714. Best Time to Buy and Sell Stock with Transaction Fee

Problem Statement

You are given an array prices where prices[i] is the price of a given stock on the ith day, and an integer fee representing a transaction fee.

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.

Note:

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

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.

Example 2:

Input: prices = [1,3,7,5,10,3], fee = 3
Output: 6

Constraints:

1 <= prices.length <= 5 * 104
1 <= prices[i] < 5 * 104
0 <= fee < 5 * 104

Intuition

Just subtract transaction fee from the II question

Links

https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/

Video Links

https://www.youtube.com/watch?v=k4eK-vEmnKg&ab_channel=takeUforward

Approach 1:

MEmoization

C++

class Solution {
public:
    int find_ans(vector<int>& prices, int index, bool buy, vector<vector<int>> &dp, int fee){

        if(index == prices.size())
            return 0;

        if(dp[index][buy] != -1)
            return dp[index][buy];

        int profit;

        if(buy){
            profit = max ( -prices[index] + find_ans(prices, index+1, false, dp, fee), 
                                find_ans(prices, index+1, true, dp, fee) );
            /*
                If buy, 
                    two cases, buy/dont buy change flag accordingly
                    
                 else if sell
                    two cases, Sell/Not sell 
            */
        }

        else{
            profit = max( prices[index] - fee + find_ans(prices, index+1, true, dp, fee) ,
                                    find_ans(prices, index+1, false, dp, fee) );
        }

        return dp[index][buy] = profit;
    }

    int maxProfit(vector<int>& prices, int fee) {
        int n = prices.size();
        vector<vector<int>> dp (n, vector<int>(2,-1));

        return find_ans(prices, 0, true, dp, fee);
    }
};

Approach 2:

Tabulation

C++

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int n = prices.size();
        vector<vector<int>> dp (n+1, vector<int>(2,0));

        //Base Case, No need here just added
        dp[n][0] = dp[n][1] = 0;

        for(int index=n-1; index>=0; index--){
            for(int buy=0; buy<=1; buy++){
                int profit;

                if(buy)
                    profit = max ( -prices[index] + dp[index+1][0], dp[index+1][1] );
                
                else
                    profit = max( prices[index] - fee + dp[index+1][1] ,dp[index+1][0] );

                dp[index][buy] = profit;
            }
        }

        return dp[0][1];
    }
};

Approach 3:

Space Optimization

C++

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int n = prices.size();
        vector<int> prev(2,0), cur(2,0);

        //Base Case, No need here just added
        cur[0] = cur[1] = 0;

        for(int index=n-1; index>=0; index--){
            for(int buy=0; buy<=1; buy++){
                int profit;

                if(buy)
                    profit = max ( -prices[index] + cur[0], cur[1] );
                
                else
                    profit = max( prices[index] - fee + cur[1] ,cur[0] );

                prev[buy] = profit;
            }

            cur = prev;
        }

        return prev[1];
    }
};

Approach 4:

C++