Maximum Profitable Triplets With Increasing Prices I

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IBM

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ArraysDynamic Programming

You are given three 0-indexed integer arrays prices, profits, and k of the same length n, representing the price, profit, and category of i^th item respectively.

You are allowed to buy at most three items, selecting them according to the following rules:

First, choose any item.
Then, choose an item with a strictly higher price than the first item.
Finally, choose an item with a strictly higher price than the second item and the same category as the third item.

Return the maximum sum of profits from buying at most three items in the described manner. If it is impossible to buy three items in such a way, return 0.

Example 1:

Input: prices = [1,3,5,4], profits = [1,2,3,4], k = [1,2,3,3]
Output: 6
Explanation:
We can choose the items with prices 1, 3, and 4.
The sum of their profits is 1 + 2 + 3 = 6.

Example 2:

Input: prices = [1,6,4,5], profits = [1,3,2,4], k = [1,2,1,2]
Output: 0
Explanation:
It is not possible to choose three items satisfying the given conditions, so we return 0.

Constraints:

3 <= prices.length <= 50
prices.length == profits.length == k.length
1 <= prices[i], profits[i], k[i] <= 100

Try coding it in LeetCode

Maximum Profitable Triplets With Increasing Prices I

Medium

IBM

0 views

Topics:

ArraysDynamic Programming

You are given three 0-indexed integer arrays prices, profits, and k of the same length n, representing the price, profit, and category of i^th item respectively.

You are allowed to buy at most three items, selecting them according to the following rules:

First, choose any item.
Then, choose an item with a strictly higher price than the first item.
Finally, choose an item with a strictly higher price than the second item and the same category as the third item.

Return the maximum sum of profits from buying at most three items in the described manner. If it is impossible to buy three items in such a way, return 0.

Example 1:

Input: prices = [1,3,5,4], profits = [1,2,3,4], k = [1,2,3,3]
Output: 6
Explanation:
We can choose the items with prices 1, 3, and 4.
The sum of their profits is 1 + 2 + 3 = 6.

Example 2:

Input: prices = [1,6,4,5], profits = [1,3,2,4], k = [1,2,1,2]
Output: 0
Explanation:
It is not possible to choose three items satisfying the given conditions, so we return 0.

Constraints:

3 <= prices.length <= 50
prices.length == profits.length == k.length
1 <= prices[i], profits[i], k[i] <= 100

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 are the possible ranges for the `prices` and `profits` values? Can they be negative?
What should I return if no such triplet exists that satisfies the increasing price condition?
Are the `prices` and `profits` arrays guaranteed to be the same length?
If multiple triplets yield the same maximum profit, is any one of them acceptable?
Are there any constraints on the size of the `prices` and `profits` arrays?

Brute Force Solution

Approach

The brute force approach means we're going to try every single possible combination to find the best one. In this problem, we're looking at all possible groups of three prices to see which one gives us the highest profit.

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

Consider the first price in the list and treat it as a potential 'first' price in our triplet.
For each possible 'first' price, look at every price that comes after it in the list.
Treat each of these later prices as a potential 'second' price, but only if it's higher than the 'first' price.
Now, for each valid combination of 'first' and 'second' prices, search for a 'third' price that appears later in the list than the 'second' price.
Only consider prices that are higher than the 'second' price as valid 'third' prices.
For every possible combination of 'first', 'second', and 'third' prices that meet the increasing price requirement, calculate the profit by adding them all together.
Compare the profit of each combination to find the combination that produces the maximum profit.
The combination with the largest profit represents the solution.

Code Implementation

def find_maximum_profitable_triplets_brute_force(prices):
    maximum_profit = 0

    for first_index in range(len(prices)):
        first_price = prices[first_index]

        for second_index in range(first_index + 1, len(prices)):
            second_price = prices[second_index]

            # Only consider second prices that are higher than the first price
            if second_price > first_price:

                for third_index in range(second_index + 1, len(prices)):
                    third_price = prices[third_index]

                    # Only consider third prices that are higher than the second price
                    if third_price > second_price:

                        current_profit = first_price + second_price + third_price

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

    return maximum_profit

Big(O) Analysis

Time Complexity

O(n^3) – The algorithm iterates through all possible triplets in the input array of size n. The outermost loop considers each element as the potential first element of the triplet. The second loop considers elements after the first as the potential second element, and the innermost loop searches for a third element after the second. Therefore, we have three nested loops, each potentially iterating up to n times. This results in a time complexity proportional to n * n * n, which simplifies to O(n^3).

Space Complexity

O(1) – The brute force approach, as described, iterates through the input list of prices to find increasing triplets and calculates their profit. It does not create any auxiliary data structures like lists, hash maps, or trees to store intermediate results. The space used is limited to a few variables to track the indices of the triplet and the maximum profit found so far, all of which require constant space regardless of the input size N (number of prices). Therefore, the space complexity is O(1).

Optimal Solution

Approach

The goal is to find the highest profit we can get by buying and selling an item three times, each time at a higher price than the last. Instead of checking every single combination, we find the best buying and selling points by working through the prices in a smart way.

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

For each price in the list, determine the highest profit you could have made buying at an earlier price and selling at the current price.
For each price in the list, determine the highest profit you could make buying at the current price and selling at a later price.
For each price in the list, add the two profits calculated above: profit from a transaction *before* the price, and a transaction *after* the price.
Find the biggest sum calculated above, which represents the maximum profit you can make from three transactions forming an increasing price sequence.

Code Implementation

def find_maximum_profitable_triplet(prices):
    maximum_profit = 0
    number_of_prices = len(prices)

    for middle_index in range(1, number_of_prices - 1):
        middle_price = prices[middle_index]
        best_price_before = 0
        best_price_after = 0

        # Find the best price before the middle price.
        for before_index in range(middle_index):
            if prices[before_index] < middle_price:
                best_price_before = max(best_price_before, prices[before_index])

        # Find the best price after the middle price.
        for after_index in range(middle_index + 1, number_of_prices):
            if prices[after_index] > middle_price:
                best_price_after = max(best_price_after, prices[after_index])

        # If there is a smaller price before and higher price after.
        if best_price_before > 0 and best_price_after > 0:
            current_profit = best_price_before + middle_price + best_price_after
            maximum_profit = max(maximum_profit, current_profit)

    return maximum_profit

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the prices array of size n to find potential middle prices for the triplet. For each of these n prices, it calculates the maximum profit achievable from a transaction before that price and a transaction after that price. Both of these profit calculations involve iterating through portions of the prices array, effectively resulting in nested loops. Therefore, the total number of operations approximates n * (average length of array portions), where the average length is proportional to n. This simplifies to a time complexity of O(n²).

Space Complexity

O(N) – The solution uses two arrays to store the maximum profit that can be made before and after each price. Both of these arrays have a size equal to the number of prices, N. Therefore, the auxiliary space used is proportional to the input size, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
prices or profits is null or empty	Return 0 immediately, as no triplet can be formed.
prices or profits array has fewer than 3 elements	Return 0 immediately, as at least three days are required.
prices array is strictly decreasing	Return 0 as no increasing triplet exists.
profits array contains negative numbers	The algorithm should correctly handle negative profits and choose the triplet maximizing total profit, even if it includes negative values.
prices array contains duplicate values, some of which are part of the optimal triplet	The algorithm should correctly identify and consider different days with duplicate prices, choosing the one with optimal profit for the triplet.
Integer overflow in profit calculation	Use a larger data type (e.g., long) for accumulating the total profit to prevent overflow.
All prices are the same	Return 0 since there are no increasing prices to form a valid triplet.
Maximum-sized input array (large N)	Ensure the algorithm's time complexity is efficient (e.g., O(N^2) or O(N log N)) to handle large inputs without exceeding time limits.