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
Solution:
The question is not so tricky. Before moving to solve the queries we will take the sum of all even numbers. Now we will move forward to solve the queries. If the index at which we need to perform the operation contains even number then we will subtract that number from our sum. After performing the operating if the index contain even number we will add it to the sum.
At last we will add the sum to our result vector. After performing on all the queries we will simply return our result vector.
At last we will add the sum to our result vector. After performing on all the queries we will simply return our result vector.
1 Comments
ReplyDelete-- CREATE TABLE STUDENT_TABLE ( SNO INT , ROLLNO INT, ENROLL INT , STUDENT_NAME varchar(50), COURSEID INT , EMAILID VARCHAR(50) , MOBILE varchar(10));
-- INSERT INTO STUDENT_TABLE VALUES (1 , 20115001 , 200561, "AADI REDHU" , 101 , "aadi@email.com" , "998587654");
-- INSERT INTO STUDENT_TABLE VALUES (2 , 20115002 , 200461, "AAYUSH" , 104 , "ayush@email.com" , "998581654");
-- INSERT INTO STUDENT_TABLE VALUES (3 , 20115003 , 200581, "ABHINAV" , 109 , "abhinav@email.com" , "978587654");
-- INSERT INTO STUDENT_TABLE VALUES (4 , 20115004 , 200961, "ajay" , 107 , "ajay@email.com" , "998587154");
-- INSERT INTO STUDENT_TABLE VALUES (5 , 20115005 , 200361, "raunit" , 102 , "raunit@email.com" , "998537654");
-- (i) ALTER TABLE STUDENT_TABLE ADD PRIMARY KEY(ENROLL);
-- (ii) SELECT LEFT(STUDENT_NAME, 2 ) FROM STUDENT_TABLE;
-- ALTER TABLE STUDENT_TABLE ADD SEMESTER INT;
-- CREATE TABLE COURSE_TABLE ( COURSEID INT , COURSE_NAME varchar(50), DURATION INT );
-- INSERT INTO COURSE_TABLE VALUES (101 , "OS" , 10);
-- INSERT INTO COURSE_TABLE VALUES (102 , "DBMS" , 17);
-- INSERT INTO COURSE_TABLE VALUES (104 , "CN" , 19 );
-- INSERT INTO COURSE_TABLE VALUES (109 , "DSA" ,14);
-- INSERT INTO COURSE_TABLE VALUES (107 , "ACA" , 12 );
-- (iii )SELECT STUDENT_TABLE.ROLLNO, STUDENT_TABLE.STUDENT_NAME, STUDENT_TABLE.SEMESTER, COURSE_TABLE.COURSE_NAME from STUDENT_TABLE RIGHT JOIN COURSE_TABLE ON STUDENT_TABLE.COURSEID= COURSE_TABLE.COURSEID;
-- (IV) select substring_index(EMAILID, '@' , 1) AS FIRSTNAME, substring_index(EMAILID, '@' , -1) AS DOMAIN FROM STUDENT_TABLE;
If you have any doubts/suggestion/any query or want to improve this article, you can comment down below and let me know. Will reply to you soon.