985. Sum of Even Numbers After Queries | leetcode solution

 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] <= 104
• 1 <= queries.length <= 104
• -104 <= vali <= 104
• 0 <= indexi < nums.length

solution:

class Solution {

public:

    vector<int> sumEvenAfterQueries(vector<int>& nums, vector<vector<int>>& queries) {

         

         vector<int>ans;

          int  sums_nums =0;

        for(int j = 0; j < nums.size() ; j++)

            {   

              if(nums[j] %2 == 0)

              sums_nums += nums[j];

            }

         for(int i = 0 ; i < nums.size() ; i++)

         {    

              int index = queries[i][1];

                int t = nums[index] ;

             nums[index] = nums[index] + queries[i][0];

                

              if( nums[index] %2 == 0)

              {   

                  if(t %2 == 0)

                  sums_nums = sums_nums - t + nums[index];

                  else

                     sums_nums = sums_nums +nums[index];

              }

             else

             {

                 if(t %2 == 0)

                     sums_nums = sums_nums - t;

             }

             

             

             ans.push_back(sums_nums);

             

         }

        return ans;

    }

};