Sum of Even Numbers After Queries
01 · Problem
You are given an integer array nums and an array queries where queries[i] = [val_i, index_i].
Process the queries in order. For query i, first add val_i to nums[index_i] (the change persists for later queries), then record the sum of all even values currently in nums. Negative even numbers and 0 count as even.
Return an array answer where answer[i] is the sum recorded after the i-th query.
02 · Examples
nums = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
[8,6,2,4]
After [1,0] nums = [2,2,3,4], even sum 8. After [-3,1] nums = [2,-1,3,4], even sum 6. After [-4,0] nums = [-2,-1,3,4], even sum 2. After [2,3] nums = [-2,-1,3,6], even sum 4.
nums = [1], queries = [[4,0]]
[0]
nums becomes [5], which has no even values, so the sum is 0.
nums = [0,-2,5], queries = [[3,2],[-1,0]]
[6,6]
After [3,2] nums = [0,-2,8], even sum 6. After [-1,0] nums = [-1,-2,8], even sum 6.
03 · Constraints
- 011 <= nums.length <= 104
- 02-104 <= nums[i] <= 104
- 031 <= queries.length <= 104
- 04-104 <= val_i <= 104
- 050 <= index_i < nums.length
04 · Optimal complexity
- Time
- O(n + q)
- Space
- O(1)
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