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 : Every skill must be mentioned.
Tip 3 : Focus on skills, projects and experiences more.
Technical interview round with questions on DSA mainly.




For the above-linked list, if k=2, then the value at the kth i.e second node from the end is ‘12’.
1.You don’t need to take any input. It has already been taken care of. Just implement the given function and return a pointer pointing to the k-th element from the last of the linked list.
2.It is guaranteed that k<=size of the linked list.
We can find the kth node from end very easily if we know the total number of nodes in the list.
We take a pointer ‘cur’ (initially pointing to head) and a variable ‘cnt’ to count total number of nodes.
We move the ‘cur’ pointer forward by one step until we reach NULL and increment the count variable by one in each move.
We will use the following property to find the kth node from the end.
Kth node from end = (cnt-k+1)th node from the beginning.
Let us store the value of cnt-k+1 in a variable ‘n’.
Now traverse the linked list again and return the pointer to the ‘nth’ node.



'S' = "{}()".
There is always an opening brace before a closing brace i.e. '{' before '}', '(' before ').
So the 'S' is Balanced.
Make use of the stack. Traverse the string and push the current character in the stack if it is an opening brace else pop from the stack If it is the corresponding starting brace for current closing brace then move to the next character of the string otherwise return false.
Technical interview round with questions on Java mainly.
Print even and odd number using two threads
JDBC connection establishment steps
Types of result set
Exception Handling
HashMap
HashSet
Hashcode
Equals
Multithreading

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Solving this problem will increase your chance to get selected in this company
What is recursion?