Algorithm
Problem Name: 985. 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] <= 1041 <= queries.length <= 104-104 <= vali <= 1040 <= indexi < nums.length
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int[] sumEvenAfterQueries(int[] nums, int[][] queries) {
int evenSum = 0;
for (int num : nums) {
evenSum += num % 2 == 0 ? num : 0;
}
int[] result = new int[nums.length];
for (int i = 0; i < queries.length; i++) {
int idx = queries[i][1];
if (nums[idx] % 2 == 0) {
evenSum -= nums[idx];
}
nums[idx] += queries[i][0];
if (nums[idx] % 2 == 0) {
evenSum += nums[idx];
}
result[i] = evenSum;
}
return result;
}
}
Input
nums = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output
[8,6,2,4]
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const sumEvenAfterQueries = function(A, queries) {
let sum = A.reduce((acc, cur) => cur%2 == 0 ? acc + cur : acc, 0);
return queries.map((q) => {
let i = q[1];
let s = A[i] + q[0];
if(s%2 === 0) {
sum += q[0];
if(A[i]%2 !== 0) sum += A[i];
} else if(A[i]%2 === 0) sum -= A[i];
A[i] = s;
return sum;
});
};
Input
nums = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output
[8,6,2,4]
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def sumEvenAfterQueries(self, A: List[int], queries: List[List[int]]) -> List[int]:
sm = sum(a for a in A if a % 2 == 0)
for i in range(len(queries)):
val, ind = queries[i]
sm -= A[ind] % 2 == 0 and A[ind]
A[ind] += val
sm += A[ind] % 2 == 0 and A[ind]
queries[i] = sm
return queries
Input
nums = [1], queries = [[4,0]]
Output
[0]
#4 Code Example with C# Programming
Code -
C# Programming
using System.Linq;
namespace LeetCode
{
public class _0985_SumOfEvenNumbersAfterQueries
{
public int[] SumEvenAfterQueries(int[] A, int[][] queries)
{
var sum = A.Where(num => num % 2 == 0).Sum();
var result = new int[queries.Length];
for (int i = 0; i < queries.Length; i++)
{
int val = queries[i][0], index = queries[i][1];
if (A[index] % 2 == 0)
sum -= A[index];
A[index] += val;
if (A[index] % 2 == 0)
sum += A[index];
result[i] = sum;
}
return result;
}
}
}
Input
nums = [1], queries = [[4,0]]
Output
[0]