SDE - Intern

You are given an undirected graph with 'N' nodes and 'M' edges. The weight of each edge in the graph is one unit.

Given a source vertex 'src', you must return an array 'answer' of length 'N', where 'answer[i]' is the shortest path length between the source vertex 'src' and 'i'th vertex.

Note:

All the nodes are zero-based.

Example:

Input:
N=5, M=5, edges=[(0, 1), (1, 4), (2, 3), (2, 4), (3, 4)], src=1 
Output: 1 0 2 2 1

Explanation: The path from vertices are:-
(1->0) = 1 -> 0, path length is 1.
(1->1) = 1 -> 1, path length is 0.
(1->2) = 1 -> 4 -> 2, the path length is 2.
(1->3) = 1 -> 4 -> 3, path length is 2.
(1->4) = 1 -> 4, the path length is 1.
Hence we return [1, 0, 2, 2, 1]

Elon is a kid who just started learning about numbers, but he always gets afraid when he sees a number with repeated digits, as it becomes hard for him to remember it as it has the same digits repeating again.

So his teacher, knowing of his fear, gave him ‘N’ tasks to cheer him up, for each task Elon have to count the numbers that have no repeated digits in a given range of [‘L’, ‘R’]. The tasks are given by a 2d array ‘TASKS’.

But Elon is again afraid that while counting he may encounter a number with repeated digits, being his friend he asked you to help him. Can you help Elon to find the count of numbers having no repeated digits in the given range for each task?.

NOTE: Both ‘L’ and ‘R’ are included in the range.

Example :

Input: ‘N’ = 1, ‘TASKS = [ [1, 20] ]

Output: 19
For the first task, the given range consists of 20 numbers present out of which ‘11’ is the only number that has repeated occurrences of ‘1’ in its decimal representation.

Design and implement a data structure for Least Recently Used (LRU) cache to support the following operations:

1. get(key) - Return the value of the key if the key exists in the cache, otherwise return -1.

2. put(key, value), Insert the value in the cache if the key is not already present or update the value of the given key if the key is already present. When the cache reaches its capacity, it should invalidate the least recently used item before inserting the new item.

You will be given ‘Q’ queries. Each query will belong to one of these two types:

Type 0: for get(key) operation.
Type 1: for put(key, value) operation.

Note :

1. The cache is initialized with a capacity (the maximum number of unique keys it can hold at a time).

2. Access to an item or key is defined as a get or a put operation on the key. The least recently used key is the one with the oldest access time.

Which scheduling algorithm may cause starvation?
a) FCFS
b) Round Robin
c) Priority Scheduling
d) FIFO

Answer: c) Priority Scheduling

You are present at point ‘A’ which is the top-left cell of an M X N matrix, your destination is point ‘B’, which is the bottom-right cell of the same matrix. Your task is to find the total number of unique paths from point ‘A’ to point ‘B’.In other words, you will be given the dimensions of the matrix as integers ‘M’ and ‘N’, your task is to find the total number of unique paths from the cell MATRIX[0][0] to MATRIX['M' - 1]['N' - 1].

To traverse in the matrix, you can either move Right or Down at each step. For example in a given point MATRIX[i] [j], you can move to either MATRIX[i + 1][j] or MATRIX[i][j + 1].

There was a detailed discussion around my URL Shortener project, focusing on High-Level Design (HLD) and real-world scalability. The interviewer asked me to explain the system architecture, including how I generate unique short URLs, handle collisions, and design the database schema.

They also explored how the system would scale with increasing traffic, how redirection latency can be minimized, and how caching mechanisms can be used for optimization. Questions were asked around trade-offs in design decisions, such as choosing between different encoding techniques or database strategies.

Additionally, there was an HR discussion covering questions like:
Challenges faced during project development
Reasons for choosing this project
Key learnings and improvements

The discussion was interactive and focused on understanding both technical depth and clarity of thought.