Maximum profit by buying and selling a share at most twice

In a trading, one buyer buys and sells the shares, at morning and the evening respectively. If at most two transactions are allowed in a day. The second transaction can only start after the first one is completed. If stock prices are given, then find the maximum profit that the buyer can make.

Input and Output

Input:
A list of stock prices. {2, 30, 15, 10, 8, 25, 80}
Output:
Here the total profit is 100. As buying at price 2 and selling at price 30.
so profit 28. Then buy at price 8 and sell it again at price 80.
So profit 72. So the total profit 28 + 72 = 100

Algorithm

findMaxProfit(pricelist, n)

Input − List of all prices, number of items in the list.

Output − Maximum Profit.

Begin
   define profit array of size n and fill with 0
   maxPrice := pricelist[n-1]          //last item is chosen

   for i := n-2 down to 0, do
      if pricelist[i] > maxPrice, then
         maxPrice := pricelist[i]
      profit[i] := maximum of profit[i+1] and maxProfit – pricelist[i]
   done

   minProce := pricelist[0]           //first item is chosen
   for i := 1 to n-1, do
      if pricelist[i] < minPrice, then
         minPrice := pricelist[i]
      profit[i] := maximum of profit[i-1] and (profit[i]+(pricelist[i] - minPrice))
   done

   return profit[n-1]
End

Example

#include<iostream>
using namespace std;

int max(int a, int b) {
   return (a>b)?a:b;
}

int findMaxProfit(int priceList[], int n) {
   int *profit = new int[n];
   for (int i=0; i<n; i++)            //initialize profit list with 0
      profit[i] = 0;

   int maxPrice = priceList[n-1];        //initialize with last element of price list

   for (int i=n-2;i>=0;i--) {
      if (priceList[i] > maxPrice)
         maxPrice = priceList[i];

      profit[i] = max(profit[i+1], maxPrice - priceList[i]);     //find the profit for selling in maxPrice
   }

   int minPrice = priceList[0];            //first item of priceList as minimum

   for (int i=1; i<n; i++) {
      if (priceList[i] < minPrice)
         minPrice = priceList[i];

      profit[i] = max(profit[i-1], profit[i] + (priceList[i]- minPrice) );
   }

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

int main() {
   int priceList[] = {2, 30, 15, 10, 8, 25, 80};
   int n = 7;
   cout << "Maximum Profit = " << findMaxProfit(priceList, n);
}

Output

Maximum Profit = 100

Chandu yadav

Published on 11-Jul-2018 08:38:33

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