eGain Communications Software Developer Interview Questions and Answers

Q1. Oops? .net core features
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1st round is coding assessment round based on graph and hashmaps

There were 5 question out of which 2 were coding Data Structure problems , 1 was SQL based , 1 was AI based Problem ( Machine Learning ) and 1 was to create UML Diagram.
First Coding question was pretty easy , other one was kinda tough , UML Diagram was also easy and SQL query was of moderate level.

1 hr and it is pretty easy and having snippets technical questions etc

I didn't selected for second round

Q1. String palindrome
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Q2. Find all anagrams in a string
Ans. 
Find all anagrams in a string
• Create a hashmap to store the frequency of characters in the input string

• Iterate through the input string and check if the frequency of characters matches the hashmap for each substring of the same length

• Store the substrings that are anagrams in an array and return it
Answered by AI
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Q1. Explain basics of c#
Ans. 
C# is a programming language developed by Microsoft for building a wide range of applications on the .NET framework.
• C# is an object-oriented language with features like classes, inheritance, and polymorphism.

• It is strongly typed, meaning variables must be declared with a specific data type.

• C# supports modern programming concepts like async/await for asynchronous programming.

• It is commonly used for developing desktop, w...read more
Answered by AI
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Q2. C# basics oops sql server
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Round duration - 120 minutes
Round difficulty - Hard

This round was very difficult. I never do these much difficulty level of coding questions. But i give my best, and just do every questions by using step by step process.

Q1. Hourglass Pattern
You are given an integer N and you have to print the following pattern.
For N=5:
For N=6:

Input Format:
The first and only line of input contains an integer, that denotes the va...read more
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Q2. BFS in Graph
You are given an undirected and disconnected graph G(V, E) having V vertices numbered from 0 to V-1 and E edges. Your task is to print its BFS traversal starting from the 0th vertex.
BFS or ...read more
Ans. Using queue
Let us consider a function BFS that accepts a list of edges, EDGES, and the number of vertices VERTEX as a parameter and do: 
1. Create ADJACENCYMATRIX from EDGES.
2. Define a boolean array VISITED to store nodes that are already visited.
3. Iterate over VISITED[i] for each 0<= i < VISITED.LENGTH, and check:
   1. If VISITED[i] is FALSE, call BFSHELPER function for ADJACENCYMATRIX, i and VISITED.
Now, the function BF...read more
Answered by CodingNinjas
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Q3. Sort a stack
Given a stack, sort the elements inside that stack in ascending order using only push and pop operation. You can use one additional stack only.
For eg.
Input Stack :  5 (top)
            ...read more
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Round duration - 90 minutes
Round difficulty - Easy

This round is also coding round in this round we have a small video call interaction too, in which hr ask few questions to only those whose coding question is correct.

Q1. Next smaller Palindrome
You are given a number 'N' in the form of a string 'S', which is a palindrome. You need to find the greatest number strictly less than 'N' which is also a pa...read more
Ans. Brute Force
• Given the fact that the input number is a palindrome itself makes this problem very trivial.
• Let us assume that the given string str is a palindrome, so we know that ‘S’ comprises of two halves  ‘S₁’ and ‘S₂’ i.e ‘S’ = ‘S₁' + ‘S₂', where ‘S₁' and ‘S₂’ are two strings, and ‘S₂’ is the reverse of ‘S₁’(in case of odd length palindromes ‘S₂’ won’t have the last character of ‘S₁’). So a change in any of the h...read more
Answered by CodingNinjas
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Q2. Rotting Oranges
You have been given a grid containing some oranges. Each cell of this grid has one of the three integers values:
Value 0 - representing an empty cell.
Value 1 - representing a fresh...read more
Ans. Naïve Solution
The idea is very simple and naive. We will process the rotten oranges second by second. Each second, we rot all the fresh oranges that are adjacent to the already rotten oranges. The time by which there are no rotten oranges left to process will be our minimum time.
 
In the first traversal of the grid, we will process all the cells with value 2 (rotten oranges). We will also mark their adjacent cells ...read more
Answered by CodingNinjas
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