Java Program to find Maximum profit by buying and selling a share atmost twice

October 14, 2022

Maximum Profit by Buying And Selling a Share Atmost twice in Java

Here, in this page we will discuss the program to find maximum profit by buying and selling a share atmost twice in Java .
In daily share trading, a buyer buys shares in the morning and sells them on the same day. If the trader is allowed to make at most 2 transactions in a day, whereas the second transaction can only start after the first one is complete Buy->Sell->Buy->Sell. We are given with stock prices throughout the day, We need to  find out the maximum profit that a share trader could have made.

Program to find Maximum profit by buying and selling a share atmost twice

Method 1 

  • Create an array say profit of size n and initialize all its value to 0.
  • Iterate price[] from right(n-1) to left(0) and update profit[i] such that profit[i] stores maximum profit achievable from one transaction in sub-array price[i..n-1].
  • Iterate price[] from left to right and update profit[i] such that profit[i] stores maximum profit such that profit[i] contains maximum achievable profit from two transactions in sub-array price[0..i].
  • After completing the iteration print value of profit[n-1].
Program to find Maximum profit by buying and selling a share atmost twice

Code to find Maximum profit by buying and selling a share atmost twice in Java

Run 
public
class Main {
    static int maxProfit(int price[], int n) {
        // Create profit array
        // and initialize it as 0
        int profit[] = new int[n];

        for (int i = 0; i < n; i++) profit[i] = 0;
        int max_price = price[n - 1];

        for (int i = n - 2; i >= 0; i--) {
            // max_price has maximum
            // of price[i..n-1]

            if (price[i] > max_price) max_price = price[i];
            profit[i] = Math.max(profit[i + 1], max_price - price[i]);
        }
        int min_price = price[0];
        for (int i = 1; i < n; i++) {

            // min_price is minimum
            // price in price[0..i]

            if (price[i] < min_price) min_price = price[i];
            profit[i] = Math.max(profit[i - 1], profit[i] + (price[i] - min_price));
        }

        int result = profit[n - 1];
        return result;
    }

    // Driver Code
    public
    static void main(String args[]) {
        int price[] = {2, 30, 15, 10, 8, 25, 80};
        int n = price.length;
        System.out.println("Maximum Profit = " + maxProfit(price, n));
    }
}
Maximum Profit = 100

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Method 2 

  • Initialize four variables for taking care of the first buy, first sell, second buy, second sell.
  • Set first buy and second buy as INT_MIN and first and second sell as 0.
  • This is to ensure to get profit from transactions.
  • Iterate through the array and return the second buy as it will store maximum profit.
Run 
public

class Main {

    static int maxtwobuysell(int arr[], int size) {

        int first_buy = Integer.MIN_VALUE;
        int first_sell = 0;

        int second_buy = Integer.MIN_VALUE;
        int second_sell = 0;

        for (int i = 0; i < size; i++) {
            first_buy = Math.max(first_buy, -arr[i]);
            first_sell = Math.max(first_sell, first_buy + arr[i]);

            second_buy = Math.max(second_buy, first_sell - arr[i]);
            second_sell = Math.max(second_sell, second_buy + arr[i]);
        }
        return second_sell;
    }

    public
    static void main(String[] args) {
        int arr[] = {2, 30, 15, 10, 8, 25, 80};
        int size = arr.length;
     System.out.print("Maximum Profit= "+maxtwobuysell(arr, size));
    }
}

Output

Maximum Profit = 100

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