Maximum Value of an Ordered Triplet II

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You are given a 0-indexed integer array nums.

Return the maximum value over all triplets of indices (i, j, k) such that i < j < k. If all such triplets have a negative value, return 0.

The value of a triplet of indices (i, j, k) is equal to (nums[i] - nums[j]) * nums[k].

Example 1:

Input: nums = [12,6,1,2,7]
Output: 77
Explanation: The value of the triplet (0, 2, 4) is (nums[0] - nums[2]) * nums[4] = 77.
It can be shown that there are no ordered triplets of indices with a value greater than 77.

Example 2:

Input: nums = [1,10,3,4,19]
Output: 133
Explanation: The value of the triplet (1, 2, 4) is (nums[1] - nums[2]) * nums[4] = 133.
It can be shown that there are no ordered triplets of indices with a value greater than 133.

Example 3:

Input: nums = [1,2,3]
Output: 0
Explanation: The only ordered triplet of indices (0, 1, 2) has a negative value of (nums[0] - nums[1]) * nums[2] = -3. Hence, the answer would be 0.

Constraints:

3 <= nums.length <= 10⁵
1 <= nums[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 values for the numbers in the `nums` array? Could there be negative numbers, zeros, or large positive numbers that might cause integer overflow?
What should I return if no such ordered triplet exists that satisfies the condition (i.e., the maximum value is negative or zero)? Should I return 0, null, or throw an exception?
Are there any restrictions on the size of the `nums` array? What is the maximum number of elements I should expect?
Are duplicate values allowed in the `nums` array, and if so, how should they be handled when considering triplets?
Is the input array guaranteed to be non-empty?

Brute Force Solution

Approach

The brute force strategy is like trying every single combination possible. We'll look at all possible ordered triplets and calculate their value to find the largest one.

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

Start by picking the first number from the list. Let's call this number 'a'.
Then, pick a second number from the list that comes after 'a'. Let's call this number 'b'.
Next, pick a third number from the list that comes after 'b'. Let's call this number 'c'.
Now, calculate the value of 'a', 'b', and 'c' using the formula: (a - b) * c.
Remember this value. If we have not seen a value before, or this value is larger than any value we have previously seen, store it as the maximum value.
Repeat this process, choosing all possible combinations of 'a', 'b', and 'c' from the list, always making sure that 'b' comes after 'a' and 'c' comes after 'b'.
After checking every single combination of 'a', 'b', and 'c', the largest value that we remember is the answer.

Code Implementation

def maximum_value_of_an_ordered_triplet_ii_brute_force(numbers):
    maximum_value = 0
    list_length = len(numbers)

    for first_index in range(list_length):

        for second_index in range(first_index + 1, list_length):
            # Ensure the second index always follows the first

            for third_index in range(second_index + 1, list_length):
                # Ensure the third index always follows the second

                first_number = numbers[first_index]
                second_number = numbers[second_index]
                third_number = numbers[third_index]

                current_value = (first_number - second_number) * third_number

                if current_value > maximum_value:
                    # Keep track of the largest value seen so far.

                    maximum_value = current_value

    return maximum_value

Big(O) Analysis

Time Complexity

O(n^3) – The provided brute force strategy iterates through all possible ordered triplets (a, b, c) in the input array of size n. The outermost loop iterates through each element as 'a'. The second loop iterates through elements after 'a' as 'b'. The innermost loop iterates through elements after 'b' as 'c'. Thus, there are three nested loops, each potentially iterating up to n times in the worst case. This results in approximately n * n * n operations, which simplifies to O(n^3).

Space Complexity

O(1) – The provided brute force strategy only uses a single variable to store the maximum value found so far. No auxiliary data structures like lists or maps are used. The space required remains constant regardless of the input array's size N, as only a single maximum value is tracked. Therefore, the space complexity is O(1).

Optimal Solution

Approach

The goal is to find the biggest possible value by combining three numbers from a list in a specific order. Instead of checking every single combination, we will cleverly track the biggest numbers to the left and the biggest differences we've seen so far to quickly find the best result.

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

For each number in the list, we want to know the biggest number that came before it.
We also want to know the biggest difference we can get by subtracting another number from a number that came before it in the list.
As we go through the list, we keep track of the biggest number we've seen so far, updating it whenever we find an even bigger one.
Also, as we go through the list, we keep track of the biggest difference we've seen so far by subtracting the current number from the biggest number we've seen before it.
Then, for each number in the list, we calculate the value of multiplying the biggest difference we've seen before it by the current number.
The biggest of all these values is our answer.

Code Implementation

def maximum_value_of_ordered_triplet(numbers):
    maximum_so_far = 0
    maximum_differences = 0
    result = 0

    for number in numbers:
        result = max(result, maximum_differences * number)

        # Update the maximum difference seen so far
        maximum_differences = max(maximum_differences, maximum_so_far - number)

        # Keep track of the maximum value seen so far
        maximum_so_far = max(maximum_so_far, number)

    return result

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n once to calculate the maximum value seen so far. It then iterates through the array a second time to calculate the maximum difference between previously seen maximums and current values. Finally, it iterates a third time to compute and track the maximum triplet value. Because the algorithm performs a fixed number (three) of iterations through the array, the time complexity is O(n).

Space Complexity

O(N) – The algorithm uses two auxiliary data structures: one list to store the maximum value seen so far to the left of each element (leftMax), and another list to store the maximum difference seen so far to the left of each element (maxDiff). Both of these lists have a size equal to the number of elements in the input list, N. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 immediately as no triplet can be formed.
Array with fewer than 3 elements	Return 0 immediately as no triplet can be formed.
Array with all identical values	The algorithm should correctly handle this case, potentially returning 0 if all values are non-positive or calculating the correct triplet value if positive.
Array containing large positive integers that could lead to integer overflow	Use a data type like `long` to store intermediate and final results to prevent overflow.
Array containing only negative integers or zeros	The algorithm should correctly handle this, returning 0 since the product will never be positive.
Array with maximum size allowed by the constraints	Ensure the solution's time complexity allows it to complete within the time limit for the maximum allowed array size, considering O(n) or O(n log n) solutions.
Array with a very skewed distribution of values (e.g., one extremely large value and many small values)	The algorithm should efficiently find the largest values even in skewed datasets by maintaining prefix max and suffix max values.
Integer overflow when calculating the triplet value	Utilize `long` data type to hold the product of (nums[i] - nums[j]) * nums[k] to mitigate potential overflow issues.

Empty or null input array

How to Handle:

Return 0 immediately as no triplet can be formed.

Array with fewer than 3 elements

How to Handle:

Return 0 immediately as no triplet can be formed.

Array with all identical values

How to Handle:

The algorithm should correctly handle this case, potentially returning 0 if all values are non-positive or calculating the correct triplet value if positive.

Array containing large positive integers that could lead to integer overflow

How to Handle:

Use a data type like `long` to store intermediate and final results to prevent overflow.

Array containing only negative integers or zeros

How to Handle:

The algorithm should correctly handle this, returning 0 since the product will never be positive.

Array with maximum size allowed by the constraints

How to Handle:

Ensure the solution's time complexity allows it to complete within the time limit for the maximum allowed array size, considering O(n) or O(n log n) solutions.

Array with a very skewed distribution of values (e.g., one extremely large value and many small values)

How to Handle:

The algorithm should efficiently find the largest values even in skewed datasets by maintaining prefix max and suffix max values.

Integer overflow when calculating the triplet value

How to Handle:

Utilize `long` data type to hold the product of (nums[i] - nums[j]) * nums[k] to mitigate potential overflow issues.