Spinny SDE-2 Interview Questions and Answers

It was a leetcode medium, related BFS,

Q1. Basic of android like fragment activity communication
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Q2. Why ViewModel is necessary, how does it save state
Ans. 
ViewModel is necessary in Android development to save and manage UI-related data during configuration changes.
• ViewModel helps in retaining data during configuration changes like screen rotations

• It separates UI-related data from UI controller (Activity/Fragment)

• ViewModel survives configuration changes and can be shared between fragments

• It helps in maintaining data consistency and prevents memory leaks

• ViewModel is lifecy...read more
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Round duration - 60 Minutes
Round difficulty - Medium

Timing was around 11 am
Environment was great, it intrigued me to join Blinkit.

Q1. Pair Sum
You are given an integer array 'ARR' of size 'N' and an integer 'S'. Your task is to return the list of all pairs of elements such that each sum of elements of each pair eq...read more
Ans. 
First approach : Loop inside Loop to figure out the sum of two numbers are equal to a constant number. But the complexity is N2 

Second approach : i used Map to store numbers. Before storing we just check rhta constant number - current number exist in map or not.

Second solution is approved
Answered by CodingNinjas
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Round duration - 60 Minutes
Round difficulty - Medium

Timing 11 am
Interviewer had good knowledge

Q1. Technical Question
Activity lifecycle
Ans. 
Tip 1 : Remember important methods and what is the used of every method
Answered by CodingNinjas
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Q2. Technical Question
Fragment lifecycle
Ans. 
Tip 1 : Remember important methods and what is the used of every method
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Q3. System Design Question
Explain MVVM pattern
Ans. 
Tip 1 : must have knowledge about MVC, mvp,mvi,mvvm patterns.
Answered by CodingNinjas
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Round duration - 30 Minutes
Round difficulty - Easy

This round is cultural fit round

Q1. Basic HR Questions
1. Why Blinkit?
2. Salary discussion
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Q1. Data Structures and Algorithms
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Q1. Machine Coding round
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Q1. What is clousers
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Q2. What is promises
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Q1. Explain javascript architecture
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Q2. What is mvvm architecture
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There were 2 questions in coding one with array and in string releated

Q1. About Selenium questions
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Q2. Other basic questions related to testing
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Q3. Management related question
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Q1. One program to solve
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Q2. Other questions in selenium
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2 question of DSA (easy, medium) in 60 minutes

Q1. Playlist Shuffle
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Q2. Deletion in min heap
Ans. 
Deletion in min heap involves removing the root node and rearranging the heap to maintain the min heap property.
• To delete the root node in a min heap, swap it with the last node in the heap.

• Remove the last node and then heapify the heap starting from the root to maintain the min heap property.

• Example: If the min heap is [10, 20, 30, 25, 40], after deletion the heap becomes [20, 25, 30, 40].
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Round duration - 45 minutes
Round difficulty - Medium

Technical Interview round with questions on DSA.
Tip : Do variants of Binary Search.

Q1. Find Square root of an integer
You are given an integer ‘A’. Your task is to find the greatest non-negative integer whose square is less than or equal to ‘A’.
Square of a number is the product of the num...read more
Ans. 
The Simple Approach is to find the floor of the square root, try with all-natural numbers starting from 1. Continue incrementing the number until the square of that number is greater than the given number.

Algorithm : 
1. Create a variable (counter) i and take care of some base cases, i.e when the given number is 0 or 1.
2. Run a loop until i*i <= n , where n is the given number. Increment i by 1.
3. The floor of t...read more
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Q2. Find the minimum element in a sorted and rotated array
You're given a sorted array that has now been rotated 'K' times, which is unknown to you. Rotation here means that every element is shifte...read more
Ans. 
The brute force solution is to traverse the entire array and find the minimum. Time Complexity : O(N)

The efficient approach is to use Binary Search. The following pattern can be observed : 
 
• The minimum element is the only element whose previous is greater than it. If there is no previous element, then there is no rotation (the first element is minimum). We check this condition for the middle element by comp...read more
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