Stock Buy Sell to Maximize Profit

• Difficulty Level : Medium
• Last Updated : 23 Aug, 2021

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.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution. 
 

Naive approach: A simple approach is to try buying the stocks and selling them on every single day when profitable and keep updating the maximum profit so far.

Below is the implementation of the above approach:

• C++
• Java
• Python3
• C#
• Javascript

// C++ implementation of the approach

#include <bits/stdc++.h>

using namespace std;

// Function to return the maximum profit

// that can be made after buying and

// selling the given stocks

int maxProfit(int price[], int start, int end)

{

    // If the stocks can't be bought

    if (end <= start)

        return 0;

    // Initialise the profit

    int profit = 0;

    // The day at which the stock

    // must be bought

    for (int i = start; i < end; i++) {

        // The day at which the

        // stock must be sold

        for (int j = i + 1; j <= end; j++) {

            // If buying the stock at ith day and

            // selling it at jth day is profitable

            if (price[j] > price[i]) {

                // Update the current profit

                int curr_profit = price[j] - price[i]

                                  + maxProfit(price, start, i - 1)

                                  + maxProfit(price, j + 1, end);

                // Update the maximum profit so far

                profit = max(profit, curr_profit);

            }

        }

    }

    return profit;

}

// Driver code

int main()

{

    int price[] = { 100, 180, 260, 310,

                    40, 535, 695 };

    int n = sizeof(price) / sizeof(price[0]);

    cout << maxProfit(price, 0, n - 1);

    return 0;

}

Output

865

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.  

1. Find the local minima and store it as starting index. If not exists, return.
2. Find the local maxima. and store it as an ending index. If we reach the end, set the end as the ending index.
3. Update the solution (Increment count of buy-sell pairs)
4. Repeat the above steps if the end is not reached.

• C++
• C
• Java
• Python3
• C#
• Javascript

// C++ Program to find best buying and selling days

#include <bits/stdc++.h>

using namespace std;

// 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;

    // 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

        int 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

        int sell = i - 1;

        cout << "Buy on day: " << buy

             << "\t Sell on day: " << sell << endl;

    }

}

// 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;

}

// This is code is contributed by rathbhupendra

Output

Buy on day: 0     Sell on day: 3Buy 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)

Valley Peak Approach:

In this approach, we just need to find the next greater element and subtract it from the current element so that the difference keeps increasing until we reach a minimum. If the sequence is a decreasing sequence so the maximum profit possible is 0.

• C++
• Java

#include <iostream>

using namespace std;

// Preprocessing helps the code run faster

#define fl(i, a, b) for (int i = a; i < b; i++)

// Function that return

int maxProfit(int* prices, int size)

{

    // maxProfit adds up the difference between

    // adjacent elements if they are in increaisng order

    int maxProfit = 0;

    // The loop starts from 1

    // as its comparing with the previous

    fl(i, 1, size) if (prices[i] > prices[i - 1]) maxProfit

        += prices[i] - prices[i - 1];

    return maxProfit;

}

// Driver Function

int main()

{

    int prices[] = { 100, 180, 260, 310, 40, 535, 695 };

    int N = sizeof(prices) / sizeof(prices[0]);

    cout << maxProfit(prices, N) << endl;

    return 0;

}

// This code is contributed by Kingshuk Deb

Output

865

Time Complexity: O(n)
Auxiliary Space: O(1)

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