Maximize Subarray Sum After Removing All Occurrences of One Element

Hard

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ArraysDynamic Programming

You are given an integer array nums.

You can do the following operation on the array at most once:

Choose any integer x such that nums remains non-empty on removing all occurrences of x.
Remove all occurrences of x from the array.

Return the maximum subarray sum across all possible resulting arrays.

For example:

nums = [-3,2,-2,-1,3,-2,3]

Here are a few possibilities.

The original array: nums = [-3, 2, -2, -1, 3, -2, 3]. The maximum subarray sum is 3 + (-2) + 3 = 4.
Deleting all occurrences of x = -3 results in nums = [2, -2, -1, 3, -2, 3]. The maximum subarray sum is 3 + (-2) + 3 = 4.
Deleting all occurrences of x = -2 results in nums = [-3, 2, -1, 3, 3]. The maximum subarray sum is 2 + (-1) + 3 + 3 = 7.
Deleting all occurrences of x = 3 results in nums = [-3, 2, -2, -1, -2]. The maximum subarray sum is 2.

The output is max(4, 4, 7, 4, 2) = 7.

Another example: nums = [1,2,3,4]. In this case the optimal solution is to not remove any numbers, thus the answer is 10.

How would you write code to solve this?

Maximize Subarray Sum After Removing All Occurrences of One Element

Hard

Google

6 views

Topics:

ArraysDynamic Programming

You are given an integer array nums.

You can do the following operation on the array at most once:

Choose any integer x such that nums remains non-empty on removing all occurrences of x.
Remove all occurrences of x from the array.

Return the maximum subarray sum across all possible resulting arrays.

For example:

nums = [-3,2,-2,-1,3,-2,3]

Here are a few possibilities.

The original array: nums = [-3, 2, -2, -1, 3, -2, 3]. The maximum subarray sum is 3 + (-2) + 3 = 4.
Deleting all occurrences of x = -3 results in nums = [2, -2, -1, 3, -2, 3]. The maximum subarray sum is 3 + (-2) + 3 = 4.
Deleting all occurrences of x = -2 results in nums = [-3, 2, -1, 3, 3]. The maximum subarray sum is 2 + (-1) + 3 + 3 = 7.
Deleting all occurrences of x = 3 results in nums = [-3, 2, -2, -1, -2]. The maximum subarray sum is 2.

The output is max(4, 4, 7, 4, 2) = 7.

Another example: nums = [1,2,3,4]. In this case the optimal solution is to not remove any numbers, thus the answer is 10.

How would you write code to solve this?

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 integers in the input array?
Can the input array be empty or null?
If all elements in the array are removed after removing all occurrences of a specific element, should I return 0?
If multiple elements result in the same maximum subarray sum after their removal, is any of them acceptable?
Is the subarray required to be contiguous?

Brute Force Solution

Approach

The brute force approach aims to find the largest possible sum of a sub-collection of numbers after removing all instances of a chosen number. We will methodically try removing each unique number in the original collection, one at a time. For each removal, we will look at all possible sub-collections and calculate the sum to keep track of the biggest sum we find.

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

First, identify all the unique numbers present in the original collection.
For each unique number, create a new collection of numbers where that unique number has been completely removed.
Now, for each of these new collections (created after removing a unique number), consider every possible sub-collection of consecutive numbers.
Calculate the sum of the numbers in each sub-collection.
Keep track of the largest sum found among all these sub-collections, across all the different removals.
After going through all the unique numbers and their corresponding collections, the largest sum you've tracked is your answer.

Code Implementation

def maximize_subarray_sum_after_removal_brute_force(numbers):
    maximum_sum = float('-inf')
    unique_numbers = set(numbers)

    # Iterate through each unique number in the original array
    for number_to_remove in unique_numbers:

        modified_numbers = [number for number in numbers if number != number_to_remove]

        # Iterate through all possible subarrays of the modified array
        for start_index in range(len(modified_numbers)):
            for end_index in range(start_index, len(modified_numbers)):
                current_sum = sum(modified_numbers[start_index:end_index+1])

                # Keep track of the maximum sum found so far
                maximum_sum = max(maximum_sum, current_sum)

    # Handles the case where all numbers are removed, resulting in an empty array.
    if maximum_sum == float('-inf'):
        return 0

    return maximum_sum

Big(O) Analysis

Time Complexity

O(n^3) – First, identifying unique numbers takes O(n) time. Then, for each unique number (at most n), we remove all its occurrences, which takes O(n) time, resulting in an array of size at most n. After the removal, finding the maximum subarray sum requires iterating through all possible subarrays, which involves two nested loops, each iterating up to n times. Therefore, the subarray sum calculation takes O(n^2) time. Since we repeat this process for each unique number, the total time complexity is O(n * n^2), which simplifies to O(n^3).

Space Complexity

O(N) – The algorithm identifies unique numbers, which could potentially all be unique, requiring a storage of up to N elements. When removing each unique number, a new collection (a list) is created without the removed number, potentially using N space in the worst case. Although these lists may be reused, the space complexity is governed by the maximum space utilized, which is dominated by the new list created, resulting in O(N) space complexity.

Optimal Solution

Approach

The most efficient strategy involves checking each distinct number in the input list. For each number, we temporarily remove all instances of it and then find the largest possible sum from a continuous group of the remaining numbers. We keep track of the largest sum found across all these removals.

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

First, identify all the unique numbers present in the original list.
For each unique number, create a modified list where all occurrences of that number are removed.
Within this modified list, find the continuous section that yields the greatest sum. If all numbers are negative, the greatest sum is zero.
Keep track of the biggest sum found so far across all the modified lists.
After checking all unique numbers and their corresponding modified lists, the biggest sum you kept track of is the final answer.

Code Implementation

def maximize_subarray_sum_after_removal(input_list):
    unique_numbers = set(input_list)
    maximum_sum = 0

    for number_to_remove in unique_numbers:

        modified_list = [number for number in input_list if number != number_to_remove]

        current_max_sum = 0
        overall_max_sum = 0

        # Kadane's algorithm to find max subarray sum
        for number in modified_list:

            current_max_sum += number

            if current_max_sum < 0:
                current_max_sum = 0

            overall_max_sum = max(overall_max_sum, current_max_sum)

        # This ensures we capture the case where the best sum is 0 (empty subarray).
        maximum_sum = max(maximum_sum, overall_max_sum)

    return maximum_sum

Big(O) Analysis

Time Complexity

O(n²) – Identifying the unique numbers in the original list takes O(n) time. For each unique number (in the worst case, all n numbers are unique), we create a modified list by removing all occurrences of that number, which takes O(n) time. Then, finding the maximum subarray sum in the modified list using Kadane's algorithm also takes O(n) time. Since we repeat these two O(n) operations for each of the (at most) n unique numbers, the overall time complexity is O(n * n) which simplifies to O(n²).

Space Complexity

O(N) – The algorithm identifies unique numbers, which in the worst case, could all be unique, requiring storage proportional to the input size N. For each unique number, a modified list is created by removing all occurrences of that number, potentially resulting in a new list of size close to N in the worst case (if the removed number appears infrequently). Thus, the space used by these temporary lists scales linearly with the size of the input list. Therefore, the overall auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0 as the maximum subarray sum when the input array is empty.
Array with only one element	If removing that single element results in an empty array, return 0; otherwise, return the value of that element if keeping it yields a valid subarray (sum >0).
Array with all elements having the same value	If all elements are the same, removing any of them results in an array of the same value repeated less, use Kadane's algorithm to calculate.
Array with all negative numbers	Return 0 if all remaining subarray sums are negative after removing one number.
Array with all zero numbers	Return 0 after removing one zero as the sum will be 0.
Large array leading to potential integer overflow	Use long long int to avoid integer overflow during sum calculation.
Array with very large positive and negative numbers	Ensure intermediate sums do not overflow by using long long int during Kadane's algorithm.
The maximum subarray sum is negative, but removing a specific element makes it zero	Keep track of whether the best subarray sum becomes 0 after an element is removed, and return 0 if the maximum is otherwise negative.