Tip 1 : Be very clear with the basics of each topic.
Tip 2 : Be thorough with the projects mentioned in the resume.
Tip 1 : Don't lie on your resume.
Tip 2 : The number of projects on the resume does not matter, what matters is that you must be thorough with your projects.
Tip 3 : Do not write too much, it should be simple and decent
The online assessment consisted of four components, a code debugging section (20 minutes), a coding test (70 minutes), a workstyles assessment (20 minutes) and a reasoning ability section (35 minutes). Code debugging questions were pretty simple and straightforward. The coding test consisted of 2 questions, first was similar to two Sum problem. Second question was similar to find the critical edges in a graph. The workstyles assessment section contained certain behavioural questions.
Reasoning ability section consisted of aptitude questions. This round was basically to check the problem solving skills of the candidate.



We cannot use the element at a given index twice.
Try to do this problem in O(N) time complexity.



If the given graph is :

Then the edge between 0 and 4 is the bridge because if the edge between 0 and 4 is removed, then there will be no path left to reach from 0 to 4.and makes the graph disconnected, and increases the number of connected components.
There are no self-loops(an edge connecting the vertex to itself) in the given graph.
There are no parallel edges i.e no two vertices are directly connected by more than 1 edge.
I was asked two coding questions. First question was related to binary tree and second question was to implement LRU cache.
I had to code both questions. The interviewer asked me the time and space complexity for both the questions.



A subtree of a node X is X, plus every node that is a descendant of X.
Look at the below example to see a Binary Tree pruning.
Input: [1, 1, 1, 0, 1, 0, 1, 0, 0, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1]
Output: [1, 1, 1, -1, 1, -1, 1, -1, -1, -1, -1]
For example, the input for the tree depicted in the below image would be :

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null (-1).
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1



1. get(key) - Return the value of the key if the key exists in the cache, otherwise return -1.
2. put(key, value), Insert the value in the cache if the key is not already present or update the value of the given key if the key is already present. When the cache reaches its capacity, it should invalidate the least recently used item before inserting the new item.
Type 0: for get(key) operation.
Type 1: for put(key, value) operation.
1. The cache is initialized with a capacity (the maximum number of unique keys it can hold at a time).
2. Access to an item or key is defined as a get or a put operation on the key. The least recently used key is the one with the oldest access time.
I was asked two coding questions and some conceptual questions about priority heap data structure. I was able to code the optimized approach for the first question. But second question was a little tricky for me as I had never heard it before.


There may be more than one Langford pair possible, you need to return anyone permutation.
For N = 4, one possible Langford pairing will be:-



1) If the theater is empty, i.e., if no one is in the theater, they sit in the first seat(seat number 0).
2) If there are multiple seats that maximize the distance from the closest person, they sit in the seat with the lowest number.

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