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I applied via Naukri.com and was interviewed in Jan 2024. There was 1 interview round.
Java is a popular programming language used for developing software applications.
Java is platform-independent, meaning it can run on any operating system.
It has a large standard library with built-in functions and classes for common tasks.
Java supports object-oriented programming, making it easier to organize and reuse code.
It provides automatic memory management through garbage collection.
Java is widely used in enterp...
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No aptitude test taken
I applied via Company Website
I applied via Company Website and was interviewed in Oct 2022. There were 4 interview rounds.
GD was about crises in India . There were 4-5 people in this round even I have participated as well. I did my level best. India as for most ancient country. Past india was the only country to find diamond. Now we hardly could not even find gold. This as an example we can conclude that India is in crises
This aptitude test was totally about general intelligence and mental fitness. We usually have a minute of time to solve
I applied via Campus Placement and was interviewed in Dec 2022. There were 4 interview rounds.
Some basic aptitude questions and technical questions basics only
Easy to medium question on data structures and algorithms
posted on 10 Aug 2023
I applied via Campus Placement and was interviewed before Aug 2022. There were 2 interview rounds.
Constructor creates and initializes an object, while method performs an action on an object.
Constructor has the same name as the class and is called when an object is created
Method has a name that describes the action it performs on an object
Constructor initializes the object's state, while method changes the object's state
Constructor doesn't have a return type, while method can have a return type
Example: public class ...
Program to form a string with first character of all words in a given string.
Split the string into an array of words
Iterate through the array and extract the first character of each word
Join the extracted characters to form the final string
Given a number n, find the number just greater than n using same digits as that of n
Convert the number to a string and sort the digits in ascending order
Starting from the rightmost digit, find the first digit that is smaller than the digit to its right
Swap this digit with the smallest digit to its right that is greater than it
Sort the digits to the right of the swapped digit in ascending order
Concatenate the digits to
Function to add two numbers represented by linked lists and return the sum list.
Traverse both linked lists and add corresponding nodes, keeping track of carry
Create a new linked list to store the sum
Handle cases where linked lists are of different lengths
Handle cases where the sum has an extra digit due to carry
Return the sum linked list
The player who picks the last coin loses the game.
The game starts with n coins.
Each player can pick 1 to 5 coins in one turn.
The player who picks the last coin loses the game.
Determine if the given n coins will result in a win or loss for the starting player.
Count the number of 0's in a sorted array of 0's and 1's.
Iterative solution: Traverse the array and count the number of 0's.
Recursive solution: Divide the array into two halves and recursively count the number of 0's in each half.
Binary search can also be used to find the first occurrence of 0 and then count the number of 0's after that index.
Reverse a linked list using recursion, handling all corner cases
Create a recursive function that takes the head of the linked list as input
Base case: if the head is null or the list contains only one node, return the head
Recursively call the function with the next node as input and set its next pointer to the current node
Set the current node's next pointer to null and return the new head
Write a function to check whether a binary tree is a sub-tree of another binary tree (Check for all corner cases)
Create a function that traverses both trees and compares them
Check if the root of the second tree matches any node in the first tree
Handle cases where one or both trees are empty
Handle cases where the trees have different structures
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