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Maximum Alternating Subarray Sum

Medium

Asked by:

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Topics:

ArraysDynamic Programming

The alternating sum of a 0-indexed array is defined as the sum of elements at even indices minus the sum of elements at odd indices.

For example, the alternating sum of [4,2,5,3] is (4 + 5) - (2 + 3) = 4.

Given an integer array nums, return the maximum alternating subarray sum of nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [4,2,5,3]
Output: 7
Explanation: It is optimal to choose the subarray [4,2,5] which has an alternating sum of (4 + 5) - 2 = 7.

Example 2:

Input: nums = [5,6,7,8]
Output: 8
Explanation: It is optimal to choose the subarray [8] which has an alternating sum of 8.

Example 3:

Input: nums = [6,2,1,2,4,5]
Output: 10
Explanation: It is optimal to choose the subarray [6,2,1,2,4,5] which has an alternating sum of (6 + 1 + 4) - (2 + 2 + 5) = 2.

Constraints:

1 <= nums.length <= 10⁵
-10⁵ <= nums[i] <= 10⁵

Try coding it in LeetCode

Maximum Alternating Subarray Sum

Medium

Asked by:

20 views

Topics:

ArraysDynamic Programming

The alternating sum of a 0-indexed array is defined as the sum of elements at even indices minus the sum of elements at odd indices.

For example, the alternating sum of [4,2,5,3] is (4 + 5) - (2 + 3) = 4.

Given an integer array nums, return the maximum alternating subarray sum of nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [4,2,5,3]
Output: 7
Explanation: It is optimal to choose the subarray [4,2,5] which has an alternating sum of (4 + 5) - 2 = 7.

Example 2:

Input: nums = [5,6,7,8]
Output: 8
Explanation: It is optimal to choose the subarray [8] which has an alternating sum of 8.

Example 3:

Input: nums = [6,2,1,2,4,5]
Output: 10
Explanation: It is optimal to choose the subarray [6,2,1,2,4,5] which has an alternating sum of (6 + 1 + 4) - (2 + 2 + 5) = 2.

Constraints:

1 <= nums.length <= 10⁵
-10⁵ <= nums[i] <= 10⁵

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:

Can the input array contain negative numbers, zeros, or only positive numbers?
What is the expected behavior if the input array is empty or null?
What is the maximum possible size of the input array?
In case of multiple subarrays with the same maximum alternating sum, is any of them acceptable?
Can you provide a more formal definition of 'alternating subarray' to avoid any ambiguity (e.g., does it start with addition or subtraction)?

Brute Force Solution

Approach

The brute force method for this problem involves checking every possible subarray. We'll look at each possible combination of numbers in the given set, calculating the alternating sum for each one. Finally, we find the largest sum we've seen.

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

Consider every possible starting point for a subarray within the given set of numbers.
For each starting point, consider every possible ending point to define a subarray.
For each subarray defined, calculate the alternating sum. This means adding the first number, subtracting the second, adding the third, and so on.
Keep track of the largest alternating sum you find as you calculate the sums of all the different subarrays.
After considering all possible subarrays, the largest alternating sum you tracked is the answer.

Code Implementation

def max_alternating_subarray_sum_brute_force(numbers):
    max_sum = float('-inf')

    # Iterate through all possible start indices
    for start_index in range(len(numbers)):

        # Iterate through all possible end indices for each start index
        for end_index in range(start_index, len(numbers)):
            current_sum = 0
            sign = 1

            # Calculate the alternating sum for the current subarray.
            for index in range(start_index, end_index + 1):
                current_sum += sign * numbers[index]
                sign *= -1

            # Update the maximum sum if the current sum is greater.
            if current_sum > max_sum:
                max_sum = current_sum

    return max_sum

Big(O) Analysis

Time Complexity

O(n³) – The algorithm iterates through all possible subarrays. The outer loop selects the starting index, which takes O(n) time. The inner loop selects the ending index, also taking O(n) time. For each subarray selected by these two loops, the alternating sum is calculated, which takes O(n) time in the worst case (when the subarray includes nearly all elements). Therefore, the total time complexity is O(n * n * n), which simplifies to O(n³).

Space Complexity

O(1) – The brute force approach, as described, iterates through all possible subarrays and calculates their alternating sums. While doing so, it only maintains a variable to store the maximum alternating sum found so far, which does not depend on the input size N (the number of elements in the input array). No additional data structures like arrays, lists, or hash maps are used to store intermediate subarray sums or track visited elements; the calculation is done on the fly. Therefore, the auxiliary space complexity is constant.

Optimal Solution

Approach

The most efficient way to solve this problem is to keep track of the best possible sum we can achieve ending with a positive number and the best sum ending with a negative number as we go through the list. At each number, we update these two best sums using the current number, either adding it or subtracting it, depending on whether we want the subarray to end on a positive or negative number. This avoids recomputing sums repeatedly.

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

Start by setting the best sum ending with a positive number and the best sum ending with a negative number to zero.
Go through each number in the list one by one.
For each number, calculate the new best sum ending with a positive number. This will either be the current number itself (if starting a new subarray), or the previous best sum ending with a negative number, plus the current number.
Similarly, calculate the new best sum ending with a negative number. This will either be zero (if starting a new subarray and not using the current number), or the previous best sum ending with a positive number, minus the current number.
Keep track of the overall maximum sum we've seen so far. Update it if either the best sum ending with a positive number or the best sum ending with a negative number is greater than the current maximum.
After going through all the numbers, the overall maximum sum will be the answer.

Code Implementation

def max_alternating_subarray_sum(numbers):
    max_ending_positive = 0
    max_ending_negative = 0
    overall_maximum = 0

    for number in numbers:
        # Calculate best sum ending positive
        new_max_ending_positive = max(number, max_ending_negative + number)

        # Calculate best sum ending negative
        # Zero handles cases where current number isn't included
        new_max_ending_negative = max(0, max_ending_positive - number)

        max_ending_positive = new_max_ending_positive
        max_ending_negative = new_max_ending_negative

        # Track overall maximum
        overall_maximum = max(overall_maximum, max_ending_positive, max_ending_negative)

    return overall_maximum

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input list of n numbers exactly once. Inside the loop, it performs a fixed number of calculations to update the best positive and negative sums, and the overall maximum. These operations take constant time. Therefore, the time complexity is directly proportional to the number of elements in the input list, resulting in O(n).

Space Complexity

O(1) – The algorithm uses a fixed number of variables: best_positive, best_negative, and max_sum. The number of variables does not depend on the size of the input list (N). Therefore, the auxiliary space used is constant.

Edge Cases

Case	How to Handle
Empty array	Return 0, as there is no subarray.
Array with a single element	Return the value of the single element, as it is the only possible subarray.
Array with all positive numbers	The alternating subarray sum will include all elements in the array.
Array with all negative numbers	The optimal alternating subarray sum will be the maximum single element (least negative).
Array with alternating positive and negative numbers	The entire array sum, calculated in an alternating fashion, is the optimal result.
Array with large positive and negative numbers that may lead to integer overflow during summation	Use a larger data type like long or consider performing modulo operations to prevent overflow.
Array with consecutive identical numbers (e.g., [1, 1, -1, 1])	The identical numbers effectively become a single number, as only the first contributes to an alternating sum.
Array with long sequences of zeros	Zeros should be skipped, as they don't affect the alternating property of the subarray.

Empty array

How to Handle:

Return 0, as there is no subarray.

Array with a single element

How to Handle:

Return the value of the single element, as it is the only possible subarray.

Array with all positive numbers

How to Handle:

The alternating subarray sum will include all elements in the array.

Array with all negative numbers

How to Handle:

The optimal alternating subarray sum will be the maximum single element (least negative).

Array with alternating positive and negative numbers

How to Handle:

The entire array sum, calculated in an alternating fashion, is the optimal result.

Array with large positive and negative numbers that may lead to integer overflow during summation

How to Handle:

Use a larger data type like long or consider performing modulo operations to prevent overflow.

Array with consecutive identical numbers (e.g., [1, 1, -1, 1])

How to Handle:

The identical numbers effectively become a single number, as only the first contributes to an alternating sum.

Array with long sequences of zeros

How to Handle:

Zeros should be skipped, as they don't affect the alternating property of the subarray.

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:

Can the input array contain negative numbers, zeros, or only positive numbers?
What is the expected behavior if the input array is empty or null?
What is the maximum possible size of the input array?
In case of multiple subarrays with the same maximum alternating sum, is any of them acceptable?
Can you provide a more formal definition of 'alternating subarray' to avoid any ambiguity (e.g., does it start with addition or subtraction)?

Brute Force Solution

Approach

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

Consider every possible starting point for a subarray within the given set of numbers.
For each starting point, consider every possible ending point to define a subarray.
For each subarray defined, calculate the alternating sum. This means adding the first number, subtracting the second, adding the third, and so on.
Keep track of the largest alternating sum you find as you calculate the sums of all the different subarrays.
After considering all possible subarrays, the largest alternating sum you tracked is the answer.

Code Implementation

def max_alternating_subarray_sum_brute_force(numbers):
    max_sum = float('-inf')

    # Iterate through all possible start indices
    for start_index in range(len(numbers)):

        # Iterate through all possible end indices for each start index
        for end_index in range(start_index, len(numbers)):
            current_sum = 0
            sign = 1

            # Calculate the alternating sum for the current subarray.
            for index in range(start_index, end_index + 1):
                current_sum += sign * numbers[index]
                sign *= -1

            # Update the maximum sum if the current sum is greater.
            if current_sum > max_sum:
                max_sum = current_sum

    return max_sum

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start by setting the best sum ending with a positive number and the best sum ending with a negative number to zero.
Go through each number in the list one by one.
For each number, calculate the new best sum ending with a positive number. This will either be the current number itself (if starting a new subarray), or the previous best sum ending with a negative number, plus the current number.
Similarly, calculate the new best sum ending with a negative number. This will either be zero (if starting a new subarray and not using the current number), or the previous best sum ending with a positive number, minus the current number.
Keep track of the overall maximum sum we've seen so far. Update it if either the best sum ending with a positive number or the best sum ending with a negative number is greater than the current maximum.
After going through all the numbers, the overall maximum sum will be the answer.

Code Implementation

def max_alternating_subarray_sum(numbers):
    max_ending_positive = 0
    max_ending_negative = 0
    overall_maximum = 0

    for number in numbers:
        # Calculate best sum ending positive
        new_max_ending_positive = max(number, max_ending_negative + number)

        # Calculate best sum ending negative
        # Zero handles cases where current number isn't included
        new_max_ending_negative = max(0, max_ending_positive - number)

        max_ending_positive = new_max_ending_positive
        max_ending_negative = new_max_ending_negative

        # Track overall maximum
        overall_maximum = max(overall_maximum, max_ending_positive, max_ending_negative)

    return overall_maximum

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty array	Return 0, as there is no subarray.
Array with a single element	Return the value of the single element, as it is the only possible subarray.
Array with all positive numbers	The alternating subarray sum will include all elements in the array.
Array with all negative numbers	The optimal alternating subarray sum will be the maximum single element (least negative).
Array with alternating positive and negative numbers	The entire array sum, calculated in an alternating fashion, is the optimal result.
Array with large positive and negative numbers that may lead to integer overflow during summation	Use a larger data type like long or consider performing modulo operations to prevent overflow.
Array with consecutive identical numbers (e.g., [1, 1, -1, 1])	The identical numbers effectively become a single number, as only the first contributes to an alternating sum.
Array with long sequences of zeros	Zeros should be skipped, as they don't affect the alternating property of the subarray.

Empty array

How to Handle:

Return 0, as there is no subarray.

Array with a single element

How to Handle:

Return the value of the single element, as it is the only possible subarray.

Array with all positive numbers

How to Handle:

The alternating subarray sum will include all elements in the array.

Array with all negative numbers

How to Handle:

The optimal alternating subarray sum will be the maximum single element (least negative).

Array with alternating positive and negative numbers

How to Handle:

The entire array sum, calculated in an alternating fashion, is the optimal result.

Array with large positive and negative numbers that may lead to integer overflow during summation

How to Handle:

Use a larger data type like long or consider performing modulo operations to prevent overflow.

Array with consecutive identical numbers (e.g., [1, 1, -1, 1])

How to Handle:

The identical numbers effectively become a single number, as only the first contributes to an alternating sum.

Array with long sequences of zeros

How to Handle:

Zeros should be skipped, as they don't affect the alternating property of the subarray.