Tip 1 : Practice at least 350 questions to have good hold on data structures
Tip 2 : Practice at least 100 - 150 company specific coding questions from geeksforgeeks
Tip 3 : Participates in online contest to solve questions within time bound
Tip 1 : Have at least one project in the resume.
Tip 2 : Must mention every skill and certificate in the resume.
Tip 3 : Must have known every strong and intermediate skill mentioned in resume.
This was online coding test on the hackerrank platform. The test consists of three coding questions



Consider the array be 1,6,4,6,2,4,2
The integers 2, 4, 6 occur twice and only the integer 1 occurs once.
Solution using Python
class Solution:
def singleNumber(self, nums: List[int]) -> int:
nums.sort()
memo = {}
for count, num in enumerate(nums):
if num not in memo:
# insert unique num
# into hash map
memo[num] = num
else:
# remove if duplicate
del memo[num]
# the last remaining one must be the answer
return list(memo.keys())[0]



Solution using python
# intution: we can't close a paranthesis without opening it so whenever we are closing a parathesis its counter part should already be opened at some
# point so we carry the count of number of parathesis opened and closed on every recursion. once opened parathesis reaches point n the string can't have
# any more opening parathesis it can only have closing paranthesis from then on. if opening paranthesis is greater than closed paranthesis and opening
# hasn't reached its max limit we can form both opening and closing paranthesis as next character. if opened is not greater than closed then we have to
# open a paranthesis as we cant close without its counter open already existing.
def generateParenthesis(self, n: int) -> List[str]:
def para(opened,closed,l,k,n):
if closed>n:
l.append(k)
return l
if opened<=n:
if opened>closed:
l=para(opened+1,closed,l,k+'(',n)
l=para(opened,closed+1,l,k+')',n)
else:
l=para(opened+1,closed,l,k+'(',n)
else:
l=para(opened,closed+1,l,k+')',n)
return l
opened=1
closed=1
k=''
l=[]
l=para(opened,closed,l,k,n)
return l



A majority element is an element that occurs more than floor('N' / 2) times in the array.



An array ‘B’ is a subarray of an array ‘A’ if ‘B’ that can be obtained by deletion of, several elements(possibly none) from the start of ‘A’ and several elements(possibly none) from the end of ‘A’.



Consider ‘arr’ = [1,2,-2,-1], the largest value of ‘K’ is 2, since a negative of 2 is also present in the array. Hence, the answer is 2.



‘num’ does not have leading zeros except when ‘num’ equals zero.
Input: ‘num’ = ‘141’ , ‘k’ = 1.
Output: ‘11’
Explanation: By removing only 1 digit from ‘num’, 3 numbers can be formed: 14, 11, and 41. Out of which 11 is the smallest number.
You don’t have to print anything. It has already been taken care of. Just implement the given function.
This was 2nd technical round taken by the senior technical manager from Patym. He started with his own introduction and then asked me to introduce myself and then alot of questions related to my skills and projects.
Introduction
What are your hobbies
Explain your projects in detail ?
How many users your project can handle at a time
Low Level design of the facebook
What are Higher-Order Components (HOC) in React?
What are the differences between a Class component and Functional component?
What is Callback Hell and what is the main cause of it?
How to avoid Callback Hell in Node.js?
What do you know about Asynchronous API?

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
To make an AI less repetitive in a long paragraph, you should increase: