Tip 1 : Must do Previously asked Interview as well as Online Test Questions.
Tip 2 : Go through all the previous interview experiences from Codestudio and Leetcode.
Tip 3 : Do at-least 2 good projects and you must know every bit of them.
Tip 1 : Have at-least 2 good projects explained in short with all important points covered.
Tip 2 : Focus on skills, projects and experiences more.
Group discussion conducted, they give one topic and give 2 min to prepare. You should speak with fluent English. They judge your communication and pronunciation.
Topic : Talk about the situation of the city after it rains.
I cleared the round.
This was the second round. It was a technical round with questions based on DSA.



String 'S' is NOT case sensitive.
Let S = “c1 O$d@eeD o1c”.
If we ignore the special characters, whitespaces and convert all uppercase letters to lowercase, we get S = “c1odeedo1c”, which is a palindrome. Hence, the given string is also a palindrome.
A simple and intuitive approach could be to compare the given string with its reversed form and check whether they are equal or not. Ignoring the symbols, whitespaces and case during the comparison.



153 = 1^3 + 5^3 + 3^3.
Therefore 153 is an Armstrong number.
The first step will be to find the length of the number.
Initialize an integer k = 0, it will be used to find the length of the number.
Initialise an integer sum = 0.
Run a while loop till num is not equal to 0 (to calculate the length of number):
Increment k by 1.
Run a while loop till num is not equal to 0:
The current digit will be num%10.
Add (current digit) ^ k into sum.
Make num = num/10.
Check if ‘sum’ is equal to the initial number.

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
What is recursion?