Paytm (One97 Communications Limited) interview experience Real time questions & tips from candidates to crack your interview

Software Developer

Paytm (One97 Communications Limited)

2 rounds | 4 Coding problems

Anonymous

Experienced | Oct 2021

Rejected

Interview preparation journey

Preparation

Duration: 6 months

Topics: Data structures, algorithms, oops, databases,Dynamic programming, recursion, backtracking, trees, graphs,SQL

Tip

Tip 1 : Prepare good for DS & Algo as most companies have a separate round for it.
Tip 2 : Don't lie over your resume
Tip 3 : Instead of doing bulk on weekends, prefer doing little everyday.Be consistent in coding.

Application process

Where: Naukri

Eligibility: good knowledge of DSA

Resume tip

Tip 1 : Only mention skills that you are good at.

Tip 2 : Have some projects on resume

Tip 3 : Try to add all your achievements in a crisp

Interview rounds

Round

Medium

Video Call

Duration60 mins

Interview date9 Oct 2021

Coding problem2

I was asked 2 DSA questions only.It was purely based on coding.Interviewer asked me approach first.

1. Rotated sorted array search

Easy

0/40

Asked in companies

You have been given an array of ‘N’ distinct integers which is sorted in ascending order and then rotated to the left by an unknown which you don’t know beforehand. For a given integer ‘X’, your task is to find the index of ‘X’ in the given array if it exists.

Please note that the sorted array A : [2, 3, 6, 8, 9, 11, 15] might become [6, 8, 9, 11, 15, 2, 3] after rotating it twice to the left.

Problem approach

I applied binary search approach.Interviewer told it is correct and start coding.Time Complexity: O(logN), using binary search
Auxiliary Space: O(1)
The minimum element is the only element whose previous is greater than it. If there is no previous element, then there is no rotation (the first element is minimum).
We check this condition for the middle element by comparing it with (mid-1)th and (mid+1)th elements.
If the minimum element is not at the middle (neither mid nor mid + 1), then:
If the middle element is smaller than the last element, then the minimum element lies in the left half
Else minimum element lies in the right half.

Try solving now

2. Right View

Moderate

35m average time

65% success

0/80

Asked in companies

You have been given a Binary Tree of integers.

Your task is to print the Right view of it.

The right view of a Binary Tree is a set of nodes visible when the tree is viewed from the Right side and the nodes are printed from top to bottom order.

Problem approach

Right View of a Binary Tree using Level Order Traversal:
The idea is to use Level Order Traversal as the last node every level gives the right view of the binary tree.

Follow the steps below to solve the problem:

Perform level order traversal on the tree
At every level print the last node of that level

Try solving now

Round

Medium

Video Call

Duration60 minutes

Interview date9 Oct 2021

Coding problem2

It was based on coding ,DSA round.Interviewer was helpful.Interview took place in afternoon.2 medium level,standard DSA questions were asked.

1. Smallest string

Easy

15m average time

85% success

0/40

Asked in companies

You are given two integers, ‘N’ and ‘V’. You need to construct a string having length ‘N’ and numeric value ‘V’.

The numeric value of a character is determined by its position in alphabets( 1- indexed)

For Example-

The numeric value of ‘a’ is 1, ‘b’ is 2.

The numeric value of a string is determined by adding the numeric value of all its characters.

For Example -

The numeric value of “bde” is 2 + 4 + 5, i.e. 11.

Example:

If 'N' = 3 and 'V' = 10, you need to construct string as “aah".

Problem approach

Initialize a string res to store the resultant string.
Store the frequency of each character present in the given string in an array freq[].
Maintain an array vis[] for marking the characters that are already present in the resultant string res.
Traverse the given string S and for each character S[i], perform the following:
Decrease the frequency of the current character by 1.
If the current character is not marked visited in the array vis[], then perform the following:
If the last letter of res is less than S[i], add S[i] to res.
If the last letter of res is greater than S[i] and the count of the last letter of res exceeds 0, then remove that character and mark visit[S[i]] as 0 and continue this step till the above condition is satisfied.
After breaking out from the above condition, add S[i] to res and mark visit[S[i]] as 1.
After completing the above steps, print the string res as the result.

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2. Reverse Linked List

Moderate

15m average time

85% success

0/80

Asked in companies

Given a singly linked list of integers. Your task is to return the head of the reversed linked list.

For example:

The given linked list is 1 -> 2 -> 3 -> 4-> NULL. Then the reverse linked list is 4 -> 3 -> 2 -> 1 -> NULL and the head of the reversed linked list will be 4.

Follow Up :

Can you solve this problem in O(N) time and O(1) space complexity?

Problem approach

I reversed using stack.
Store the nodes(values and address) in the stack until all the values are entered.
Once all entries are done, Update the Head pointer to the last location(i.e the last value).
Start popping the nodes(value and address) and store them in the same order until the stack is empty.
Update the next pointer of last Node in the stack by NULL.

Try solving now