985. Sum of Even Numbers After Queries

Problem

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You are given an integer array nums and an array queries where ueries[i] = [val$_i$, index$_i$].

For each query i, first, apply nums[indexi] = nums[indexi] + val$_i$, then print the sum of the even values of nums.

Return an integer array answer where answer[i] is the answer to the i$^t$$^h$ query.

Example 1:

Input: nums = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: [8,6,2,4]
Explanation: At the beginning, the array is [1,2,3,4].
After adding 1 to nums[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to nums[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to nums[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to nums[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.

Example 2:

Input: nums = [1], queries = [[4,0]]
Output: [0]

Constraints:

• 1 <= nums.length <= 10$^4$
• -10$^4$ <= nums[i] <= 10$^4$
• 1 <= queries.length <= 10$^4$
• -10$^4$ <= val$_i$ <= 10$^4$
• 0 <= index$_i$ < nums.length

Solution

/**
 * @param {number[]} nums
 * @param {number[][]} queries
 * @return {number[]}
 */
var sumEvenAfterQueries = function(nums, queries) {
  let result = [];
  let num;

  for (let i = 0; i < queries.length; i++) {
    nums[queries[i][1]] += queries[i][0]; 
    num = nums.reduce((prev, arr) => arr % 2 == 0 ? prev + arr : prev + 0, 0)
    result.push(num)
  }
  return result;
};