Stock Buy Sell to Maximize Profit

3.1

The cost of a stock on each day is given in an array, find the max profit that you can make by buying and selling in those days. For example, if the given array is {100, 180, 260, 310, 40, 535, 695}, the maximum profit can earned by buying on day 0, selling on day 3. Again buy on day 4 and sell on day 6. If the given array of prices is sorted in decreasing order, then profit cannot be earned at all.

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 algorithm for this problem.
1. Find the local minima and store it as starting index. If not exists, return.
2. Find the local maxima. and store it as ending index. If we reach the end, set the end as ending index.
3. Update the solution (Increment count of buy sell pairs)
4. Repeat the above steps if 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 sell schedule for maximum profit
void stockBuySell(int price[], int n)
{
    // Prices must be given for at least two days
    if (n == 1)
        return;

    int count = 0; // count of solution pairs

    // 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 program to test above functions
int main()
{
    // stock prices on consecutive days
    int price[] = {100, 180, 260, 310, 40, 535, 695};
    int n = sizeof(price)/sizeof(price[0]);

    // fucntion call
    stockBuySell(price, n);

    return 0;
}

Java

// Program to find best buying and selling days
import java.util.ArrayList;

// Solution structure
class Interval 
{
    int buy, sell;
}

class StockBuySell 
{
    // This function finds the buy sell schedule for maximum profit
    void stockBuySell(int price[], int n) 
    {
        // Prices must be given for at least two days
        if (n == 1)
            return;
        
        int count = 0;

        // solution array
        ArrayList<Interval> sol = new ArrayList<Interval>();

        // 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;

            Interval e = new Interval();
            e.buy = i++;
            // Store the index of minima
            

            // 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
            e.sell = i-1;
            sol.add(e);
            
            // Increment number of buy/sell
            count++;
        }

        // print solution
        if (count == 0)
            System.out.println("There is no day when buying the stock "
                                                  + "will make profit");
        else 
            for (int j = 0; j < count; j++)
                System.out.println("Buy on day: " + sol.get(j).buy
                        +"        " + "Sell on day : " + sol.get(j).sell);
        
        return;
    }

    public static void main(String args[]) 
    {
        StockBuySell stock = new StockBuySell();
        
        // stock prices on consecutive days
        int price[] = {100, 180, 260, 310, 40, 535, 695};
        int n = price.length;

        // fucntion call
        stock.stockBuySell(price, n);
    }
}

// This code has been contributed by Mayank Jaiswal


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 becomes n-1. The inner two loops increment value of i in every iteration. So overall time complexity is O(n)

This article is compiled by Ashish Anand and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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