<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 findFewestByRecursion(vector<int>& coins, int amount){

        /// base case
        if(amount == 0) return 0;
        if( amount < 0) return INT_MAX;


        int mini = INT_MAX;
        for(int i = 0; i < coins.size(); i++){
            int ans = findFewestByRecursion(coins,amount-coins[i]);
            if(ans != INT_MAX){
                mini = min(mini, 1+ans);
            }
        }
        return mini;


    }
    int findFewestByMemo(vector<int>& coins, int amount,vector<int>&dp){

        /// base case
        if(amount == 0) return 0;
        if( amount < 0) return INT_MAX;

        // Check value present in DP 
        if(dp[amount] != -1){
            return dp[amount];
        }

        int mini = INT_MAX;
        for(int i = 0; i < coins.size(); i++){
            int ans = findFewestByMemo(coins,amount-coins[i],dp);
            if(ans != INT_MAX){
                mini = min(mini, 1+ans);
            }
        }
        dp[amount]= mini;
        return dp[amount];


    }

    int coinChange(vector<int>& coins, int amount) {
        //if(amount == 0) return 0; 
        vector<int> dp(amount+1, -1); // step1
        int ans = findFewestByMemo(coins,amount,dp);
        if(ans == INT_MAX)return -1;
        else return ans;
    }
};
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