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: 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
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
#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); }
Maximum Profit = 100