Tip 1 : Practice DSA questions
Tip 2 : Prepare one project on microservices
Tip 3 : Practice Java programs
Tip 1 : To the point detail, not very long description of projects
Tip 2 : No more then 2 pages
Online round was having 2 DSA problems, that need to be solved in 1 hour only.



Suppose given input is "abacb", then the length of the longest substring without repeating characters will be 3 ("acb").



1. There might be duplicates present in the array.
2. The order of the permutations in the output does not matter.
3. Do not use any kind of in-built library functions to find the answer.
Once I cleared the first online test exam, then I was eligible for second round. In this round they were focused on DSA problems and High level design of one project. The duration of this round in 60 mins.



If ‘S’ = “beaninjacoder” and ‘ROW’ = 4
Then the zig-zag pattern is:
b j r
e n a e
a i c d
n o
Therefore, we will print “bjrenaeaicdno”.
This was also the same as second round. for one hour and some DSA and High level design questions.



Input:
'num1' : 1 -> 2 -> 3 -> NULL
'num2' : 4 -> 5 -> 6 -> NULL
Output: 5 -> 7 -> 9 -> NULL
Explanation: 'num1' represents the number 321 and 'num2' represents 654. Their sum is 975.
Just like how you would sum two numbers on a piece of paper, we begin by summing the least-significant digits, which is the head of l1l1 and l2l2. Since each digit is in the range of 0 \ldots 90…9, summing two digits may "overflow". For example 5 + 7 = 125+7=12. In this case, we set the current digit to 22 and bring over the carry = 1carry=1 to the next iteration. carrycarry must be either 00 or 11 because the largest possible sum of two digits (including the carry) is 9 + 9 + 1 = 199+9+1=19.
The pseudocode is as following:
Initialize current node to dummy head of the returning list.
Initialize carry to 00.
Loop through lists l1l1 and l2l2 until you reach both ends and crarry is 00.
Set xx to node l1l1's value. If l1l1 has reached the end of l1l1, set to 00.
Set yy to node l2l2's value. If l2l2 has reached the end of l2l2, set to 00.
Set sum = x + y + carrysum=x+y+carry.
Update carry = sum / 10carry=sum/10.
Create a new node with the digit value of (sum \bmod 10)(summod10) and set it to current node's next, then advance current node to next.
Advance both l1l1 and l2l2.
Return dummy head's next node.
Note that we use a dummy head to simplify the code. Without a dummy head, you would have to write extra conditional statements to initialize the head's value.
This was also same as third round, for one hour. Some DSA problems asked here.



Symbol Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000
3 is written as III in Roman numeral, just three ones added together. 13 is written as XIII, which is simply X + III. The number 25 is written as XXV, which is XX + V
Steps:
Create a array of all integers for which we have a distinct roman literal, and a matching array with their respetive roman literal
Run a loop for iterate through all integer value from previously created integer array
Inside this for loop, give a while loop to subtract the input number from array number until it becomes lesser than the array number, append the respective roman literal to a string buffer
This was a managerial round. It went for and hour and two people were there in the panel. They were asking the process and some system design problems.
We need to design one application were we have lot of data. And we need to handle the user concurrency. What is the way we can do that?
Tip 1 : First understand the problem first.
Tip 2 : Relate this problem to real life problem.
Tip 3 : Answer it wisely.
This was HR round, where they only asked about your current and expected salary.
What is your current and expected CTC.
What are your hobbies and talents?
Tip 1 : You can tell the current CTC.
Tip 2 : You can ask CTC according to company standard.
Tip 3 : You can ask for company's share also.

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
To make an AI less repetitive in a long paragraph, you should increase: