LeetCode in Kotlin

985. Sum of Even Numbers After Queries

Medium

You are given an integer array nums and an array queries where queries[i] = [vali, indexi].

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

Return an integer array answer where answer[i] is the answer to the ith 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 <= 104
• -104 <= nums[i] <= 104
• 1 <= queries.length <= 104
• -104 <= vali <= 104
• 0 <= indexi < nums.length

Solution

class Solution {
    fun sumEvenAfterQueries(nums: IntArray, queries: Array<IntArray>): IntArray {
        val result = IntArray(queries.size)
        var res = 0
        for (num in nums) {
            res += if (num and 1 == 0) num else 0
        }
        for ((k, query) in queries.withIndex()) {
            res -= if (nums[query[1]] and 1 == 0) nums[query[1]] else 0
            nums[query[1]] += query[0]
            if (nums[query[1]] and 1 == 0) {
                res += nums[query[1]]
            }
            result[k] = res
        }
        return result
    }
}

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