How do you solve Compute Alternating Sum on LeetCode?

Compute Alternating Sum (LeetCode #3701) 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: Understanding even and odd index manipulation is crucial.

What is the time complexity of Compute Alternating Sum?

The optimal solution for Compute Alternating Sum runs in O(n) time and uses O(1) space. The time complexity is O(n) due to a single loop through the array. Space complexity is O(1) as we only use one variable.

What is the brute force solution for Compute Alternating Sum?

The brute force approach for Compute Alternating Sum is Brute Force, with O(n) time complexity and O(1) space complexity. We can iterate through the array and separately sum elements at even and odd indices, then compute the result. This straightforward approach is easy to understand but not efficient. The optimal Optimal Solution approach improves this to O(n).

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

Compute Alternating Sum uses the following concepts: Array, Simulation. The recommended patterns to study are: Array, Simulation.

What are the interview tips for Compute Alternating Sum?

Clarify the problem requirements before coding. Discuss your thought process while coding. Test edge cases after implementation.

What are common mistakes when solving Compute Alternating Sum?

Confusing index parity when adding/subtracting. Not initializing variables properly.

#3701

Compute Alternating Sum

Easy

ArraySimulationArraySimulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

We can achieve the same result in a single pass through the array, maintaining a running total. This avoids the need for separate sums.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable total to 0.
2Step 2: Loop through the array, adding nums[i] if i is even and subtracting nums[i] if i is odd.
3Step 3: Return the total.

solution.py5 lines

1def alternatingSum(nums):
2    total = 0
3    for i in range(len(nums)):
4        total += nums[i] if i % 2 == 0 else -nums[i]
5    return total

ℹ

Complexity note: The time complexity is O(n) due to a single loop through the array. Space complexity is O(1) as we only use one variable.

1Understanding even and odd index manipulation is crucial.
2Single-pass algorithms can often simplify problems.

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