42Gears Mobility Systems Mts1 Interview Questions, Process, and Tips

Round duration - 180 Minutes
Round difficulty - Easy

It was from 8 to 11 in the morning. Basic aptitude MCQs were there and 2 coding questions were there.

Q1. 
Expression Equality Checker
Given two strings representing expressions in variables, determine if they are equivalent. Return 'YES' if the expressions are identical and 'NO' if they are different. Each ex...read more
Ans. Optimal Solution
• We keep track of the current sign, whether it is + or -
• For example, the expression: a-(-b+c) evaluates to a+b-c. Here the sign of b and c are changed in the final expression.
• We maintain a boolean stack for doing so. If it is true, that means character is added and if it is false, that means the character is removed.
• We maintain a count array of size 26(since there are 26 lowercase alphabets). We keep a c...read more
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Q2. 
Longest Increasing Subsequence Problem Statement
Given an array of integers with 'N' elements, determine the length of the longest subsequence where each element is greater than the previous element. This...read more
Ans. Recursive Approach
• We will write a recursive algorithm that will try all the possibilities.
• The argument of recursive function will be the current index and what was the previous number used for LIS. Initially, we will be passing the previous number as the minimum number of an integer and starting index as 0.
• The base case would be when we reach index n,i.e we have exhausted the array elements so return 0 because there is...read more
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Round duration - 32 Minutes
Round difficulty - Easy

It was my first face to face technical round. DSA questions of easy-medium type were asked. And some discussion on the project was there.

Q1. 
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. 
It is a standard problem in the linked list we can solve it using two pointers one moves one step at a time and the other two steps at a time. For each time we check whether they are the same. And this goes up to NULL is reached.
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Q2. 
Convert Binary Tree to Mirror Tree
Convert a given binary tree into its mirror tree, where the left and right children of all non-leaf nodes are interchanged.
Input:
An integer ‘T’ denoting the number o...read more
Ans. Recursive approach.
The idea is to traverse the binary tree and swap the left and right subtrees.
 
The steps are as follows:
 
1. Call mirror function as mirror(node->left)  to access the left subtree.
2. Call mirror function as mirror(node->right) to access the right subtree.
3. Swap left and right subtrees using:
• ‘TEMP’ = leftSubtree
• leftSubtree = rightSubtree
• rightSubtree = ‘TEMP’
Space Complexity: O(n)Explanation...read more
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Round duration - 45 Minutes
Round difficulty - Medium

It was technical plus HR round

Q1. 
Matrix Element Cube Sum Problem
For a given M x N sized 2D array 'MATRIX', find and return the value of (i * i + j * j) for elements where the sum of the cubes of its digits equals the element itself. Her...read more
Ans. Brute Force approach
The idea is to traverse the matrix of size M x N. For each element in the matrix, we find the sum of the cube of their digits recursively. If the sum of the cube of digits is equal to the element itself, calculate (i*i + j*j).
 
1. Traverse the given matrix.
2. For each element in the matrix, store the value of the element in variable ‘DUPLICATE’ and call a function that calculates the sum of the cubes ...read more
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