How do you solve Maximum Alternating Subsequence Sum on LeetCode?

Maximum Alternating Subsequence Sum (LeetCode #1911) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The alternating sum can be maximized by strategically choosing elements based on their indices.

What is the brute force solution for Maximum Alternating Subsequence Sum?

The brute force approach for Maximum Alternating Subsequence Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries all possible subsequences of the array to calculate their alternating sums. This method guarantees that we explore every combination, but it is inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Alternating Subsequence Sum?

Maximum Alternating Subsequence Sum uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Maximum Alternating Subsequence Sum?

Always clarify the problem and constraints before jumping into coding. Think about edge cases, such as arrays with only one element. Explain your thought process clearly while coding, as this shows your understanding.

What are common mistakes when solving Maximum Alternating Subsequence Sum?

Not considering both even and odd sums during the iteration. Failing to update the sums correctly based on the current number.

#1911

Maximum Alternating Subsequence Sum

Medium

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution uses dynamic programming to keep track of two states: the maximum alternating sum when we include the current number and when we exclude it. This allows us to efficiently compute the maximum sum without exploring all subsequences.

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Algorithm

4 steps

1Step 1: Initialize two variables, evenSum and oddSum, to track the maximum sums for even and odd indexed elements respectively.
2Step 2: Iterate through each number in the array, updating evenSum and oddSum based on the current number.
3Step 3: For each number, calculate the new evenSum as the maximum of the current evenSum and oddSum + current number, and update oddSum as the maximum of the current oddSum and evenSum - current number.
4Step 4: Return evenSum as the result since it represents the maximum alternating sum.

solution.py9 lines

1# Full working Python code
2
3def maxAlternatingSum(nums):
4    evenSum = 0
5    oddSum = 0
6    for num in nums:
7        evenSum = max(evenSum, oddSum + num)
8        oddSum = max(oddSum, evenSum - num)
9    return evenSum

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array, and the space complexity is O(1) since we are using only a constant amount of extra space.

1The alternating sum can be maximized by strategically choosing elements based on their indices.
2Dynamic programming allows us to efficiently compute the maximum sum without generating all subsequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.