Maximum Alternating Subsequence Sum

Medium

Asked by:

39 views

Topics:

ArraysDynamic ProgrammingGreedy Algorithms

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

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

Given an array nums, return the maximum alternating sum of any subsequence of nums (after reindexing the elements of the subsequence).

A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements' relative order. For example, [2,7,4] is a subsequence of [4,2,3,7,2,1,4] (the underlined elements), while [2,4,2] is not.

Example 1:

Input: nums = [4,2,5,3]
Output: 7
Explanation: It is optimal to choose the subsequence [4,2,5] with alternating sum (4 + 5) - 2 = 7.

Example 2:

Input: nums = [5,6,7,8]
Output: 8
Explanation: It is optimal to choose the subsequence [8] with alternating sum 8.

Example 3:

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

Constraints:

1 <= nums.length <= 10⁵
1 <= 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 `nums` contain negative numbers, zeros, or only positive numbers?
What are the constraints on the size of the input array `nums`? Will it always be a valid array?
If multiple alternating subsequences yield the maximum sum, is any one of them acceptable?
Is an empty subsequence allowed, and if so, what should the return value be in that case (e.g., 0)?
By 'alternating', do you strictly mean that the sign of the difference between consecutive elements in the subsequence must alternate (e.g., strictly increasing then strictly decreasing, or vice versa), or is a non-strict increase/decrease followed by a non-strict decrease/increase acceptable?

Brute Force Solution

Approach

The brute force way to find the maximum alternating sum is like trying every single combination of numbers from the original list. We carefully pick and choose numbers, alternating between adding and subtracting, to find the best possible sum.

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

First, consider all possible sub-lists you can make from the original list of numbers. For example, you could choose just the first number, the first two numbers, or any other combination.
For each sub-list you create, calculate its alternating sum. Start by adding the first number, then subtracting the second, then adding the third, and so on.
As you calculate each alternating sum, remember to keep track of the largest one you've found so far.
Once you have gone through all the possible sub-lists and calculated their alternating sums, the largest sum you tracked is the answer.

Code Implementation

def max_alternating_subsequence_sum_brute_force(numbers):
    maximum_sum = 0

    # Iterate through all possible subsequences
    for i in range(1 << len(numbers)):
        subsequence = []
        for j in range(len(numbers)):
            if (i >> j) & 1:
                subsequence.append(numbers[j])

        # Calculate the alternating sum of the subsequence
        alternating_sum = 0
        if subsequence:
            alternating_sum = subsequence[0]

            # Alternate between adding and subtracting
            for k in range(1, len(subsequence)):

                # The goal of this if/else is to check the index
                if k % 2 == 0:
                    alternating_sum += subsequence[k]

                else:
                    alternating_sum -= subsequence[k]

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

    return maximum_sum

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach involves generating all possible subsequences of the input array. Generating all subsequences of an array of size n takes O(2^n) time because each element can either be included or excluded from a subsequence. For each subsequence, calculating the alternating sum takes O(n) time in the worst case (when the subsequence is nearly as long as the original array), although the actual subsequences have sizes from 1 to n. Therefore, the time complexity for generating the subsequences dominates the time complexity for the alternating sum calculation, giving a time complexity of O(2^n).

Space Complexity

O(1) – The provided brute force approach iterates through all possible sublists and calculates their alternating sum. While calculating the alternating sum for each sublist, it only requires storing the current maximum alternating sum found so far and a few temporary variables to track the addition/subtraction. The amount of extra memory used does not scale with the input size N, which represents the number of elements in the original list. Therefore, the space complexity is constant.

Optimal Solution

Approach

The best approach is to find the biggest sum we can make by carefully choosing numbers that alternate between adding and subtracting. We can achieve this by keeping track of the best possible sum if we ended on an added number, and the best possible sum if we ended on a subtracted number, building our solution as we go.

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

Start by assuming our best sums are zero, both when adding last and when subtracting last.
Go through the list of numbers one at a time.
For each number, consider what happens if we add it to our sum that ended on subtraction: update the best sum when adding last to be the larger of the old value or adding the new number to the 'subtract last' sum.
Also, consider what happens if we subtract it from our sum that ended on addition: update the best sum when subtracting last to be the larger of the old value or subtracting the new number from the 'add last' sum.
After going through all the numbers, the larger of the two best sums ('add last' or 'subtract last') is the biggest alternating sum you can create.

Code Implementation

def max_alternating_subsequence_sum(numbers):
    add_last_sum = 0
    subtract_last_sum = 0

    for number in numbers:
        # Update best sum ending with addition
        new_add_last_sum = max(add_last_sum, subtract_last_sum + number)

        # Update best sum ending with subtraction
        # Uses previous add_last_sum to ensure correct alternation
        new_subtract_last_sum = max(subtract_last_sum, add_last_sum - number)

        add_last_sum = new_add_last_sum
        subtract_last_sum = new_subtract_last_sum

    # Return the greater of the two, which is the max sum
    return max(add_last_sum, subtract_last_sum)

Big(O) Analysis

Time Complexity

O(n) – The provided solution iterates through the input array nums exactly once. Inside the loop, it performs a constant number of operations: updating the 'add last' and 'subtract last' sums based on the current number. Since the number of operations is directly proportional to the size of the input array (n), the time complexity is O(n).

Space Complexity

O(1) – The algorithm uses only two variables, 'add last' and 'subtract last', to store the best sums. The space used by these variables remains constant regardless of the input array's size N. No other data structures or recursion are employed. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0, as there's no subsequence.
Single element array	Return the single element if it's positive; otherwise, return 0.
Array with only negative numbers	Return 0, as no positive contribution is possible.
Array with alternating positive and negative numbers starting with negative	Return 0, or the first positive number if it exists, after removing the initial negative number.
Array with large positive and negative numbers (potential overflow)	Use long data type to prevent integer overflow during summation.
Array with all identical positive numbers	Return the single number if the length of the array is odd, and 0 otherwise.
Array with all identical negative numbers	Return 0, as no element contributes positively to the alternating sum.
Maximum-sized array with mixed positive and negative numbers	Ensure the solution has O(n) time complexity (e.g., dynamic programming) to avoid exceeding time limits.

Null or empty input array

How to Handle:

Return 0, as there's no subsequence.

Single element array

How to Handle:

Return the single element if it's positive; otherwise, return 0.

Array with only negative numbers

How to Handle:

Return 0, as no positive contribution is possible.

Array with alternating positive and negative numbers starting with negative

How to Handle:

Return 0, or the first positive number if it exists, after removing the initial negative number.

Array with large positive and negative numbers (potential overflow)

How to Handle:

Use long data type to prevent integer overflow during summation.

Array with all identical positive numbers

How to Handle:

Return the single number if the length of the array is odd, and 0 otherwise.

Array with all identical negative numbers

How to Handle:

Return 0, as no element contributes positively to the alternating sum.

Maximum-sized array with mixed positive and negative numbers

How to Handle:

Ensure the solution has O(n) time complexity (e.g., dynamic programming) to avoid exceeding time limits.