Hexaware Technologies Software Developer Interview Questions, Process, and Tips

Round duration - 60 minutes
Round difficulty - Easy

Q1. Can you explain Bipolar Junction Transistors (BJTs)?
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Q2. 
Graph Coloring Problem
You are given a graph with 'N' vertices numbered from '1' to 'N' and 'M' edges. Your task is to color this graph using two colors, such as blue and red, in a way that no two adjacen...read more
Ans. DFS
We can approach this problem by running a DFS starting from vertex-1.
We have a total of ‘m’ edges.
When can we achieve the configuration when we see in reference to vertex-1? 
1. Let's say vertex-1 is ‘Blue’
   1. If all the vertices that connect to vertex-1 are ‘Red’, then we can say that the configuration is possible.
2. If this would be the negation of the last one. Say, vertex-1 is ‘Red’
   1. If all the vertices that connect to ...read more
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Round duration - 30 minutes
Round difficulty - Easy

Q1. Can you explain in detail about Stacks and Graphs?
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Q2. 
Rat in a Maze: All Paths Problem
You are provided with an N * N maze where a rat is positioned at starting cell MAZE[0][0]. The goal is to determine and print all possible paths that the rat can take to r...read more
Ans. Backtracking Approach
Initialize, all the cells of the solution matrix used to print the path matrix to 0. First, you cannot make use of the existing maze to print the solution maze as you have to distinguish b/w 1 of maze or 1 of ‘SOLUTION matrix.
 
Form a recursive function, which will follow a path and check if the path reaches the destination or not. If the path does not reach the destination then backtrack and t...read more
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Round duration - 50 Minutes
Round difficulty - Medium

first round was aptitude round + coding round.

Q1. 
Ninja Competition Problem Statement
Ninja is organizing a coding competition where two teams compete at a time. To keep it fair and interesting, both teams must have an equal number of members. Ninja’s ta...read more
Ans. Brute Force
We will iterate from 1 to ‘N’ and count the number of even divisors and odd divisors. If they are equal, we will return true, else false.
 
Algorithm:
 
1. Declare two variables, ‘Odd_Count’ and ‘Even_Count’, and initialize them with zero.
2. Start iterating from 1 to ‘N’.
3. If the current iterator ( say i ) is a divisor of ‘N’ ( N % i == 0 ):
   1. If ‘i’ is odd, increase the ‘Odd_Count’ by 1.
   2. If ‘i’ is even, increase ...read more
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