Accolite interview experience Real time questions & tips from candidates to crack your interview

SDE - 1

Accolite
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3 rounds | 3 Coding problems

Interview preparation journey

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Journey
It was an excellent session throughout the interview journey, during which I answered adequately and received an offer from Accolite Digital.
Application story
I got this opportunity on campus, which was only for girls, and it was a significant advantage for all of us. I answered the interview questions very well and was truthful in my explanation of the project. Don’t try to copy and paste your project if you haven’t completed it on your own.
Why selected/rejected for the role?
I was selected for the software development role because I answered the questions well, and the interviewer was very satisfied with all my responses. Additionally, if I didn't know the answer, I stated directly that I didn't know.
Preparation
Duration: 10 months
Topics: Data structure and algorithm, Operating System, DBMS, OOPS, Computer Network, Web Development, Cyber Security
Tip
Tip

Tip 1: Be very clear with your project explanation.
Tip 2: Try to cover all the CS core subjects explanation.

Application process
Where: Campus
Eligibility: 6.5 CGPA
Resume Tip
Resume tip

Tip 1: Make it clear by demonstrating appropriate knowledge.
Tip 2: Provide accurate information and avoid lying when you are unsure.

Interview rounds

01
Round
Easy
Video Call
Duration60 minutes
Interview date25 Aug 2021
Coding problem1

It was morning, around 9:00 a.m. The onboarding process took place via Google Meet and went very smoothly. The interviewer had an amicable nature.

1. Swap Nodes in Pairs

Moderate
40m average time
60% success
0/80
Asked in companies
Dell TechnologiesWalmartOLX Group

You are given a singly linked list of integers.

Your task is to swap every two adjacent nodes, and return the head of the modified, linked list.

For Example:

We have a linked list 1->2->3->4->5->6->7 and so on. You are supposed to swap pairs of a linked list like swap (1,2), (3,4), (5,6), and so on.
Note:
1. You may not modify the data in the list’s nodes; only nodes themselves may be changed. Because imagine a case where a node contains many fields, so there will be too much unnecessary swap.

2. If a pair of a node does not exist, then leave the node as it is.
Problem approach

class Node:

# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None


class LinkedList:

# Function to initialize head
def __init__(self):
self.head = None

# Function to pairwise swap elements of a linked list
def pairwiseSwap(self):
temp = self.head

# There are no nodes in linked list
if temp is None:
return

# Traverse furthethr only if there are at least two
# left
while(temp and temp.next):

# If both nodes are same,
# no need to swap data
if(temp.data != temp.next.data):

# Swap data of node with its next node's data
temp.data, temp.next.data = temp.next.data, temp.data

# Move temp by 2 to the next pair
temp = temp.next.next

# Function to insert a new node at the beginning
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node

# Utility function to print the linked LinkedList
def printList(self):
temp = self.head
while(temp):
print temp.data,
temp = temp.next


# Driver program
llist = LinkedList()
llist.push(5)
llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)

print "Linked list before calling pairWiseSwap() "
llist.printList()
llist.pairwiseSwap()

print "\nLinked list after calling pairWiseSwap()"
llist.printList

Try solving now
02
Round
Easy
Video Call
Duration60 minutes
Interview date2 Sep 2021
Coding problem1

It was the morning session again, and the interviewer was very supportive, providing hints to help me think through the approach. Finally, I came up with the solution he was expecting.

1. Closest Binary Search Tree Value

Easy
15m average time
85% success
0/40
Asked in companies
AdobeAccoliteJosh Technology Group

You have been given a binary search tree of integers with ‘N’ nodes and a target integer value ‘K’. Your task is to find the closest element to the target ‘K’ in the given binary search tree.

A node in BST is said to be the closest to the target if its absolute difference with the given target value ‘K’ is minimum. In the case of more than one closest element, return the element with a minimum value.

A binary search tree (BST) is a binary tree data structure with the following properties.

• The left subtree of a node contains only nodes with data less than the node’s data.

• The right subtree of a node contains only nodes with data greater than the node’s data.

• Both the left and right subtrees must also be binary search trees.
For Example:

example

For the given BST and target value ‘K’ =  32, the closest element is 30 as the absolute difference between 30 and 32 (|32 - 30|) is the minimum among all other possible node-target pairs.
Problem approach

# Recursive Python program to find key
# closest to k in given Binary Search Tree.

# Utility that allocates a new node with the
# given key and NULL left and right pointers.
class newnode:

# Constructor to create a new node
def __init__(self, data):
self.key = data
self.left = None
self.right = None

# Function to find node with minimum
# absolute difference with given K
# min_diff --> minimum difference till now
# min_diff_key --> node having minimum absolute
# difference with K
def maxDiffUtil(ptr, k, min_diff, min_diff_key):
if ptr == None:
return

# If k itself is present
if ptr.key == k:
min_diff_key[0] = k
return

# update min_diff and min_diff_key by
# checking current node value
if min_diff > abs(ptr.key - k):
min_diff = abs(ptr.key - k)
min_diff_key[0] = ptr.key

# if k is less than ptr->key then move
# in left subtree else in right subtree
if k < ptr.key:
maxDiffUtil(ptr.left, k, min_diff,
min_diff_key)
else:
maxDiffUtil(ptr.right, k, min_diff,
min_diff_key)

# Wrapper over maxDiffUtil()
def maxDiff(root, k):

# Initialize minimum difference
min_diff, min_diff_key = 999999999999, [-1]

# Find value of min_diff_key (Closest
# key in tree with k)
maxDiffUtil(root, k, min_diff, min_diff_key)

return min_diff_key[0]

# Driver Code
if __name__ == '__main__':
root = newnode(9)
root.left = newnode(4)
root.right = newnode(17)
root.left.left = newnode(3)
root.left.right = newnode(6)
root.left.right.left = newnode(5)
root.left.right.right = newnode(7)
root.right.right = newnode(22)
root.right.right.left = newnode(20)
k = 18
print(maxDiff(root, k))

Try solving now
03
Round
Easy
Face to Face
Duration60 minutes
Interview date7 Sep 2021
Coding problem1

It was late evening and happened in a very smooth way.

1. OS Questions

What is Kernel, and what are its main functions? (Learn)
 

Problem approach

The kernel is basically a computer program usually considered as a central component or module of OS. It is responsible for handling, managing, and controlling all operations of computer systems and hardware. Whenever the system starts, the kernel is loaded first and remains in the main memory. It also acts as an interface between user applications and hardware.

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