Ruckus Networks Interview Questions and Answers

Q1. Python basics and modules
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Q2. Networking protocols basic networking
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Q1. Array Linked list BST
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Q1. Array LL BST Stack
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Coding test was comprising of mcq based questions on db, oops, c++ and java

Q1. It was a high intense interview. Interview starts with introduction and some resume based questions. Then, followed by deep networking and os based questions. Then, dsa questions were asked specifically on...read more
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Q1. It was also a tough one. Questions were specifically on project based of the resume and other questions were on btech curriculum, subjects, and experience.
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Q1. HR was easy. Questions were only about relocation and package breakup. But remember that they can put you in waiting and when needed they can call for another interview. So mark this and all the best.
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90 min MCQ +coding test on hackerrank.

Q1. Sort colors DSA question
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Q2. Detect cycle in a linked list
Ans. 
Use Floyd's Tortoise and Hare algorithm to detect cycle in a linked list.
• Initialize two pointers, slow and fast, at the head of the linked list.

• Move slow pointer by one step and fast pointer by two steps.

• If they meet at some point, there is a cycle in the linked list.
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Q1. Was asked about why you want to join the company,my experience etc.
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Round duration - 90 minutes
Round difficulty - Medium

Q1. 
Cycle Detection in a Singly Linked List
Determine if a given singly linked list of integers forms a cycle or not.
A cycle in a linked list 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 a Binary Tree
Determine the minimum depth of an integer-based binary tree. The minimum depth is defined as the number of nodes present along the shortest path from the root node down to t...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 Problem Statement
Given two arrays A and B with sizes N and M respectively, both sorted in non-decreasing order, determine their intersection.
The intersection of two arrays in...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. 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