**Problem link** — https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/

**Problem Description:**

Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee.

You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)

Return the maximum profit you can make.

**Example 1:** Input: prices = [1, 3, 2, 8, 4, 9], fee = 2

Output: 8 Explanation: 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.

**Note:** 0 < prices.length <= 50000; 0 < prices[i] < 50000; 0 <= fee < 50000.

**My Solution** : I have come up with a backtrack solution, which is working fine with smaller inputs, but gives TLE with bigger inputs. I am unable to memoize it. Need Help !!!

What I am doing is, I am checking if the we have already purchased a share, If Yes, either we can sell the share now or later. If No, we either purchase it now or later.

~~~~ class Solution { public:

int solve (vector<int> &prices, bool hasPurchased, int lastPurchasedVal, int i, int fee) { if (i == prices.size()) return 0; // if (dp[hasPurchased][i] != -1) return dp[hasPurchased][i]; if (hasPurchased) { int sold = INT_MIN; int profit = prices[i] - lastPurchasedVal - fee; if (profit > 0) sold = profit + solve (prices, !hasPurchased, 0, i+1, fee); int notsold = solve(prices, hasPurchased, lastPurchasedVal, i+2, fee); return max(sold, notsold); } else { int purchased = solve (prices, !hasPurchased, prices[i], i+1, fee); int notPurchased = solve(prices, hasPurchased, lastPurchasedVal, i+1, fee); return max(purchased, notPurchased); } } int maxProfit(vector<int>& prices, int fee) { //int dp[2][50001]; //memset(dp, -1, sizeof dp); return solve (prices, false, 0, 0, fee); }

}; ~~~~~

Thank you.