<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>

/*
coins -> [1 to 2^31]
amount -> [0 to 10000] 
 [2,5] amount = 13
 [5] amount => 8 len 5 
 [1,2,5] => 11
 [1,2,5] => 6
 1. if amount == currSumOfCoins I have reached return len;
 2. if idx > coins.size() return INT_MAX; 

    prune sort the coins then if you know the currentLeftAmount<coins[idx] [idx,n)

 [5,5] len => 3
*/


class Solution {
public:

    int minCoinsToSum(int index,vector<int> &coins,int amount,vector<vector<int>> &dp){
        if(amount == 0){
            return dp[index][amount] = 0;
        }

        if(index>=coins.size()){
            return dp[index][amount] = 1e9;
        }
        if(dp[index][amount]!=-1) return dp[index][amount];
        // check if I can take it or not
        if(amount<coins[index]){
            return dp[index][amount] = 1e9;
        }

        return dp[index][amount] = min({1+minCoinsToSum(index+1,coins,amount-coins[index],dp),minCoinsToSum(index+1,coins,amount,dp),1+minCoinsToSum(index,coins,amount-coins[index],dp)});

    }

    int coinChange(vector<int>& coins, int amount) {
        sort(coins.begin(),coins.end());

        vector<vector<int>> dp(coins.size()+1,vector<int>(amount+1,-1));

        int ans = minCoinsToSum(0,coins,amount,dp);
        if(ans>=1e9) return -1;
        return ans;
    }
};
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