USHA International Java Developer Interview Questions and Answers for Experienced

Round duration - 120 Minutes
Round difficulty - Medium

The round had 2 coding problems to solve with varying difficulty. Each candidate had a different set of questions. The round was around 2 pm. The webcam was turned on to keep an eye on candidates.

Q1. 
Bursting Balloons Problem
Given an array ARR of size N, where each element represents the height of a balloon. The task is to destroy all balloons by shooting arrows from left to right. When an arrow hits...read more
Ans. 
Find the minimum number of arrows needed to burst all balloons by shooting arrows from left to right.
• Sort the array in non-decreasing order to make it easier to calculate the minimum number of arrows needed.

• Iterate through the sorted array and count the number of times the height decreases compared to the previous balloon.

• The count of decreases + 1 will give the minimum number of arrows needed to burst all balloons.

• Exa...read more
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Q2. 
Count Leaf Nodes in a Binary Tree
Count the number of leaf nodes present in a given binary tree. A binary tree is a data structure where each node has at most two children, known as the left child and the...read more
Ans. 
Count the number of leaf nodes in a binary tree.
• Traverse the binary tree and check if both left and right children are NULL to identify leaf nodes.

• Use recursion to traverse the tree efficiently.

• Keep track of the count of leaf nodes as you traverse the tree.

• Handle base cases where the current node is NULL or a leaf node.
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Round duration - 60 Minutes
Round difficulty - Medium

This round had 2 questions related to DSA. I was first asked to explain my approach with proper complexity analysis and then code the soution in any IDE that I prefer.

Q1. 
Rod Cutting Problem Statement
Given a rod of a certain length, the rod can be divided into different sizes, each with an associated cost. Your task is to determine the maximum cost that can be obtained by...read more
Ans. 
The Rod Cutting Problem involves maximizing the profit obtained by cutting a rod into smaller pieces and selling them.
• Use dynamic programming to solve this problem efficiently.

• Create a table to store the maximum profit for each sub-length of the rod.

• Iterate through the rod lengths and update the table with the maximum profit.

• The final answer will be the maximum profit for the total length of the rod.
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Q2. 
Count Subarrays with Given XOR Problem Statement
You are given an array of integers ARR and an integer X. Your task is to determine the number of subarrays of ARR whose bitwise XOR is equal to X.
Example...read more
Ans. 
Count the number of subarrays in an array whose XOR is equal to a given value.
• Iterate through the array and keep track of XOR values and their frequencies using a hashmap.

• For each element in the array, calculate the XOR with the current element and check if the required XOR value exists in the hashmap.

• Increment the count of subarrays whenever the required XOR value is found in the hashmap.
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Round duration - 60 Minutes
Round difficulty - Medium

This round had 2 questions of DSA of Easy-Medium difficulty and at the end I was asked a Puzzle to check my general problem solving ability.

Q1. 
Rotting Oranges Problem Statement
You are given a grid containing oranges where each cell of the grid can contain one of the three integer values:
• 0 - representing an empty cell
• 1 - representing a fre...read more
Ans. 
Find the minimum time required to rot all fresh oranges in a grid.
• Use Breadth First Search (BFS) to simulate the rotting process of oranges.

• Keep track of the time taken to rot all oranges and the count of fresh oranges remaining.

• If all fresh oranges are not rotten after simulation, return -1.

• Handle edge cases like empty grid or no fresh oranges present.

• Example: For the given grid, the minimum time required is 4 seconds...read more
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Q2. 
Detect and Remove Loop in Linked List
For a given singly linked list, identify if a loop exists and remove it, adjusting the linked list in place. Return the modified linked list.
Expected Complexity:
A...read more
Ans. 
Detect and remove loop in a singly linked list in place with O(n) time complexity and O(1) space complexity.
• Use Floyd's Cycle Detection Algorithm to identify the loop in the linked list.

• Once the loop is detected, use two pointers approach to find the start of the loop.

• Adjust the pointers to remove the loop and return the modified linked list.

• Example: For input 5 2 and 1 2 3 4 5, return 1 2 3 4 5 without the loop.
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Q3. You have two wires of different lengths that are both capable of burning for exactly one hour when ignited at both ends. How can you measure a time interval of 45 minutes using these two wires?
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Round duration - 50 Minutes
Round difficulty - Medium

This round had 2 Algorithmic questions wherein I was supposed to code both the problems after discussing their
approaches and respective time and space complexities . After that , I was grilled on some OOPS concepts related to C++.

Q1. 
Longest Palindromic Substring Problem Statement
You are provided with a string STR of length N. The task is to find the longest palindromic substring within STR. If there are several palindromic substring...read more
Ans. 
Given a string, find the longest palindromic substring within it.
• Iterate through the string and expand around each character to find palindromes

• Keep track of the longest palindrome found so far

• Return the longest palindromic substring
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Q2. 
Quick Sort Problem Statement
You are provided with an array of integers. The task is to sort the array in ascending order using the quick sort algorithm.
Quick sort is a divide-and-conquer algorithm. It ...read more
Ans. 
Yes, the quick sort algorithm can be enhanced to achieve NlogN complexity in the worst case by using a randomized version of the algorithm.
• Randomized quick sort involves randomly selecting the pivot element to reduce the chances of worst-case scenarios.

• By choosing a random pivot, the algorithm becomes less predictable and more likely to achieve the desired time complexity.

• This enhancement helps in avoiding the worst-ca...read more
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Q3. What are friend functions in C++?
Ans. 
Friend functions in C++ are functions that are not members of a class but have access to its private and protected members.
• Friend functions are declared inside a class with the keyword 'friend'.

• They can access private and protected members of the class.

• They are not member functions of the class, but have the same access rights as member functions.

• Friend functions are useful for implementing operators that are not part ...read more
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Q4. What is the difference between Early Binding and Late Binding in C++?
Ans. 
Early binding is resolved at compile time while late binding is resolved at runtime in C++.
• Early binding is also known as static binding, where the function call is resolved at compile time based on the type of the object.

• Late binding is also known as dynamic binding, where the function call is resolved at runtime based on the actual type of the object.

• Early binding is faster as the function call is directly linked to ...read more
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Round duration - 60 minutes
Round difficulty - Easy

This was coding round and was conducted on Cocubes platform. It is not IDE but kind of white board coding platform. C/C++, Java and Python were only allowed languages.

Round duration - 30 minutes
Round difficulty - Easy

This was pen and paper round. Total 40 shortlisted candidates were grouped into 8 groups (each of 5). Each group was given a coding question which they have to solve on paper and explain it to the recruiter. 2 to 3 from each group were selected for the next round.

Round duration - 60 minutes
Round difficulty - Easy

This was general face to face Data Structures and Algorithms based round.

Q1. How do you perform a spiral order traversal of a binary tree? Please provide an explanation or code to print the binary tree in spiral order.
Ans. 
Spiral order traversal of a binary tree involves printing nodes level by level alternating between left to right and right to left.
• Start by pushing the root node into a queue.

• While the queue is not empty, pop a node, print its value, and push its children into the queue.

• For each level, alternate between printing nodes from left to right and right to left.

• Repeat until all nodes are printed in spiral order.
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Round duration - 45 minutes
Round difficulty - Easy

It was the last round.

Round duration - 120 Minutes
Round difficulty - Medium

This was an online coding round where we had 3 questions to solve under 120 minutes. The questions were of medium to hard difficulty level.

Q1. 
Minimum Time in Wormhole Network
Determine the minimum time required to travel from a starting point to a destination point in a two-dimensional coordinate system, considering both direct movement and the...read more
Ans. 
Find the minimum time to travel from a starting point to a destination point using direct movement and wormholes.
• Calculate the time taken for direct movement from source to destination.

• Consider using each wormhole to see if it reduces the total travel time.

• Choose the path with the minimum total time to reach the destination.
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Q2. 
Substrings Differ by One Problem Statement
Ninja needs help in a battle against the string man. Given two strings, 'S' and 'T', the task is to find the number of substrings in 'S' that differ from some su...read more
Ans. 
The task is to find the number of substrings in 'S' that differ from some substrings of 'T' by exactly one character.
• Iterate through all substrings of 'S' and 'T' and compare them character by character to find the ones that differ by exactly one character.

• Use nested loops to generate all possible substrings of 'S' and 'T'.

• Count the number of substrings that differ by exactly one character and return the total count.
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Q3. 
Power of Two Problem Statement
Determine whether a given integer N is a power of two. Return true if it is, otherwise return false.
Explanation
An integer 'N' is considered a power of two if it can be e...read more
Ans. 
Check if a given integer is a power of two or not.
• Check if the given integer is greater than 0.

• Check if the given integer has only one bit set in its binary representation.

• Return true if the above conditions are met, else return false.
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Round duration - 60 Minutes
Round difficulty - Medium

This round had 2 questions related to DSA where I was first expected to explain my approaches and then discuss the time and space complexities of my solution. After that , I was asked some core concepts related to OS.

Q1. 
Reverse Linked List Problem Statement
Given a singly linked list of integers, return the head of the reversed linked list.
Example:
Initial linked list: 1 -> 2 -> 3 -> 4 -> NULL
Reversed link...read more
Ans. 
Reverse a singly linked list of integers and return the head of the reversed linked list.
• Iterate through the linked list and reverse the pointers to point to the previous node instead of the next node.

• Use three pointers - prev, current, and next to reverse the linked list in O(N) time and O(1) space complexity.

• Update the head of the reversed linked list as the last node encountered during reversal.
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Q2. 
Cycle Detection in Undirected Graph Problem Statement
You are provided with an undirected graph containing 'N' vertices and 'M' edges. The vertices are numbered from 1 to 'N'. Your objective is to determi...read more
Ans. 
Detect cycles in an undirected graph.
• Use Depth First Search (DFS) to detect cycles in the graph.

• Maintain a visited array to keep track of visited vertices.

• If a visited vertex is encountered again during DFS, a cycle exists.

• Check for cycles in each connected component of the graph.

• Example: For input N=3, Edges=[[1, 2], [2, 3], [1, 3]], output is Yes.
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Q3. What are the differences between a mutex and a semaphore?
Ans. 
Mutex is used for exclusive access to a resource by only one thread at a time, while semaphore can allow multiple threads to access a resource simultaneously.
• Mutex is binary semaphore with ownership, used for mutual exclusion.

• Mutex is typically used to protect critical sections of code.

• Semaphore is a signaling mechanism, used for synchronization between multiple threads.

• Semaphore can have a count greater than 1, allowi...read more
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Q4. What is meant by multitasking and multithreading in operating systems?
Ans. 
Multitasking refers to the ability of an operating system to run multiple tasks concurrently, while multithreading involves executing multiple threads within a single process.
• Multitasking allows multiple processes to run simultaneously on a single processor, switching between them quickly to give the illusion of parallel execution.

• Multithreading involves breaking a process into multiple threads that can be executed ind...read more
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Round duration - 60 Minutes
Round difficulty - Medium

This round had 2 Algorithmic questions wherein I was supposed to code both the problems after discussing their
approaches and respective time and space complexities . After that , I was grilled on some OOPS concepts related to C++.

Q1. 
Trapping Rain Water Problem Statement
You are given a long type array/list ARR of size N, representing an elevation map. The value ARR[i] denotes the elevation of the ith bar. Your task is to determine th...read more
Ans. 
Calculate the total amount of rainwater that can be trapped between given elevations in an array.
• Iterate through the array and calculate the maximum height on the left and right of each bar.

• Calculate the amount of water that can be trapped at each bar by taking the minimum of the maximum heights on the left and right.

• Sum up the trapped water at each bar to get the total trapped water for the entire array.
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Q2. 
Find The Repeating And Missing Number Problem Statement
You are provided with an array nums which contains the first N positive integers. In this array, one integer appears twice, and one integer is missi...read more
Ans. 
Given an array of first N positive integers with one number repeating and one missing, find the repeating and missing numbers.
• Iterate through the array and keep track of the sum of elements and sum of squares to find the missing and repeating numbers.

• Use a set to identify the repeating number and calculate the missing number based on the sum of elements.

• Example: For nums = [1, 2, 3, 4, 4, 5], the repeating number is 4 ...read more
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Q3. What is the Diamond Problem in C++ and how can it be resolved?
Ans. 
Diamond Problem in C++ is a common issue in multiple inheritance where a class inherits from two classes that have a common base class.
• Diamond Problem occurs when a class inherits from two classes that have a common base class, leading to ambiguity in accessing members.

• It can be resolved in C++ using virtual inheritance, where the common base class is inherited virtually to avoid duplicate copies of base class members.

• ...read more
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Q4. What are friend functions in C++?
Ans. 
Friend functions in C++ are functions that are not members of a class but have access to its private and protected members.
• Friend functions are declared inside a class with the 'friend' keyword.

• They can access private and protected members of the class.

• They are not member functions of the class, but have the same access rights as member functions.

• Friend functions are useful for implementing operators that are not membe...read more
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