C Program For Stock Buy Sell To Maximize Profit
Efficient approach: If we are allowed to buy and sell only once, then we can use following algorithm. Maximum difference between two elements. Here we are allowed to buy and sell multiple times.
Following is the algorithm for this problem.
- Find the local minima and store it as starting index. If not exists, return.
- Find the local maxima. and store it as an ending index. If we reach the end, set the end as the ending index.
- Update the solution (Increment count of buy-sell pairs)
- Repeat the above steps if the end is not reached.
C
// Program to find best buying and // selling days #include <stdio.h> // Solution structure struct Interval { int buy; int sell; }; // This function finds the buy and sell // schedule for maximum profit void stockBuySell( int price[], int n) { // Prices must be given for at // least two days if (n == 1) return ; // Count of solution pairs int count = 0; // Solution vector Interval sol[n / 2 + 1]; // Traverse through given price array int i = 0; while (i < n - 1) { // Find Local Minima. Note that the // limit is (n-2) as we are comparing // present element to the next element. while ((i < n - 1) && (price[i + 1] <= price[i])) i++; // If we reached the end, break as no // further solution possible if (i == n - 1) break ; // Store the index of minima sol[count].buy = i++; // Find Local Maxima. Note that the // limit is (n-1) as we are comparing // to previous element while ((i < n) && (price[i] >= price[i - 1])) i++; // Store the index of maxima sol[count].sell = i - 1; // Increment count of buy/sell pairs count++; } // Print solution if (count == 0) printf ( "There is no day when buying the stock will make profitn" ); else { for ( int i = 0; i < count; i++) printf ( "Buy on day: %dt Sell on day: %dn" , sol[i].buy, sol[i].sell); } return ; } // Driver code int main() { // Stock prices on consecutive days int price[] = {100, 180, 260, 310, 40, 535, 695}; int n = sizeof (price) / sizeof (price[0]); // Function call stockBuySell(price, n); return 0; } |
Output:
Buy on day: 0 Sell on day: 3 Buy on day: 4 Sell on day: 6
Time Complexity: The outer loop runs till I become n-1. The inner two loops increment value of I in every iteration. So overall time complexity is O(n)
Auxiliary Space: O(1) since using constant variables
Please refer complete article on Stock Buy Sell to Maximize Profit for more details!