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C Program For Stock Buy Sell To Maximize Profit

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  • Last Updated : 24 Jul, 2022
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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.


// 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)
    // 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]))
        // If we reached the end, break as no
        // further solution possible
        if (i == n - 1)
        // 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]))
        // Store the index of maxima
        sol[count].sell = i - 1;
        // Increment count of buy/sell pairs
    // Print solution
    if (count == 0)
        "There is no day when buying the stock will make profitn");
        for (int i = 0; i < count; i++)
            "Buy on day: %dt Sell on day: %dn", sol[i].buy, sol[i].sell);
// 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;


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!

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