Choice International DPT.Manager Interview Questions and Answers

Q1. Basic to intermediate.net, EF, SQL, C#, OOP
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Q2. Basics of .Net Core
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Q1. Deep into.net and sql
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Q1. About yourself and background only
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Q1. Array vs string
Ans. 
Arrays are collections of multiple values, while strings are sequences of characters.
• Arrays can store multiple values of different data types, while strings are specifically for storing sequences of characters.

• Arrays can be accessed and manipulated using index positions, while strings have specific methods for working with characters.

• Example: array = ['apple', 'banana', 'cherry'], string = 'hello world'
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Q2. String reverse , tree traversal
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Q3. Read trie topic
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Q1. Work experience and concern work
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Q2. 18 Plus in infrastructure field
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Q3. Positive approach
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Q4. Punctuality and accessible
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Q5. Clear vision on working area ,valuated risk,
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Tableau intensive assignment with few questions on SQL and basic Python

Q1. Discussion on submitted assignment and current job role
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Q2. What kind of analysis have you done in your current role?
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Q1. Basic discussion on why I am switching from my current company and salary discussion
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Q1. What is HashMap?
Ans. 
HashMap is a data structure that stores key-value pairs and provides constant time complexity for basic operations.
• HashMap allows quick access to values based on their keys

• Keys must be unique and values can be duplicated

• HashMap is not thread-safe and requires synchronization for concurrent access

• Java's HashMap implementation uses hashing to distribute keys across buckets
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Q2. It’s a collection framework.
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Round duration - 90 minutes
Round difficulty - Medium

Q1. Cycle Detection in a Singly Linked List
You have given a Singly Linked List of integers, determine if it forms a cycle or not.
A cycle occurs when a node's next points back to a previous node in the ...read more
Ans. 
The task is to determine if a given singly linked list forms a cycle or not.
• A cycle occurs when a node's next points back to a previous node in the list.

• To solve this problem, we can use the Floyd's Cycle-Finding Algorithm.

• The algorithm uses two pointers, one moving at a normal pace and the other moving twice as fast.

• If there is a cycle, the fast pointer will eventually catch up to the slow pointer.

• If the fast pointer ...read more
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Round duration - 60 minutes
Round difficulty - Medium

Q1. Minimum Depth Of Binary Tree
You have been given a Binary Tree of integers, find the minimum depth of this Binary Tree. The minimum depth of a Binary Tree is the number of nodes along the shortest path fro...read more
Ans. 
The minimum depth of a binary tree is the number of nodes along the shortest path from the root node down to the nearest leaf node.
• The minimum depth can be found by performing a breadth-first search (BFS) traversal of the binary tree

• During the BFS traversal, keep track of the current level and increment the depth by 1 for each level

• Stop the BFS traversal when a leaf node is encountered and return the depth as the minim
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Q2. Intersection Of Two Arrays
You are given two arrays 'A' and 'B' of size 'N' and 'M' respectively. Both these arrays are sorted in non-decreasing order. You have to find the ...read more
Ans. 
The problem is to find the intersection of two sorted arrays.
• Use two pointers to iterate through the arrays.

• Compare the elements at the current pointers and move the pointers accordingly.

• If the elements are equal, add it to the intersection array and move both pointers.

• If the element in the first array is smaller, move the first pointer.

• If the element in the second array is smaller, move the second pointer.

• Repeat until...read more
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Q3. Operating System Question
What is the difference between a mutex and a semaphore
Ans. 
A mutex is a binary semaphore used for mutual exclusion, while a semaphore is a generalized synchronization primitive.
• Mutex is used to protect a critical section of code, allowing only one thread to access it at a time.

• Semaphore is used to control access to a shared resource, allowing multiple threads to access it simultaneously.

• Mutex has ownership, meaning the thread that locks it must unlock it.

• Semaphore does not hav...read more
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Round duration - 60 minutes
Round difficulty - Easy

Round duration - 25 minutes
Round difficulty - Easy