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By haribhatt34, history, 12 months ago, In English,

Problem linkhttps://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.

 
 
 
 
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