Maximum Linear Stock Score

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ArraysDynamic Programming

You are given an array prices of length n representing the daily prices of a particular stock over n days. The linear stock score on day i is calculated as follows:

If the price on day i is greater than the price on day i-1, add i * price[i] to the score.
If the price on day i is less than or equal to the price on day i-1, subtract i * price[i] from the score.

The linear stock score for the first day (day 0) is always prices[0].

Your task is to find the maximum linear stock score you can achieve by reordering the elements of the prices array.

Example 1:

Input: prices = [5,2,8,1]
Output: 45
Explanation:
One possible reordering is [1,2,5,8].
- Day 0: 1
- Day 1: 2 * 2 = 4
- Day 2: 3 * 5 = 15
- Day 3: 4 * 8 = 32
Total score = 1 + 4 + 15 + 32 = 52

Another possible reordering is [8,5,2,1].
- Day 0: 8
- Day 1: -2 * 5 = -10
- Day 2: -3 * 2 = -6
- Day 3: -4 * 1 = -4
Total score = 8 - 10 - 6 - 4 = -12

After trying all possible reorderings, we can determine that the arrangement [1, 5, 2, 8] gives the maximum score, which is 45.

Example 2:

Input: prices = [10,1,2,3,4]
Output: 72
Explanation:
One possible reordering is [1,2,3,4,10].
- Day 0: 1
- Day 1: 2 * 2 = 4
- Day 2: 3 * 3 = 9
- Day 3: 4 * 4 = 16
- Day 4: 5 * 10 = 50
Total score = 1 + 4 + 9 + 16 + 50 = 80

However, the arrangement [1, 3, 2, 4, 10] yields a higher score of 72.

Constraints:

1 <= n <= 10⁵
1 <= prices[i] <= 10⁶

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible stock prices? Can prices be negative, zero, or floating-point numbers?
If no buy/sell combination results in a positive score, what should I return? Should I return 0, negative infinity, or some other sentinel value?
Are there any constraints on the size of the input array (number of days)? Is it possible the input array is empty or null?
Are we looking for the maximum *linear* score only, or are other non-linear score calculations possible/relevant? (I want to confirm I understand 'linear' correctly in this context.)
If multiple buy/sell combinations yield the same maximum score, is any one of them acceptable, or is there a specific tie-breaking rule (e.g., prefer the earliest buy date)?

Brute Force Solution

Approach

The brute force approach is like trying every single possible way to buy and sell a stock within a given period to see which one makes the most money. We'll check every combination of buying and selling days.

Here's how the algorithm would work step-by-step:

First, pick a day to buy the stock. This could be any day.
Then, for that buying day, try every day after it as a potential selling day.
Calculate the profit for buying on the first day and selling on each subsequent day.
Record the profit for each of those buy-sell combinations.
Now, repeat the process, but this time, pick the second day as the buying day, and again try every day after it as a potential selling day.
Continue this process, considering each day as a potential buying day and trying all the days after it as potential selling days.
Finally, compare all the profits you recorded and choose the highest one. That represents the maximum profit you could have made.

Code Implementation

def max_linear_stock_score_brute_force(stock_prices):
    maximum_profit = 0

    # Iterate through all possible buying days
    for buying_day in range(len(stock_prices)):

        # Iterate through all possible selling days after the buying day
        for selling_day in range(buying_day + 1, len(stock_prices)):

            # Calculate the profit for this buy-sell combination
            profit = stock_prices[selling_day] - stock_prices[buying_day]

            # Update maximum profit if the current profit is higher
            if profit > maximum_profit:
                maximum_profit = profit

    return maximum_profit

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each day as a potential buying day. For each buying day, it iterates through all subsequent days as potential selling days to calculate the profit. This creates a nested loop structure where the outer loop runs n times (where n is the number of days/stock prices). The inner loop, on average, runs n/2 times. Therefore, the time complexity is proportional to n * (n/2), which simplifies to O(n²).

Space Complexity

O(1) – The brute force approach described in the plain English explanation calculates profits on the fly and only stores the maximum profit found so far. It doesn't explicitly create or store lists or other data structures to hold intermediate results or visited states. Only a constant number of variables (like a variable to store the current max profit) are used regardless of the number of prices (N). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The goal is to find a starting and ending point in the list of stock values that gives you the biggest difference, representing the best possible profit. Instead of checking every single possible start and end combination, we will use a running calculation to identify the biggest profit more efficiently.

Here's how the algorithm would work step-by-step:

Start by assuming the first stock value is the best possible buying point.
Go through each remaining stock value one by one.
For each value, calculate the potential profit if you were to sell at that value, using your current best buying point.
If this potential profit is better than any profit you've seen so far, remember this new maximum profit.
Also, if the current stock value is lower than your current best buying point, update your best buying point to this new lower value.
Continue this process until you've examined all stock values.
The maximum profit you've kept track of during this process is your answer.

Code Implementation

def max_linear_stock_score(stock_prices):
    max_profit = 0
    best_buying_price = stock_prices[0]

    for current_price in stock_prices:
        potential_profit = current_price - best_buying_price

        # Update max_profit if we found a bigger profit
        if potential_profit > max_profit:
            max_profit = potential_profit

        # Track the lowest buying price so far
        if current_price < best_buying_price:
            best_buying_price = current_price

    return max_profit

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the list of stock values once. For each stock value, it performs a constant number of operations: calculating the potential profit, comparing it with the current maximum profit, and potentially updating the best buying point. The number of operations scales linearly with the number of stock values (n), thus resulting in a time complexity of O(n).

Space Complexity

O(1) – The algorithm uses a fixed number of variables: one to store the best buying point and another to store the maximum profit seen so far. Regardless of the number of stock values (N), these variables occupy constant space. No auxiliary data structures like arrays, hash maps, or recursion stacks are used. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty price array	Return 0 or throw an exception, depending on the requirements, as no profit can be made without prices.
Price array with only one element	Return 0, as we need at least two prices to buy and sell.
Price array with prices in descending order	Return 0, as no profitable transaction is possible.
Price array with all identical prices	Return 0, as no profit can be made.
Price array with extremely large prices (potential integer overflow)	Use a data type that can accommodate large numbers, such as long, to avoid integer overflow when calculating profit.
Maximum sized input array	Ensure the solution's time complexity is efficient enough to handle large inputs, potentially O(n) using dynamic programming.
Price array contains zero values	The algorithm should handle zero prices correctly, without causing division by zero or incorrect calculations.
Array contains negative prices	If negative prices are allowed, the algorithm should correctly identify the maximum profit even with negative prices; otherwise, reject array.

Empty price array

How to Handle:

Return 0 or throw an exception, depending on the requirements, as no profit can be made without prices.

Price array with only one element

How to Handle:

Return 0, as we need at least two prices to buy and sell.

Price array with prices in descending order

How to Handle:

Return 0, as no profitable transaction is possible.

Price array with all identical prices

How to Handle:

Return 0, as no profit can be made.

Price array with extremely large prices (potential integer overflow)

How to Handle:

Use a data type that can accommodate large numbers, such as long, to avoid integer overflow when calculating profit.

Maximum sized input array

How to Handle:

Ensure the solution's time complexity is efficient enough to handle large inputs, potentially O(n) using dynamic programming.

Price array contains zero values

How to Handle:

The algorithm should handle zero prices correctly, without causing division by zero or incorrect calculations.

Array contains negative prices

How to Handle:

If negative prices are allowed, the algorithm should correctly identify the maximum profit even with negative prices; otherwise, reject array.