How do you solve Sum of Even Numbers After Queries on LeetCode?

Sum of Even Numbers After Queries (LeetCode #985) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(1) space complexity. Key insight: Maintaining a running total of even sums can significantly reduce computation time.

What is the brute force solution for Sum of Even Numbers After Queries?

The brute force approach for Sum of Even Numbers After Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we handle each query one by one. For each query, we update the array and then calculate the sum of even numbers from scratch. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Sum of Even Numbers After Queries?

Sum of Even Numbers After Queries uses the following concepts: Array, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sum of Even Numbers After Queries?

Always consider edge cases, such as when all numbers are odd or when queries lead to all even numbers. Think about how to minimize repeated calculations; maintaining state can often help. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Sum of Even Numbers After Queries?

Not checking if the updated number is even or odd before adjusting the sum. Forgetting to initialize the sum of even numbers before processing queries.

#985

Sum of Even Numbers After Queries

Medium

ArraySimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n + m)Space O(1)

Instead of recalculating the sum of even numbers from scratch for each query, we can maintain a running total of the sum of even numbers. This allows us to update the sum efficiently when we modify an element.

⚙️

Algorithm

6 steps

1Step 1: Calculate the initial sum of even numbers in the nums array.
2Step 2: Initialize an empty list to store results.
3Step 3: For each query, update the specified index in the nums array.
4Step 4: Check if the updated number is even or odd and adjust the even sum accordingly.
5Step 5: Append the current even sum to the results list.
6Step 6: Return the results list.

solution.py13 lines

1# Full working Python code
2nums = [1, 2, 3, 4]
3queries = [[1, 0], [-3, 1], [-4, 0], [2, 3]]
4even_sum = sum(x for x in nums if x % 2 == 0)
5result = []
6for val, index in queries:
7    if nums[index] % 2 == 0:
8        even_sum -= nums[index]
9    nums[index] += val
10    if nums[index] % 2 == 0:
11        even_sum += nums[index]
12    result.append(even_sum)
13print(result)

ℹ

Complexity note: This complexity arises because we initially compute the sum of even numbers (O(n)), and then for each query (O(m)), we perform constant time operations to update the sum.

1Maintaining a running total of even sums can significantly reduce computation time.
2Understanding the impact of adding or subtracting values on evenness is crucial.

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