Maximum profit after buying and selling the stocks

Given an array of positive integers containing the price of stocks and transaction fee, the task is to find the maximum profit and the difference of days on which you are getting the maximum profit.
Examples:

Input: arr[] = {6, 1, 7, 2, 8, 4}
      transactionFee = 2
Output: 8 1

Input: arr[] = {7, 1, 5, 3, 6, 4}
       transactionFee = 1
Output: 5 1


Explanation: Considering the first example: arr[] = {6, 1, 7, 2, 8, 4}, transactionFee = 2

  1. If we buy and sell on the same day, we will not get any profit that’s why the difference between the buying and selling must be at least 1.
  2. With the difference of 1 day, if we buy a stock of rupees 1 and sell it rupees 7 with the difference of day 1 which mean purchase on day 2 and sell it next day,then after paying the transaction fee of rupees 2 i.e. 7-1-2=4, we will get profit of 4 rupees, same as if we purchase on day 4 and sell it on day 5 with the difference of day 1 then we get profit of 4 rupees. So the total profit is 8 rupees.
  3. With the difference of 2 days, we will not get any profit.
  4. With the difference of 3 days, if we buy stock of rupees 1 and sell it rupees 8 with the difference of 3 days which mean purchase on day 2 and sell it after 3 days then maximum profit after paying the transaction fee of rupees 2 i.e.8-1-2=5 we will get the profit of 5 rupees.
  5. With the difference of 4 days, if we buy stocks of rupees 1 and sell it rupees 4 with the difference of 4 days which mean purchase on day 2 and sell it after 4 days then after paying the transaction fee of rupees 2 i.e. 4-1-2=1, we will get profit of 1 rupees.
  6. With the difference of 5 days, we will not get any profit.

Approach:

  1. Traverse the whole array with the difference of each day.
  2. Check the profit by subtracting the price of each day including transaction fee.
  3. Trace the maximum profit and store the diff_days on which we are getting the maximum profit.
  4. Repeat the above steps until the loop terminates.

Below is the implementation of above approach:

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
int max_profit(int a[],int b[],int n,int fee)
{
int i, j, profit;
  
int l, r, diff_day = 1, sum = 0;
  
//b[0] will contain the maximum profit
    b[0]=0;                 
//b[1] will contain the day
//on which we are getting the maximum profit
    b[1]=diff_day;
for(i=1;i<n;i++)
{
    l=0;
    r=diff_day;
        sum=0;     
  
    for(j=n-1;j>=i;j--)
        {
        //here finding the max profit
            profit=(a[r]-a[l])-fee; 
      
        //if we get less then or equal to zero
        // it means we are not getting the profit 
            if(profit>0)     
                {
                sum=sum+profit;
                }
            l++; 
      
            r++;
            }
//check if sum is greater then maximum then store the new maximum
    if(b[0] < sum) 
{
    b[0] = sum;
      
    b[1] = diff_day; 
  
    }
diff_day++;
}
  
return 0;
}
  
// Driver code
int main()
{
    int arr[] = { 6, 1, 7, 2, 8, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int b[2];
    int tranFee = 2;
  
    max_profit(arr, b, n, tranFee);
  
    cout << b[0] << ", " << b[1] << endl;
  
    return 0;
}

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Java














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// Java implementation of above approach 


import java.util.*; 


  


class solution 


{ 


  


static int max_profit(int a[],int b[],int n,int fee) 


{ 


int i, j, profit; 


  


int l, r, diff_day = 1, sum = 0; 


  


//b[0] will contain the maximum profit 


    b[0]=0;                  


//b[1] will contain the day 


//on which we are getting the maximum profit 


    b[1]=diff_day; 


for(i=1;i<n;i++) 


{ 


    l=0; 


    r=diff_day; 


        sum=0;      


  


    for(j=n-1;j>=i;j--) 


        { 


        //here finding the max profit 


            profit=(a[r]-a[l])-fee;  


      


        //if we get less then or equal to zero 


        // it means we are not getting the profit  


            if(profit>0)      


                { 


                sum=sum+profit; 


                } 


            l++;  


      


            r++; 


            } 


//check if sum is greater then maximum then store the new maximum 


    if(b[0] < sum)  


{ 


    b[0] = sum; 


      


    b[1] = diff_day;  


  


    } 


diff_day++; 


} 


  


return 0; 


} 


  


// Driver code 


public static void main(String args[]) 


{ 


    int arr[] = { 6, 1, 7, 2, 8, 4 }; 


    int n = arr.length; 


    int[] b = new int[2]; 


    int tranFee = 2; 


  


    max_profit(arr, b, n, tranFee); 


  


    System.out.println(b[0]+", "+b[1]); 


  


} 


} 


  


//This code is contributed by Surendra_Gangwar 



















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Python3














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# Python3 implementation of above approach 


def max_profit(a, b, n, fee): 


  


    i, j, profit = 1, n - 1, 0


      


    l, r, diff_day = 0, 0, 1 


      


    # b[0] will contain the maximum profit 


    b[0] = 0    


      


    # b[1] will contain the day on which  


    # we are getting the maximum profit 


    b[1] = diff_day 


  


    for i in range(1, n): 


        l = 0


        r = diff_day 


        Sum = 0


      


        for j in range(n - 1, i - 1, -1): 


              


            # here finding the max profit 


            profit = (a[r] - a[l]) - fee  


          


            # if we get less then or equal to zero 


            # it means we are not getting the profit  


            if(profit > 0): 


                Sum = Sum + profit 


                      


            l += 1


            r += 1


                  


        # check if Sum is greater then maximum  


        # then store the new maximum 


        if(b[0] < Sum): 


            b[0] = Sum


            b[1] = diff_day  


      


    diff_day += 1


      


    return 0


      


# Driver code 


arr = [6, 1, 7, 2, 8, 4]  


n = len(arr) 


b = [0 for i in range(2)] 


tranFee = 2


  


max_profit(arr, b, n, tranFee) 


  


print(b[0], ",", b[1]) 


  


# This code is contributed by 


# Mohit kumar 29 



















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C#














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// C# implementation of above approach 


using System; 


  


class GFG 


{ 


      


static int max_profit(int []a, int []b,  


                      int n, int fee)  


{  


int i, j, profit;  


  


int l, r, diff_day = 1, sum = 0;  


  


// b[0] will contain the  


// maximum profit  


b[0] = 0; 


  


// b[1] will contain the day on which 


// we are getting the maximum profit  


b[1] = diff_day;  


for(i = 1; i < n; i++)  


{  


    l = 0; r = diff_day; sum = 0;  


  


    for(j = n - 1; j >= i; j--)  


        {  


            // here finding the max profit  


            profit = (a[r] - a[l]) - fee;  


      


            // if we get less then or equal  


            // to zero it means we are not  


            // getting the profit  


            if(profit > 0)  


            {  


                sum = sum + profit;  


            }  


            l++;  


      


            r++;  


        }  


          


    // check if sum is greater then maximum 


    // then store the new maximum  


    if(b[0] < sum)  


    {  


        b[0] = sum;  


          


        b[1] = diff_day;  


      


    }  


    diff_day++;  


}  


  


return 0;  


}  


  


// Driver code  


static public void Main () 


{ 


    int []arr = { 6, 1, 7, 2, 8, 4 };  


    int n = arr.Length;  


    int[] b = new int[2];  


    int tranFee = 2;  


      


    max_profit(arr, b, n, tranFee);  


      


    Console.WriteLine(b[0] + ", " + b[1]);  


}  


}  


  


// This code is contributed by Sachin 



















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PHP














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<?php 


// PHP implementation of above approach 


  


function max_profit(&$a, &$b, $n, $fee) 


{ 


    $diff_day = 1; 


    $sum = 0; 


      


    // b[0] will contain the maximum profit 


    $b[0] = 0; 


      


    // b[1] will contain the day on which we  


    // are getting the maximum profit 


    $b[1] = $diff_day; 


      


    for($i = 1; $i < $n; $i++) 


    { 


        $l = 0; 


        $r = $diff_day; 


        $sum = 0;  


      


        for($j = $n - 1; $j >= $i; $j--) 


        { 


            // here finding the max profit 


            $profit = ($a[$r] - $a[$l]) - $fee;  


      


            // if we get less then or equal to zero 


            // it means we are not getting the profit  


            if($profit > 0)      


            { 


                $sum = $sum + $profit; 


            } 


            $l++;  


      


            $r++; 


        } 


          


        // check if sum is greater then maximum 


        // then store the new maximum 


        if($b[0] < $sum)  


        { 


            $b[0] = $sum; 


              


            $b[1] = $diff_day;  


        } 


        $diff_day++; 


    } 


  


} 


  


// Driver code 


$arr = array(6, 1, 7, 2, 8, 4 ); 


$n = sizeof($arr); 


$b = array(); 


$tranFee = 2; 


  


max_profit($arr, $b, $n, $tranFee); 


echo($b[0]); 


echo(", "); 


echo($b[1]); 


  


// This code is contributed 


// by Shivi_Aggarwal 


?> 



















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Output:


8, 1





Time complexity: O(N2)


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