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985. Sum of Even Numbers After Queries

Description

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

Solutions

  • class Solution {
    public int[] sumEvenAfterQueries(int[] nums, int[][] queries) {
        int s = 0;
        for (int x : nums) {
            if (x % 2 == 0) {
                s += x;
            }
        }
        int m = queries.length;
        int[] ans = new int[m];
        int k = 0;
        for (var q : queries) {
            int v = q[0], i = q[1];
            if (nums[i] % 2 == 0) {
                s -= nums[i];
            }
            nums[i] += v;
            if (nums[i] % 2 == 0) {
                s += nums[i];
            }
            ans[k++] = s;
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<int> sumEvenAfterQueries(vector<int>& nums, vector<vector<int>>& queries) {
        int s = 0;
        for (int x : nums) {
            if (x % 2 == 0) {
                s += x;
            }
        }
        vector<int> ans;
        for (auto& q : queries) {
            int v = q[0], i = q[1];
            if (nums[i] % 2 == 0) {
                s -= nums[i];
            }
            nums[i] += v;
            if (nums[i] % 2 == 0) {
                s += nums[i];
            }
            ans.push_back(s);
        }
        return ans;
    }
};

  • class Solution:
    def sumEvenAfterQueries(
        self, nums: List[int], queries: List[List[int]]
    ) -> List[int]:
        s = sum(x for x in nums if x % 2 == 0)
        ans = []
        for v, i in queries:
            if nums[i] % 2 == 0:
                s -= nums[i]
            nums[i] += v
            if nums[i] % 2 == 0:
                s += nums[i]
            ans.append(s)
        return ans


  • func sumEvenAfterQueries(nums []int, queries [][]int) (ans []int) {
	s := 0
	for _, x := range nums {
		if x%2 == 0 {
			s += x
		}
	}
	for _, q := range queries {
		v, i := q[0], q[1]
		if nums[i]%2 == 0 {
			s -= nums[i]
		}
		nums[i] += v
		if nums[i]%2 == 0 {
			s += nums[i]
		}
		ans = append(ans, s)
	}
	return
}

  • function sumEvenAfterQueries(nums: number[], queries: number[][]): number[] {
    let s = 0;
    for (const x of nums) {
        if (x % 2 === 0) {
            s += x;
        }
    }
    const ans: number[] = [];
    for (const [v, i] of queries) {
        if (nums[i] % 2 === 0) {
            s -= nums[i];
        }
        nums[i] += v;
        if (nums[i] % 2 === 0) {
            s += nums[i];
        }
        ans.push(s);
    }
    return ans;
}


  • /**
 * @param {number[]} nums
 * @param {number[][]} queries
 * @return {number[]}
 */
var sumEvenAfterQueries = function (nums, queries) {
    let s = 0;
    for (const x of nums) {
        if (x % 2 === 0) {
            s += x;
        }
    }
    const ans = [];
    for (const [v, i] of queries) {
        if (nums[i] % 2 === 0) {
            s -= nums[i];
        }
        nums[i] += v;
        if (nums[i] % 2 === 0) {
            s += nums[i];
        }
        ans.push(s);
    }
    return ans;
};

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