To start, try writing an example value for stock_prices and finding the maximum profit "by hand." What's your process for figuring out the maximum profit?

The brute force ↴

  def get_max_profit(stock_prices):
    max_profit = 0

    # Go through every time
    for outer_time in range(len(stock_prices)):

        # For every time, go through every other time
        for inner_time in range(len(stock_prices)):
            # For each pair, find the earlier and later times
            earlier_time = min(outer_time, inner_time)
            later_time   = max(outer_time, inner_time)

            # And use those to find the earlier and later prices
            earlier_price = stock_prices[earlier_time]
            later_price   = stock_prices[later_time]

            # See what our profit would be if we bought at the
            # earlier price and sold at the later price
            potential_profit = later_price - earlier_price

            # Update max_profit if we can do better
            max_profit = max(max_profit, potential_profit)

    return max_profit

But that will take $O(n^2)$ time, ↴

Well, we’re doing a lot of extra work. We’re looking at every pair twice. We know we have to buy before we sell, so in our inner for loop we could just look at every price after the price in our outer for loop.

That could look like this:

  def get_max_profit(stock_prices):
    max_profit = 0

    # Go through every price (with its index as the time)
    for earlier_time, earlier_price in enumerate(stock_prices):

        # And go through all the LATER prices
        for later_time in range(earlier_time + 1, len(stock_prices)):
            later_price = stock_prices[later_time]

            # See what our profit would be if we bought at the
            # earlier price and sold at the later price
            potential_profit = later_price - earlier_price

            # Update max_profit if we can do better
            max_profit = max(max_profit, potential_profit)

    return max_profit

What’s our runtime now?

Well, our outer for loop goes through all the times and prices, but our inner for loop goes through one fewer price each time. So our total number of steps is the sum $n + (n - 1) + (n - 2) ... + 2 + 1$ ↴

The sum of integers $1..n$ is $\approx \frac{n^2}{2}$, which is $O(n^2)$

Series like this actually come up quite a bit:

$1 + 2 + 3 + ... + (n-1) + n$

Or, equivalently, the other way around:

$n + (n-1) + ... + 3 + 2 + 1$

And sometimes the last $n$ is omitted, but as we'll see it doesn't affect the big O:

$(n-1) + (n-2) + ... + 3 + 2 + 1$

Let's draw this out. Let's say $n=10$, so we'll represent $n-1$ as nine circles:

We can continue the pattern with $n-2$

And $n-3$, $n-4$, etc, all the way down to 1:

Notice both the top and right "sides" of our set of circles have $n-1$ items:

In fact, we could imagine our circles inside of a square with sides of length n-1:

Notice that we've filled in just about half of the square!

Of course, the area of the square is $(n-1) * (n-1)$, which is $O(n^2)$. Our total number of circles is about half of that, so $O(n^2/2)$, which is still $O(n^2)$. Remember: with big O notation, we throw out the constants.

If we had started from $n$ instead of $n-1$ we'd have $O(n^2 + n)$, which again is still $O(n^2)$ since in big O notation we drop the less significant terms.

We can do better!

If we're going to do better than $O(n^2)$, we're probably going to do it in either $O(n\lg{n})$ or $O(n)$. $O(n\lg{n})$ comes up in sorting and searching algorithms where we're recursively cutting the list in half. It's not obvious that we can save time by cutting the list in half here. Let's first see how well we can do by looping through the list only once.

Since we're trying to loop through the list once, let's use a greedy ↴

At each iteration, our max_profit is either:

1. the same as the max_profit at the last time step, or
2. the max profit we can get by selling at the current_price

How do we know when we have case (2)?

The max profit we can get by selling at the current_price is simply the difference between the current_price and the min_price from earlier in the day. If this difference is greater than the current max_profit, we have a new max_profit.

So for every price, we’ll need to:

• keep track of the lowest price we’ve seen so far
• see if we can get a better profit

Here’s one possible solution:

  def get_max_profit(stock_prices):
    min_price  = stock_prices[0]
    max_profit = 0

    for current_price in stock_prices:
        # Ensure min_price is the lowest price we've seen so far
        min_price = min(min_price, current_price)

        # See what our profit would be if we bought at the
        # min price and sold at the current price
        potential_profit = current_price - min_price

        # Update max_profit if we can do better
        max_profit = max(max_profit, potential_profit)

    return max_profit

We’re finding the max profit with one pass and constant space!

Are we done? Let’s think about some edge cases. What if the price stays the same? What if the price goes down all day?

If the price doesn't change, the max possible profit is 0. Our function will correctly return that. So we're good.

But if the value goes down all day, we’re in trouble. Our function would return 0, but there’s no way we could break even if the price always goes down.

How can we handle this?

Well, what are our options? Leaving our function as it is and just returning zero is not a reasonable option—we wouldn't know if our best profit was negative or actually zero, so we'd be losing information. Two reasonable options could be:

1. return a negative profit. “What’s the least badly we could have done?”
2. raise an exception. “We should not have purchased stocks yesterday!”

In this case, it’s probably best to go with option (1). The advantages of returning a negative profit are:

• We more accurately answer the challenge. If profit is "revenue minus expenses", we’re returning the best we could have done.
• It's less opinionated. We'll leave decisions up to our function’s users. It would be easy to wrap our function in a helper function to decide if it’s worth making a purchase.
• We allow ourselves to collect better data. It matters if we would have lost money, and it matters how much we would have lost. If we’re trying to get rich, we’ll probably care about those numbers.

How can we adjust our function to return a negative profit if we can only lose money? Initializing max_profit to 0 won’t work...

Well, we started our min_price at the first price, so let’s start our max_profit at the first profit we could get—if we buy at the first time and sell at the second time.

  min_price  = stock_prices[0]
max_profit = stock_prices[1] - stock_prices[0]

But we have the potential for an index out of bounds error here, if stock_prices has fewer than 2 prices.

We do want to raise an exception in that case, since profit requires buying and selling, which we can't do with less than 2 prices. So, let's explicitly check for this case and handle it:

  if len(stock_prices) < 2:
    raise ValueError('Getting a profit requires at least 2 prices')

min_price  = stock_prices[0]
max_profit = stock_prices[1] - stock_prices[0]

Ok, does that work?

No! max_profit is still always 0. What’s happening?

If the price always goes down, min_price is always set to the current_price. So current_price - min_price comes out to 0, which of course will always be greater than a negative profit.

When we’re calculating the max_profit, we need to make sure we never have a case where we try both buying and selling stocks at the current_price.

To make sure we’re always buying at an earlier price, never the current_price, let’s switch the order around so we calculate max_profit before we update min_price.

We'll also need to pay special attention to time 0. Make sure we don't try to buy and sell at time 0.