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Mentor Graphics interview questions for popular designations
I applied via Naukri.com and was interviewed before Jan 2020. There were 3 interview rounds.
To check if a directed graph is cyclic or not
Use Depth First Search (DFS) algorithm to traverse the graph
Maintain a visited set to keep track of visited nodes
Maintain a recursion stack to keep track of nodes in the current DFS traversal
If a node is visited and is already in the recursion stack, then the graph is cyclic
If DFS traversal completes without finding a cycle, then the graph is acyclic
Return a random byte from a stream of bytes with equal probability.
Create a variable to store the count of bytes read
Create a variable to store the current random byte
For each byte read, generate a random number between 0 and the count of bytes read
If the random number is 0, store the current byte as the random byte
Return the random byte
Check if a binary tree is a binary search tree or not.
Traverse the tree in-order and check if the values are in ascending order.
For each node, check if its value is greater than the maximum value of its left subtree and less than the minimum value of its right subtree.
Use recursion to check if all nodes in the tree satisfy the above condition.
Algorithm to find Nth-to-Last element in a singly linked list of unknown length
Traverse the list and maintain two pointers, one at the beginning and one at Nth node from beginning
Move both pointers simultaneously until the second pointer reaches the end of the list
The first pointer will be pointing to the Nth-to-Last element
If N=0, return the last element
Parse the list only once
Print all possible permutations of an array of integers
Use recursion to swap elements and generate permutations
Start with the first element and swap it with each subsequent element
Repeat the process for the remaining elements
Stop when all elements have been swapped with the first element
Print each permutation as it is generated
Design a stack that prints the minimum element pushed in O(1)
Use two stacks, one for storing elements and another for storing minimums
When pushing an element, compare it with the top of minimum stack and push the smaller one
When popping an element, pop from both stacks
To get the minimum element, just return the top of minimum stack
To find the starting point of a loop in a linked list, use Floyd's cycle-finding algorithm.
Use two pointers, one moving at twice the speed of the other.
When they meet, move one pointer to the head of the list and keep the other at the meeting point.
Move both pointers one step at a time until they meet again, which is the starting point of the loop.
To find a number in a matrix where all rows and columns are sorted non-decreasingly. Complexity of the solution.
Use binary search to find the number in each row and column
Start from the top-right corner or bottom-left corner to optimize search
Time complexity: O(m log n) or O(n log m) depending on the starting corner
posted on 14 Apr 2017
I was interviewed in Feb 2017.
SPI and I2C are communication protocols used for connecting devices, while latch and FF are different types of sequential logic elements.
SPI (Serial Peripheral Interface) and I2C (Inter-Integrated Circuit) are both serial communication protocols used for connecting devices in embedded systems.
SPI uses separate lines for data input and output, while I2C uses a shared bus for both input and output.
SPI supports higher dat...
To find the average speed for the whole trip, we need to know the distance and the time taken for the trip.
Average speed = Total distance / Total time
If the distance and time for the trip are known, the average speed can be calculated.
If only the average speed while going is known, we cannot determine the speed of the whole trip without additional information.
The probability of getting heads maximum two times in a series of coin flips.
To calculate the probability, divide the number of favorable outcomes by the total number of possible outcomes.
In this case, the favorable outcomes are 0, 1, or 2 heads, and the total possible outcomes are 2^n, where n is the number of coin flips.
For example, if we flip a coin 3 times, the favorable outcomes are HHT, HTH, THH, HHH, TTH, THT, a...
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