MagicBricks Software Developer Interview Questions, Process, and Tips

Q1. Launch modes on android
Ans. 
Launch modes on Android determine how a new instance of an activity is associated with the current task.
• Standard: Creates a new instance of the activity in the task's stack

• SingleTop: If an instance of the activity already exists at the top of the stack, it will not be recreated

• SingleTask: Activity is always at the root of the task and no other activities can be on top of it

• SingleInstance: Similar to SingleTask, but the
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Questios based on Activity backstacks

Round duration - 60 minutes
Round difficulty - Easy

It comprised of general aptitude questions and two coding questions. It was an offline test.

Q1. 
N Queens Problem
Given an integer N, find all possible placements of N queens on an N x N chessboard such that no two queens threaten each other.
Explanation:
A queen can attack another queen if they ar...read more
Ans. 
DFS can be used to solve this problem. The idea is to place queens in different columns one by one, starting from the leftmost column. As we cannot place 2 queens in the same column, when we place the queen in the nth column and if its valid position, dfs again starting n+1 th column and try placing the next queen in all the rows of n+1th column and find a valid position. If no valid queen row found in n+1th row, backt...read more
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Q2. 
Sort 0 1 2 Problem Statement
Given an integer array arr of size 'N' containing only 0s, 1s, and 2s, write an algorithm to sort the array.
Input:
The first line contains an integer 'T' representing the n...read more
Ans. 
The simple approach is to simply count the number of 0’s, 1’s, and 2’s. Then, place all 0’s at the beginning of the array followed by 1’s and 2's. The time complexity of this approach would be O(n) and space complexity O(1). 
However, the approach requires two traversals of the array. The question can also be solved in a single scan of the array by maintaining the correct order of 0’s, 1’s, and 2’s using variables....read more
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Round duration - 60 minutes
Round difficulty - Easy

After having a technical discussion about my CV. He gave me two questions to code.

Q1. 
Ninja and Substrings Problem Statement
Ninja has to determine all the distinct substrings of size two that can be formed from a given string 'STR' comprising only lowercase alphabetic characters. These su...read more
Ans. 
The brute force solution is to check for every index in the string whether the sub-string can be formed at that index or not. To implement this, run a nested loop traversing the given string and check for sub-string from every index in the inner loop. 
Time Complexity : O(n^2)
The solution can be optimised by using KMP Algorithm. The KMP matching algorithm uses degenerating property (pattern having same sub-patterns...read more
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Q2. 
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. 
Floyd's algorithm can be used to solve this question.
Define two pointers slow and fast. Both point to the head node, fast is twice as fast as slow. There will be no cycle if it reaches the end. Otherwise, it will eventually catch up to the slow pointer somewhere in the cycle.
Let X be the distance from the first node to the node where the cycle begins, and let X+Y be the distance the slow pointer travels. To catch up, t...read more
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Round duration - 30 minutes
Round difficulty - Easy

This was supposed to be the HR round but out of surprise the interviewer started by giving me a question to code. 
After I approached this question with the right solution he just asked about my family. After that he said to wait. After half an hour the results were announced. A total of three students were hired and I was amongst one of them.

Q1. 
Balanced Parentheses Combinations
Given an integer N representing the number of pairs of parentheses, find all the possible combinations of balanced parentheses using the given number of pairs.
Explanati...read more
Ans. 
A stack can be used to solve this question. 
We traverse the given string s and if we:
1. see open bracket we put it to stack
2. see closed bracket, then it must be equal to bracket in the top of our stack, so we check it and if it is true, we remove this pair of brackets.
3. In the end, if and only if we have empty stack, we have valid string.
Complexity: time complexity is O(n): we put and pop each element of string ...read more
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