Tip 1: Must do Previously asked Interviews as well as Online Test Questions.
Tip 2: Must have good knowledge of DSA
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.
Those who cleared the Backend Assignment(Backend for a shopping site) for CityMall were called for interview. The interview asked pretty good questions for testing knowledge of coding. lots of questions were asked on projects, resume, Assignment.



An irreducible fraction is a fraction in which the numerator and denominator are integers that have no other common divisors than 1.
Example: 2/3, 8/9, 0/1 are irreducible fractions, where 8/12, 9/12 are not irreducible fractions.
Ayush expects your answer to be an irreducible fraction, so if your answer is an integer, convert it into an irreducible fraction by putting 1 in the denominator.
For example, if your answer is "9", then print "9/1" (without quotes).



You are given ‘WORDS’ = [“cat”, “mat”, “tac”]. Then the answer will be [(0, 2), (2, 0)}, because “cat” + “tac” = “cattac” which is a palindrome and “tac” + “cat” = “taccat” which is also a palindrome.
Build a HashMap to store the String-idx pairs.
Case1: If s1 is a blank string, then for any string that is palindrome s2, s1+s2 and s2+s1 are palindrome.
Case 2: If s2 is the reversing string of s1, then s1+s2 and s2+s1 are palindrome.
Case 3: If s1[0:cut] is palindrome and there exists s2 is the reversing string of s1[cut+1:] , then s2+s1 is palindrome.
Case 4: Similiar to case3. If s1[cut+1: ] is palindrome and there exists s2 is the reversing string of s1[0:cut] , then s1+s2 is palindrome.
This round was more of a hiring manager round where my backend knowledge is tested and question on my my past project tech stack was asked followed by an easy coding question.



The Linked Lists, where a1, a2, c1, c2, c3 is the first linked list and b1, b2, b3, c1, c2, c3 is the second linked list, merging at node c1.

In the first iteration, we will reset the pointer of one linkedlist to the head of another linkedlist after it reaches the tail node. In the second iteration, we will move two pointers until they points to the same node. Our operations in first iteration will help us counteract the difference. So if two linkedlist intersects, the meeting point in second iteration must be the intersection point. If the two linked lists have no intersection at all, then the meeting pointer in second iteration must be the tail node of both lists, which is null

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