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Maximum Erasure Value

Medium

Amazon

10 views

Topics:

ArraysSliding Windows

You are given an array of positive integers nums and want to erase a subarray containing unique elements. The score you get by erasing the subarray is equal to the sum of its elements.

Return the maximum score you can get by erasing exactly one subarray.

An array b is called to be a subarray of a if it forms a contiguous subsequence of a, that is, if it is equal to a[l],a[l+1],...,a[r] for some (l,r).

For example:

If nums = [4,2,4,5,6], the output should be 17 because the optimal subarray here is [2,4,5,6].
If nums = [5,2,1,2,5,2,1,2,5], the output should be 8 because the optimal subarray here is [5,2,1] or [1,2,5].

Could you provide an efficient algorithm to solve this problem? What is the time and space complexity of your solution?

Maximum Erasure Value

Medium

Amazon

10 views

Topics:

ArraysSliding Windows

You are given an array of positive integers nums and want to erase a subarray containing unique elements. The score you get by erasing the subarray is equal to the sum of its elements.

Return the maximum score you can get by erasing exactly one subarray.

An array b is called to be a subarray of a if it forms a contiguous subsequence of a, that is, if it is equal to a[l],a[l+1],...,a[r] for some (l,r).

For example:

If nums = [4,2,4,5,6], the output should be 17 because the optimal subarray here is [2,4,5,6].
If nums = [5,2,1,2,5,2,1,2,5], the output should be 8 because the optimal subarray here is [5,2,1] or [1,2,5].

Could you provide an efficient algorithm to solve this problem? What is the time and space complexity of your solution?

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 value ranges for the numbers in the input array?
Can the input array be empty or null?
Are there any constraints on the size of the input array?
What should I return if all numbers in the array are negative?
Are duplicate numbers allowed in the array, and if so, how should they be handled when calculating the sum of a subarray?

Brute Force Solution

Approach

The brute force approach to this problem is all about trying every possible combination. We'll look at every possible sub-group of numbers to find the one with the largest sum where all the numbers within the group are unique.

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

Start by considering just the first number in the list.
Then, consider the first two numbers, the first three, and so on, all the way to including all the numbers.
For each of these sub-groups, check if any number appears more than once. If it does, it's not a valid group.
If all the numbers are unique, calculate the sum of the numbers in that group.
Keep track of the largest sum you've seen so far.
Next, start from the second number and repeat the process, considering the second number alone, the second and third, and so on.
Do this for every starting position in the original list, always checking for uniqueness and updating the largest sum if you find a better one.
Finally, the largest sum you've kept track of is your answer.

Code Implementation

def maximum_unique_subarray_brute_force(numbers):
    maximum_sum = 0

    for start_index in range(len(numbers)): 
        current_sum = 0
        seen_numbers = set()

        for end_index in range(start_index, len(numbers)): 
            current_number = numbers[end_index]

            # If we've seen this number before, the subarray is invalid
            if current_number in seen_numbers:
                break

            seen_numbers.add(current_number)
            current_sum += current_number

            # Update the maximum sum if the current sum is larger
            maximum_sum = max(maximum_sum, current_sum)

    return maximum_sum

Big(O) Analysis

Time Complexity

O(n³) – The outer loop iterates through each of the n elements of the input array nums. The inner loop iterates from the current element to the end of the array, creating subarrays. Inside the inner loop, checking for uniqueness of elements within each subarray takes O(n) time in the worst case (e.g., using a set or another loop to compare each element). Therefore, the overall time complexity is O(n * n * n), which simplifies to O(n³).

Space Complexity

O(N) – The described brute force approach requires a data structure to check for uniqueness within each sub-group. In the worst case, this data structure, such as a hash set or a frequency map, might need to store all N numbers from the input list to ensure that no number appears more than once in the current sub-group. Thus the auxiliary space required grows linearly with the input size N. Therefore, the space complexity is O(N).

Optimal Solution

Approach

The best way to solve this is to efficiently find the largest possible sum of a continuous group of numbers where each number appears only once. We achieve this by using a sliding window approach: expanding the window to include new unique numbers and shrinking it when a duplicate is encountered.

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

Imagine a window that can slide across the list of numbers.
Start with an empty window at the beginning of the list.
Slowly expand the window to the right, one number at a time.
As you add numbers to the window, keep track of the sum of the numbers in the window.
Also, keep track of which numbers are already inside the window.
If you encounter a number that's already in the window (a duplicate), you need to shrink the window from the left until that duplicate is removed.
Each time the window changes, calculate the sum of the numbers in the current window.
Keep track of the largest sum you've seen so far.
Continue sliding the window to the right until you reach the end of the list.
The largest sum you've kept track of is the answer.

Code Implementation

def maximum_unique_subarray(numbers):
    maximum_score = 0
    current_score = 0
    window_start = 0
    numbers_in_window = set()

    for window_end in range(len(numbers)):
        # Shrink window if duplicate found
        while numbers[window_end] in numbers_in_window:
            current_score -= numbers[window_start]
            numbers_in_window.remove(numbers[window_start])

            window_start += 1

        current_score += numbers[window_end]
        numbers_in_window.add(numbers[window_end])
        # Update maximum score found so far
        maximum_score = max(maximum_score, current_score)

    return maximum_score

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size 'n' using a sliding window approach. While the window expands to the right by one element in each outer loop iteration, the window may shrink from the left within the while loop. However, each element is added to the window and removed from the window at most once. Therefore, both the left and right pointers traverse the array a maximum of once. Thus, the time complexity is O(n).

Space Complexity

O(N) – The algorithm uses a data structure to keep track of the numbers already present inside the window. In the worst-case scenario, all N numbers in the input list are unique, requiring us to store all of them. This tracking is typically achieved using a hash set or a hash map, which would then grow linearly with the size of the input list. Therefore, the auxiliary space used is proportional to the input size N, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0, as no subarray can be formed and thus no score calculated.
Array contains only one element	Return the single element's value, as it is the only possible subarray.
Array contains all identical elements	Return the value of a single element since we must erase all but one.
Array with maximum size (e.g., 10^5) and widely varying numbers	Sliding window approach handles large arrays efficiently with O(N) time complexity.
Array containing negative numbers	The sliding window and sum calculation should function correctly with negative numbers without modification.
Array containing zero values	Zero values are handled the same as any other number in the sliding window and sum.
Integer overflow when calculating the sum of a large subarray with large numbers	Use a larger data type (e.g., long) to store the sum to prevent potential overflow.
All elements must be erased to maximize score	The sliding window expands until all unique elements are included and then continues expanding to maintain uniqueness and maximize sum.

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 value ranges for the numbers in the input array?
Can the input array be empty or null?
Are there any constraints on the size of the input array?
What should I return if all numbers in the array are negative?
Are duplicate numbers allowed in the array, and if so, how should they be handled when calculating the sum of a subarray?

Brute Force Solution

Approach

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

Start by considering just the first number in the list.
Then, consider the first two numbers, the first three, and so on, all the way to including all the numbers.
For each of these sub-groups, check if any number appears more than once. If it does, it's not a valid group.
If all the numbers are unique, calculate the sum of the numbers in that group.
Keep track of the largest sum you've seen so far.
Next, start from the second number and repeat the process, considering the second number alone, the second and third, and so on.
Do this for every starting position in the original list, always checking for uniqueness and updating the largest sum if you find a better one.
Finally, the largest sum you've kept track of is your answer.

Code Implementation

def maximum_unique_subarray_brute_force(numbers):
    maximum_sum = 0

    for start_index in range(len(numbers)): 
        current_sum = 0
        seen_numbers = set()

        for end_index in range(start_index, len(numbers)): 
            current_number = numbers[end_index]

            # If we've seen this number before, the subarray is invalid
            if current_number in seen_numbers:
                break

            seen_numbers.add(current_number)
            current_sum += current_number

            # Update the maximum sum if the current sum is larger
            maximum_sum = max(maximum_sum, current_sum)

    return maximum_sum

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine a window that can slide across the list of numbers.
Start with an empty window at the beginning of the list.
Slowly expand the window to the right, one number at a time.
As you add numbers to the window, keep track of the sum of the numbers in the window.
Also, keep track of which numbers are already inside the window.
If you encounter a number that's already in the window (a duplicate), you need to shrink the window from the left until that duplicate is removed.
Each time the window changes, calculate the sum of the numbers in the current window.
Keep track of the largest sum you've seen so far.
Continue sliding the window to the right until you reach the end of the list.
The largest sum you've kept track of is the answer.

Code Implementation

def maximum_unique_subarray(numbers):
    maximum_score = 0
    current_score = 0
    window_start = 0
    numbers_in_window = set()

    for window_end in range(len(numbers)):
        # Shrink window if duplicate found
        while numbers[window_end] in numbers_in_window:
            current_score -= numbers[window_start]
            numbers_in_window.remove(numbers[window_start])

            window_start += 1

        current_score += numbers[window_end]
        numbers_in_window.add(numbers[window_end])
        # Update maximum score found so far
        maximum_score = max(maximum_score, current_score)

    return maximum_score

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0, as no subarray can be formed and thus no score calculated.
Array contains only one element	Return the single element's value, as it is the only possible subarray.
Array contains all identical elements	Return the value of a single element since we must erase all but one.
Array with maximum size (e.g., 10^5) and widely varying numbers	Sliding window approach handles large arrays efficiently with O(N) time complexity.
Array containing negative numbers	The sliding window and sum calculation should function correctly with negative numbers without modification.
Array containing zero values	Zero values are handled the same as any other number in the sliding window and sum.
Integer overflow when calculating the sum of a large subarray with large numbers	Use a larger data type (e.g., long) to store the sum to prevent potential overflow.
All elements must be erased to maximize score	The sliding window expands until all unique elements are included and then continues expanding to maintain uniqueness and maximize sum.