# Stock Buy Sell to Maximize Profit

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.

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 sell;
};

// This function finds the buy sell schedule for maximum profit
{
// 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

// 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++)
}

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

return 0;
}
```

## Java

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

// Solution structure
class Interval
{
}

{
// This function finds the buy sell schedule for maximum profit
{
// 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();
// 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;

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++)
+"        " + "Sell on day : " + sol.get(j).sell);

return;
}

public static void main(String args[])
{

// stock prices on consecutive days
int price[] = {100, 180, 260, 310, 40, 535, 695};
int n = price.length;

// fucntion call
}
}

// 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```

### Asked in: Accolite,Amazon,Directi,Flipkart,Goldman-Sachs,Intuit ,MakeMyTrip,Microsoft,Ola-Cabs,Oracle,Paytm,Pubmatic,Quikr,Sapinet,Swiggy,Walmart labs

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)

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