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Software Developer

RedBus
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2 rounds | 4 Coding problems

Interview preparation journey

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Journey
I started preparation from my first year of the college itself. Then i started practicing the DSA at the regular basic from the coding Ninjas platform CodeStudio
Application story
This is a on-campus opportunity this company visited our campus for the selection where i have applied
Why selected/rejected for the role?
I was rejected for this role because I am not able to give a optimize solution for my one of the questions
Preparation
Duration: 3
Topics: Data Structures, Pointers, OOPS, System Design, Algorithms, Dynamic Programming
Tip
Tip

Tip 1 - Practice at Atleast 250 Questions
Tip 2 - Ex- Do at least 2 projects

Application process
Where: Campus
Eligibility: Above 7 CGPA
Resume Tip
Resume tip

Tip 1 : Have some projects on resume.
Tip 2 : Do not put false things on resume.

Interview rounds

01
Round
Hard
Video Call
Duration60 mins
Interview date29 Sep 2022
Coding problem2

1. Maximum Coins

Hard
16m average time
78% success
0/120
Asked in companies
ZSPayPalSamsung

You are given a two-dimensional matrix of integers of dimensions N*M, where each cell represents the number of coins in that cell. Alice and Bob have to collect the maximum number of coins. The followings are the conditions to collect coins:

Alice starts from top left corner, i.e., (0, 0) and should reach left bottom corner, i.e., (N-1, 0). Bob starts from top right corner, i.e., (0, M-1) and should reach bottom right corner, i.e., (N-1, M-1).

From a point (i, j), Alice and Bob can move to (i+1, j+1) or (i+1, j-1) or (i+1, j)

They have to collect all the coins that are present at a cell. If Alice has already collected coins of a cell, then Bob gets no coins if goes through that cell again.

For example :
If the matrix is 
0 2 4 1
4 8 3 7
2 3 6 2
9 7 8 3
1 5 9 4

Then answer is 47. As, Alice will collect coins 0+8+3+9+1 = 21 coins. Bob will collect coins 1+7+6+8+4 = 26 coins. Total coins is 21+26 = 47 coins.
Problem approach

Solved through recursion first.
Applied DP next.

Try solving now

2. Reverse List In K Groups

Hard
15m average time
85% success
0/120
Asked in companies
SAP LabsSamsungIBM

You are given a linked list of 'n' nodes and an integer 'k', where 'k' is less than or equal to 'n'.


Your task is to reverse the order of each group of 'k' consecutive nodes, if 'n' is not divisible by 'k', then the last group of nodes should remain unchanged.


For example, if the linked list is 1->2->3->4->5, and 'k' is 3, we have to reverse the first three elements, and leave the last two elements unchanged. Thus, the final linked list being 3->2->1->4->5.


Implement a function that performs this reversal, and returns the head of the modified linked list.


Example:
Input: 'list' = [1, 2, 3, 4], 'k' = 2

Output: 2 1 4 3

Explanation:
We have to reverse the given list 'k' at a time, which is 2 in this case. So we reverse the first 2 elements then the next 2 elements, giving us 2->1->4->3.


Note:
All the node values will be distinct.


Problem approach

You are given a linked list of 'N' nodes and an integer 'K'. You have to reverse the given linked list in groups of size K i.e if the list contains x nodes numbered from 1 to x, then you need to reverse each of the groups (1,K),(K+1,2*K), and so on.

Try solving now
02
Round
Hard
Video Call
Duration60 mins
Interview date29 Sep 2022
Coding problem2

1. Paint House

Hard
20m average time
80% success
0/120
Asked in companies
AppleAmazonSamsung

You have been given ‘N’ houses, each house can be painted with any of three colours: green, red and yellow. You are also given a “cost” matrix of ‘N’ * 3 dimension which represents the cost of painting an i-th house (0-th based indexing) with j-th colour. The colour code is as follows: green - 0, red - 1 and yellow - 2. Now, you are supposed to find the minimum cost of painting all houses such that no adjacent houses are painted with the same colour.

Problem approach

You have been given ‘N’ houses, each house can be painted with any of three colours: green, red and yellow. You are also given a “cost” matrix of ‘N’ * 3 dimension which represents the cost of painting an i-th house (0-th based indexing) with j-th colour. The colour code is as follows: green - 0, red - 1 and yellow - 2. Now, you are supposed to find the minimum cost of painting all houses such that no adjacent houses are painted with the same colour.

Try solving now

2. DBMS Questions

What do you mean by Acid Properties?

What is normalisation?

What is Data abstraction?

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?

Choose another skill to practice
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