# C Program For Stock Buy Sell To Maximize Profit

• Last Updated : 24 Jul, 2022

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

 `// Program to find best buying and ` `// selling days` `#include `   `// 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!

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