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

Software Engineer

Accolite
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3 rounds | 3 Coding problems

Interview preparation journey

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Journey
It was really wonderful session throughout the interview journey and in which I answered properly and hence i got the offer from Accolite Digital company . Also it was only for the girls in campus placement.
Application story
I got this opportunity via on campus and it was only for girls which was a great advantages for all of us. I answered very well in my interview and i was very truthful about my project explaination and don't try to copy paste your project if u haven't done it by your own.
Why selected/rejected for the role?
I got selected for the software development role because I gave my answer very well and the interviewer was very satisfied with all of my answers. Also if was not knowing the answer i told directly i don't know .
Preparation
Duration: 10 months
Topics: Data structure and algorithms, operating system, DBMS, oops, computer network, web development, CYBER SECURITY
Tip
Tip

Tip 1 : Be very clear with your project explanation.

Tip 2 : Also try to cover all the cs core subjects explanation clearly.

Application process
Where: Campus
Eligibility: 6.5 cgpa and 65% 12th marks minimun requirement
Resume Tip
Resume tip

Tip 1 : Make it clean with appropriate kowledge.
Tip 2 : Provide true information and avoid telling lie in which You are not sure.

Interview rounds

01
Round
Medium
Video Call
Duration60 minutes
Interview date23 Aug 2021
Coding problem1

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
Medium
Video Call
Duration60 minutes
Interview date26 Aug 2021
Coding problem1

It was again the morning session and the interviewer was very supportive in nature and gave me the hint too to think the approach and finally i came up the solution which he was expecting.

1. Minimum Difference

Moderate
20m average time
80% success
0/80
Asked in companies
AmazonSoft SuaveJupiter Money

Goku is fighting Jiren in “Tournament of Power”. If Goku loses then Universe 7 will get destroyed by Zeno but if Jiren loses then Universe 11 will get destroyed and Universe 7 will survive. Both Goku and Jiren launched ‘N’ attacks towards each other. Each attack has some energy. Jiren has a special ability through which he can absorb the attacks of Goku. If Goku launches an attack with energy ‘A’ and Jiren launches an attack with energy ‘B’ then

If ‘A’ > ‘B’, then Goku’s attack will destroy Jiren’s attack but Jiren will absorb Goku’s attack and absorb energy = ‘A’ – ‘B’.

If ‘A’ < ‘B’, then Jiren’s attack will destroy Goku’s attack and Jiren will absorb the remaining energy of his attack which is equal to = ‘B’ – ‘A’.

If ‘A’ == ‘B’, then both Goku’s and Jiren’s attack will get destroyed and Jiren will not absorb any energy.

Goku finds out the energy of all attacks of Jiren and the order in which Jiren is going to attack with the help of Whis. As Jiren becomes powerful by absorbing energy, Goku wants to minimize the total energy absorbed by Jiren by rearranging the order of his attacks.

As Goku is busy fighting with Jiren, he called you for help.

The fate of Universe 7 lies in your hand.

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
Medium
Video Call
Duration60 minutes
Interview date3 Sep 2021
Coding problem1

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

1. Operating System Question

What is Kernel and write its main functions?
 

Here's your problem of the day

Solving this problem will increase your chance to get selected in this company

Skill covered: Programming

How do you remove whitespace from the start of a string?

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