
            }
        }

        int ans = dp[m][amount];

        return ans == INT_MAX-1 ? -1 : ans;
    }
};
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