<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 amount,int idx,vector<vector<int>>& dp){

        if(idx==0){

            if(amount % coins[0]==0){

                return amount / coins[0];
            }

            else return INT_MAX;

        }

        if(dp[idx][amount]!=-1){

            return dp[idx][amount];
        }

        int take = INT_MAX;

        int nottake = solve(coins,amount,idx-1,dp);

        if(amount-coins[idx]>=0){

            take = solve(coins,amount-coins[idx],idx,dp);

            if(take!=INT_MAX){

                take = 1 + take;
            }
        }

        return dp[idx][amount] = min(take,nottake);

    }


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

        vector<vector<int>> dp(coins.size(),vector<int>(amount+1,0));

        vector<int> prev(amount+1,0) , curr(amount+1,0);

        // int ans =  solve(coins,amount,coins.size()-1,dp);

        // if(ans==INT_MAX) return -1;

        // return ans;

        for(int i=0;i<=amount;i++){

            if(i % coins[0]==0) prev[i] = (i / coins[0]);
            else prev[i] = INT_MAX;

        }

        for(int idx=1;idx<coins.size();idx++){

            for(int target = 0;target<=amount;target++){

                int take = INT_MAX;

                int nottake = prev[target];

                if(target-coins[idx]>=0){

                    take = curr[target-coins[idx]];

                    if(take!=INT_MAX){

                        take = 1 + take;
                    }
            }

                curr[target] = min(take,nottake);
            }

            prev = curr;
        }

        
        if(prev[amount]==INT_MAX){

            return -1;
        }
        
        return prev[amount];
        


    }
};
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