Tip 1 : Revise data structures and algorithms from time to time
Tip 1 : Keep content of your resume as relevant and specific to the profile to which you are applying as possible.
The round consisted of 3 sections, Aptitude (28 mins), Technical Section (17 mins), and a Coding Section (two questions, one for 20 mins and the other for 30 mins). There was negative marking. An incorrect answer carried deduction of 33% of the total marks of that question. There was a time window of 12 hours within which the test was to be taken.



1. Declare a 3D array with each dimension being equal to the length of one of the strings.
2. Fill the 3D array using tabulation, based on the matching characters.

Input: 'X' = 2, ‘Y’=3
Output: "6"
As “6” is the smallest number that is divisible by both 2 and 3.
1. Define a function to calculate HCF of two numbers.
2. Find the LCM of the first two numbers using the relation, LCM(a, b) = a*b / HCF(a, b).
3. Pass the calculated LCM in the above step and the third number to the HCF function, LCM(a, b, c) = LCM(a, b) * c / HCF(LCM(a, b), c).
I was asked to suggest the day and time of the interview by the recruiter. I had to postpone the initial date of interview that I had suggested to which the recruiting team was accommodating. The interview took place on HackerRank where there was a collaborative code editor with video chat.



I gave two approaches — the naïve approach is to sort and traverse the array and check for the missing number. A better approach would be to create a hash map containing elements from 0 to n, where n is the length of the array (the smallest missing number would be at most equal to the length of the array). Traverse the array and increment the count of each element that is encountered. Traverse the hash map, the first element to have its value as 0 would be the answer.
You are going from point A to point B and encounter 2 red lights but stop at one of them. While returning, i.e. going from point B to point A through the same route, you stop twice. Draw the route for such a scenario to be legally possible.
Tip 1 : It can be solved on the idea of free left turn. One need not stop when they want to turn left even if the traffic signal is red.
Tip 2 : So, while going from point A to point B, the person had to turn left, so they did not stop, but while going from B to A, there would be a right turn for which they would have to stop.



Input : 1 -> 2 -> 3 -> 4 -> 'NULL' and 'K' = 2
Output: 1 -> 2 -> 4 -> 'NULL'
Explanation:
After removing the second node from the end, the linked list become 1 -> 2 -> 4 -> 'NULL'.

1. Traverse the linked list to find the length of the linked list and then traverse again to visit the kth node.
2. Define a temp pointer equal to head.
3. Move the head pointer from (k-1) to (k+1)th node.
4. Move the head pointer to kth node and free it.
Another way to solve it without the use of a head pointer would be to replace the data for every node, starting from the kth node.

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Which type of comments is used to comment on single lines of Java code?