Synopsys Interview Questions, Process, and Tips

The left view of a binary tree shows the nodes that are visible when the tree is viewed from the left side.

• Traverse the tree recursively from left to right

• Print the first node encountered at each level

• Example: For tree with root node A, left view is A, B, D, G

C++ is an extension of C with object-oriented programming features.

• C++ supports classes and objects while C does not.

• C++ has better support for polymorphism and inheritance.

• C++ has a standard template library (STL) for data structures and algorithms.

• C++ allows function overloading while C does not.

• C++ has exception handling while C does not.

To find the middle of a linked list, use two pointers - one moving at twice the speed of the other.

• Use two pointers - slow and fast

• Slow pointer moves one node at a time, fast pointer moves two nodes at a time

• When fast pointer reaches the end, slow pointer will be at the middle

Semaphores, OSI layers, deadlock and ACID protocols in DBMS

• Semaphores are used for process synchronization and communication

• OSI model has 7 layers: Physical, Data Link, Network, Transport, Session, Presentation, Application

• Deadlock protocols prevent processes from entering a deadlock state

• ACID protocols ensure database transactions are reliable and consistent

Boundary traversal of a binary tree is the process of visiting the nodes on the boundary of the tree.

• Start with the root node and traverse the left boundary nodes from top to bottom.

• Traverse the leaf nodes from left to right.

• Traverse the right boundary nodes from bottom to top.

• Exclude the leaf nodes that are already traversed in the second step.

• Repeat the above steps for all the subtrees of the binary tree.

Linear data structures have a sequential arrangement of elements while non-linear data structures have a hierarchical arrangement.

• Linear data structures have a fixed number of elements and are easy to traverse.

• Non-linear data structures have an arbitrary number of elements and are difficult to traverse.

• Arrays, linked lists, and stacks are examples of linear data structures.

• Trees, graphs, and heaps are examples of non-l

Steps for client server connection request.

• Client sends a connection request to the server.

• Server receives the request and sends a response.

• Client receives the response and establishes a connection.

• Connection is maintained until terminated by either party.

Producer and consumer problem is a synchronization problem in computer science where multiple processes share a common resource.

• It occurs when multiple processes try to access a shared resource simultaneously.

• The producer produces data and the consumer consumes it.

• The problem is to ensure that the producer does not produce data when the buffer is full and the consumer does not consume data when the buffer is empty.

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Processes are independent programs while threads are lightweight processes within a program.

• Processes have their own memory space while threads share memory space.

• Processes communicate through inter-process communication while threads communicate directly.

• Processes are slower to create and terminate than threads.

• Examples of processes are web browsers, word processors, etc. while examples of threads are spell checkers,

Signing off using formal involves thorough verification and validation of the design against specifications.

• Ensure all design properties are formally verified against specifications

• Review and analyze formal verification results for completeness and correctness

• Document the formal verification process and results for sign off

• Obtain approval from stakeholders based on the formal verification results

Verilog code for counters

• Use always blocks to create synchronous counters

• Use reg data types to store the count values

• Use if statements to implement the counting logic

• Example: module counter (input clk, reset, output reg [3:0] count); always @(posedge clk) begin if (reset) count <= 4'b0000; else count <= count + 1; end endmodule