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.
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:
<script> // Javascript program to implement // the above approach // Function to return the maximum profit // that can be made after buying and // selling the given stocks function maxProfit(price, start, end)
{ // If the stocks can't be bought
if (end <= start)
return 0;
// Initialise the profit
let profit = 0;
// The day at which the stock
// must be bought
for (let i = start; i < end; i++)
{
// The day at which the
// stock must be sold
for (let 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
let curr_profit = price[j] - price[i] +
maxProfit(price,
start, i - 1) +
maxProfit(price,
j + 1, end);
// Update the maximum profit
// so far
profit = Math.max(profit,
curr_profit);
}
}
}
return profit;
} // Driver code let price = [100, 180, 260, 310, 40, 535, 695];
let n = price.length; document.write(maxProfit( price, 0, n - 1));
</script> |
Output:
865
Time Complexity: O(N2)
Auxiliary Space: O(1)
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.
- Find the local minima and store it as starting index. If not exists, return.
- Find the local maxima. and store it as an ending index. If we reach the end, set the end as the ending index.
- Update the solution (Increment count of buy-sell pairs)
- Repeat the above steps if the end is not reached.
<script> // JavaScript program to implement // the above approach // This function finds the buy sell // schedule for maximum profit function stockBuySell(price, n)
{ // Prices must be given for at
// least two days
if (n == 1)
return ;
// Traverse through given price array
let 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
let 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
let sell = i - 1;
document.write(`Buy on day: ${buy}
Sell on day: ${sell}<br>`);
}
} // Driver code // Stock prices on consecutive days let price = [100, 180, 260, 310, 40, 535, 695];
let n = price.length; // Function call stockBuySell(price, n); // This code is contributed by Potta Lokesh </script> |
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)
Space Complexity: O(1) since using constant variables
Please refer complete article on Stock Buy Sell to Maximize Profit for more details!