Problem Description

In this problem, you are provided with two pieces of information: 1) an integer array nums, which contains some integers, and 2) an array of arrays queries, where each subarray contains exactly two integers representing a value (val_i) and an index (index_i). The goal is to process each subarray in queries sequentially. The processing steps for each query are as follows: a) Add val_i to nums[index_i] (modify the element in nums at the specified index by adding the value). b) Calculate the sum of all even numbers currently in the nums array.

You need to perform this two-step process for each query and record the result of the second step. The final output should be an array where each element corresponds to the sum of even numbers after the corresponding query has been processed.

Intuition

The naive approach to solving this problem would be to compute the sum of all even numbers in the nums array after applying each query. However, this would be inefficient because it would require iterating through the entire nums array to recalculate the sum after each query, resulting in a time complexity that would be burdensome for large arrays.

The solution instead maintains a running total sum s of all even numbers in the nums array. This sum is updated incrementally. Here's the thought process for this approach:

1. Calculate and store the sum of all even numbers in the nums array before processing any queries.
2. For each query (val_i, index_i):
   • If the number at nums[index_i] is even before the query, subtract it from s to temporarily remove it from the even sum. This is done because the addition might make it odd.
   • Apply the update nums[index_i] = nums[index_i] + val_i.
   • After the update, check if nums[index_i] is now even. If it is, add it back to s, as the change either kept it even or turned it from odd to even.
3. Record the current value of s after each query to build the answer array, which represents the sum of even numbers after each modification.

This way, we avoid recalculating the sum of even numbers from scratch after each query, effectively reducing complexity and speeding up the algorithm.

Solution Approach

The implementation of the solution uses an approach that efficiently updates and calculates the sum of even numbers after each query applied to the nums array. The initial setup and steps followed in the code are as follows:

1. Initialize the Even Sum: Before processing any queries, the solution calculates the initial sum s of all the even numbers in nums. This is done using a list comprehension and the built-in sum function:

   s = sum(x for x in nums if x % 2 == 0)

2. Process Queries: The implementation then iterates over each query in queries. For each query [v, i], where v is the value to be added and i is the index, the solution:

   • Checks if nums[i] is even before the update. If it is, the value of nums[i] is subtracted from the running even sum s, as the update might turn it into an odd number:

     if nums[i] % 2 == 0:
         s -= nums[i]

   • Applies the update to nums[i] by adding the value v to it:

     nums[i] += v

   • Checks if after the update, nums[i] is even. If it has become or remained even, it is added to the running even sum s:

     if nums[i] % 2 == 0:
         s += nums[i]

   • Append the current running sum s (which now reflects the sum of even values after the update) to an answer list ans:

     ans.append(s)

3. Return Results: After processing all the queries, the list ans, which has recorded the sum of even numbers after each query, is returned as the final output.

By maintaining and updating the running even sum s with each query, the algorithm achieves an efficient and dynamic way to keep track of changes without recomputing the entire sum after each update. This greatly improves the time efficiency, especially when there are many queries or the nums array is large.

Example Walkthrough

Let's go through a small example to illustrate the solution approach.

Suppose we are given the integer array nums = [1, 2, 3, 4] and queries [[1, 0], [-3, 2], [5, 3], [4, 1]]. We need to perform the queries on nums and track the sum of even numbers in nums after each query.

1. Initialize the Even Sum:

   Before any queries, we find the sum of all even numbers in nums:

   nums = [1, 2, 3, 4]
   s = sum(x for x in nums if x % 2 == 0) # s = 2 + 4 = 6

   The initial sum s of even numbers in nums is 6.

2. Process Queries:

   First query [1, 0]: We add 1 to nums[0], which results in nums becoming [2, 2, 3, 4].

   Since nums[0] was not even before the query (1 is odd), we don't subtract anything from s. After the update, nums[0] becomes even (2), so we add it to s.

   s += nums[0] # s = 6 + 2 = 8
   ans.append(s) # ans = [8]

   Second query [-3, 2]: We subtract 3 from nums[2] (nums[2] - 3), and nums becomes [2, 2, 0, 4].

   Before the query, nums[2] was odd, so the sum s remains unchanged. After the update, nums[2] is even (0), and it's added to the sum s.

   s += nums[2] # s = 8 + 0 = 8 (unchanged)
   ans.append(s) # ans = [8, 8]

   Third query [5, 3]: We now add 5 to nums[3], resulting in nums becoming [2, 2, 0, 9].

   nums[3] was even before (4), so we subtract it from the sum s first. After the update (4 + 5 = 9), nums[3] is odd, so we don't add it back to s.

   s -= nums[3] # Subtracting the old value of `nums[3]`, s = 8 - 4 = 4
   nums[3] += 5 # nums[3] becomes odd after the addition
   ans.append(s) # ans = [8, 8, 4]

   Fourth query [4, 1]: The last query adds 4 to nums[1], changing nums to [2, 6, 0, 9].

   Since nums[1] was even before (2), we remove it from the sum s. After the update (2 + 4 = 6), nums[1] is still even, so we add the new value back to s.

   s -= nums[1] # Subtracting the initial value of `nums[1]`, s = 4 - 2 = 2
   nums[1] += 4 # Updating `nums[1]`
   s += nums[1] # Adding the updated value to `s`, s = 2 + 6 = 8
   ans.append(s) # ans = [8, 8, 4, 8]

3. Return Results:

   After processing all the queries, the ans list, which contains the sum of even numbers after each query, is [8, 8, 4, 8]. This list is the final output.

Thus, by following this approach, we efficiently kept track of the sum of even numbers in nums after each query without recalculating the entire sum each time.

Solution Implementation

Time and Space Complexity

The provided code snippet consists of an initial computation that sums the even values of the input nums list, followed by an iteration through each query in queries. Each query modifies a single element of nums and conditionally updates the sum of even numbers.

Time Complexity:

The initial sum computation has a time complexity of O(N), where N is the number of elements in nums.

The for loop runs for every query in queries, which we can denote as Q, where Q is the number of queries. Inside the loop, we have constant-time operations. Therefore, the loop runs in O(Q) time.

The total time complexity of the function is a result of the initial sum computation and the queries loop, which we can express as O(N) + O(Q). Therefore, the overall time complexity is O(N + Q).

Space Complexity:

The space complexity is determined by the additional space used by the algorithm aside from the input itself. The space used includes the sum s and the list ans that accumulates the resulting sums after each query.

• s is a single integer, so it requires O(1) space.
• ans will hold the result after every query, which means it will grow to the size of Q. Therefore, it requires O(Q) space.

Considering the additional space used by the function, the total space complexity is O(Q) since Q may vary independently of N and the size of ans directly depends on the number of queries.

Learn more about how to find time and space complexity quickly using problem constraints.