# https://leetcode.com/problems/sum-of-even-numbers-after-queries/
"""
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
"""


# brute force
def sum_even_after_queries(nums: list[int], queries: list[list[int]]) -> list[int]:
    def sum_even(nums: list[int]):
        return sum([v for v in nums if v % 2 == 0])

    ans = []
    for val, idx in queries:
        nums[idx] += val
        ans.append(sum_even(nums))

    return ans


def sum_even_after_queries(nums: list[int], queries: list[list[int]]) -> list[int]:
    ans = []
    cur_sum_even = sum([v for v in nums if v % 2 == 0])
    for val, idx in queries:
        old, new = nums[idx], nums[idx] + val
        if old % 2 == 0:
            cur_sum_even -= old
        if new % 2 == 0:
            cur_sum_even += new
        nums[idx] = new
        ans.append(cur_sum_even)

    return ans
