<p>You are given an array <code>prices</code> where <code>prices[i]</code> is the price of a given stock on the <code>i<sup>th</sup></code> day, and an integer <code>fee</code> representing a transaction fee.</p>

<p>Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.</p>

<p><strong>Note:</strong></p>

<ul>
	<li>You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).</li>
	<li>The transaction fee is only charged once for each stock purchase and sale.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> prices = [1,3,2,8,4,9], fee = 2
<strong>Output:</strong> 8
<strong>Explanation:</strong> The maximum profit can be achieved by:
- Buying at prices[0] = 1
- Selling at prices[3] = 8
- Buying at prices[4] = 4
- Selling at prices[5] = 9
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> prices = [1,3,7,5,10,3], fee = 3
<strong>Output:</strong> 6
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= prices.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= prices[i] &lt; 5 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= fee &lt; 5 * 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int n = prices.size();
        int lastSold = 0;
        int t = 0;
        int *dp = new int[n];
        dp[0] = 0;
        for(int i = 1; i < n; i++){
            dp[i] = 0;
            t = 0;
            if(prices[i]-prices[i-1]>0){
                if(lastSold==i-1&&i-1!=0){
                    dp[i] = dp[i-1]+prices[i]-prices[i-1];
                    lastSold = i;
                }
                else if(lastSold==i-1&&i-1==0){
                    if(prices[i]-prices[i-1]>fee){
                        dp[i] = dp[i-1]+prices[i]-prices[i-1]-fee;
                        lastSold = i;
                    }                    
                }
                else{
                    //cout<<lastSold<<","<<i<<endl;
                    if(prices[i]-prices[lastSold]>0&&lastSold!=0){
                        if(dp[i]<dp[lastSold]+prices[i]-prices[lastSold]){
                            dp[i] = dp[lastSold]+prices[i]-prices[lastSold];
                            t = 1;
                        }
                    }
                    else if(prices[i]-prices[lastSold]>fee&&lastSold==0){
                        if(dp[i]<dp[lastSold]+prices[i]-prices[lastSold]-fee){
                            dp[i] = dp[lastSold]+prices[i]-prices[lastSold]-fee;
                            t = 1;
                        }
                    }
                    for(int j = i-1; j > lastSold; j--){
                        if(prices[i]-prices[j]>fee){
                            //cout<<"here "<<i<<","<<j<<endl;
                            if(dp[i]<dp[j]+prices[i]-prices[j]-fee){
                                dp[i] = dp[j]+prices[i]-prices[j]-fee;
                                t = 1;
                            }
                        }
                    }
                    if(t==1) lastSold = i;
                    else dp[i] = dp[i-1];
                }
            }
            else{
                dp[i] = dp[i-1];
            }
        }
        return dp[n-1];

    }
};
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