Qualcomm Interview Questions

Table Of Contents

• Why Qualcomm is a Great Place to Work
• Qualcomm Recruitment Process: What to Expect
• Qualcomm Technical Interview Questions
• Qualcomm Interview Preparation
  • Understanding the Interview Process
  • Key Topics to Prepare
  • Behavioral & HR Questions
  • Tips for Cracking the Qualcomm Interview

Qualcomm, an American multinational company, is renowned for its innovations in wireless technology. Founded in Delaware and headquartered in San Diego, California, Qualcomm specializes in semiconductors, software, and services related to wireless technology. Its patents are vital for the mobile communication standards of 5G, 4G, CDMA2000, TD-SCDMA, and WCDMA, making it a key player in global connectivity.

In India, Qualcomm has played a pivotal role in advancing mobile communication by offering cutting-edge technological solutions. With offices dedicated to digital media networking, DSP and embedded applications, as well as wireless modems and multimedia software, Qualcomm has helped make mobile communications more accessible and affordable for the Indian market.

Kickstart your Salesforce career with our Salesforce training in Delhi, designed to provide hands-on experience and job-ready skills through real-world projects. Join us for a free demo session today!

Why Qualcomm is a Great Place to Work

Qualcomm offers excellent opportunities for career growth and development. Known for its stability and competitive benefits, it provides a work environment that fosters both personal and professional growth. The company’s inclusive culture allows employees to network with talented individuals, attend engaging seminars, and gain valuable insights into the tech industry.

In addition to professional development, Qualcomm places a strong emphasis on employee well-being. The company offers a range of benefits, including wellness programs, community involvement, and family care. Qualcomm’s work culture promotes learning, collaboration, and building soft skills, making it an ideal place to start or advance your career.

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Qualcomm Technical Interview Questions

1. What is a MAC address?

A MAC address (Media Access Control address) is a unique identifier assigned to network interfaces for communications at the data link layer of a network. This 48-bit address is typically represented in hexadecimal format and is essential for network devices to communicate within a local network. It’s crucial for devices such as network cards, routers, and switches to differentiate between each other in a network environment.

For example, a MAC address may look like: 00:14:22:01:23:45. This identifier is unique to the device and remains the same across all networks it connects to. In a scenario, when a device connects to a Wi-Fi network, the router uses the MAC address to identify the device and control traffic between the device and the network.

2. What are the HTTP and the HTTPS protocol?

The HTTP (Hypertext Transfer Protocol) is the foundation of data communication on the World Wide Web. It’s a protocol used by the web browser and server to request and deliver web pages and their resources like images, videos, and scripts. HTTP operates on the client-server model, where the client (typically a browser) sends requests, and the server responds with the requested content. However, HTTP is not secure, meaning that data, such as login credentials and personal information, is transmitted in plaintext.

For instance, when you visit http://example.com, your browser sends an HTTP request to the server, and the server responds with the website content. However, the data exchanged, like passwords or payment details, is visible to anyone intercepting the connection. On the other hand, HTTPS (Hypertext Transfer Protocol Secure) uses SSL/TLS encryption to secure this data, making it unreadable to any unauthorized party. An example would be visiting https://example.com, where the connection is encrypted, ensuring the security of any sensitive information transmitted.

3. Explain LAN (Local Area Network).

A LAN (Local Area Network) is a network of devices that are geographically close, typically within a single building or campus. It’s used to connect computers, printers, servers, and other devices to share resources like files and internet access. One of the main characteristics of a LAN is its high data transfer rate, typically ranging from 100 Mbps to 10 Gbps, depending on the technology used (Ethernet, Wi-Fi, etc.). LANs are usually private networks, which means they are not open to the public and can be securely managed.

For example, I might set up a LAN at my home by connecting multiple computers and printers using a router, and then I can share files across devices or print documents from any connected device. In a business environment, employees in different departments may be connected to the same LAN to easily access shared resources, such as databases or printers, without relying on the internet.

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4. What is a ‘frame relay’ and in which layer does it operate?

Frame Relay is a high-performance WAN protocol used to connect devices over large distances. It operates at the Data Link Layer (Layer 2) of the OSI model. Initially, Frame Relay was used to interconnect Local Area Networks (LANs) over long distances, especially before more advanced protocols like MPLS (Multiprotocol Label Switching) took over. Frame Relay is a connection-oriented protocol, which means it establishes a path for data to travel before the transmission begins. It uses virtual circuits to route data and offers a best-effort service, meaning there’s no guarantee of data delivery or congestion control.

For example, in a business setting, if I needed to connect two branch offices in different cities, I could use Frame Relay to set up a dedicated virtual circuit between them, allowing secure communication and file sharing. The Frame Relay protocol would establish a path between the devices, ensuring the data reaches the right destination, but it wouldn’t provide automatic error correction, relying instead on upper-layer protocols for that.

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5. How does a firewall work?

A firewall is a network security system designed to monitor and control incoming and outgoing traffic based on predetermined security rules. It can be hardware-based, software-based, or a combination of both. Firewalls work by filtering traffic between networks, such as between a private local area network (LAN) and the public internet. In my experience, firewalls are typically set up to block malicious traffic, unauthorized access, and potential threats, while allowing legitimate traffic to pass through.

For instance, in a business network, the firewall might block all incoming connections on certain ports (like port 80 for HTTP) but allow traffic on port 443 for secure HTTPS communication. Additionally, if I have a home firewall in place, it might block any external devices trying to access my private network, while allowing my computer to access the internet. A stateful inspection firewall can track the state of active connections, ensuring that only responses to requests initiated by my device are allowed. Here’s an example of a basic firewall rule to allow HTTP traffic:

iptables -A INPUT -p tcp --dport 80 -j ACCEPT

This rule allows incoming TCP traffic on port 80 (HTTP), ensuring my server can handle requests from clients.

6. What is IPC communications?

IPC (Inter-Process Communication) refers to a set of mechanisms that allow processes to communicate with each other and synchronize their actions. It is crucial when multiple processes need to exchange data, share resources, or coordinate activities in a computer system. IPC enables different processes, which may be running on the same machine or across different machines, to interact without disrupting the operating system’s stability or security.

For example, in a multi-threaded application, one process may need to send data to another for processing. IPC methods like message passing, shared memory, or pipes allow one process to pass information, such as the result of a computation, to another process without interfering with the independent operation of each. This is commonly used in server-client applications where the server may handle requests and responses from clients, with each client being a separate process.

7. What is Stack corruption?

Stack corruption occurs when the data on the stack (a region of memory used for storing temporary variables and function call information) becomes corrupted due to programming errors or vulnerabilities. The stack is essential for managing function calls, return addresses, and local variables. Corruption of the stack can lead to unpredictable behavior, including program crashes or security vulnerabilities such as buffer overflow attacks, where attackers overwrite the stack to gain control over the execution flow of a program.

For example, when a buffer overflow occurs, an attacker may input more data than the buffer can handle, overwriting adjacent memory locations on the stack. This can overwrite a return address or a function pointer, which can then direct the program’s execution to malicious code. Here’s a simple example where stack corruption can occur in C programming:

char buffer[10];
strcpy(buffer, "This is a very long string");  // buffer overflow, overwriting the stack

In this case, the overflow may lead to unexpected behavior or crashes due to stack corruption. The code above demonstrates a simple buffer overflow that occurs when more data is written to buffer than its capacity of 10 characters. This overwrites adjacent memory, potentially leading to stack corruption and unintended execution flow.

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8. What is Race condition?

A race condition happens in a multi-threaded or multi-process environment when the system’s behavior depends on the sequence or timing of uncontrollable events, such as thread execution order. Essentially, two or more processes or threads try to access shared resources simultaneously, and the outcome of their execution depends on the timing or order in which they access the shared resource.

For example, consider a situation where two threads are trying to update a bank account balance. If both threads read the balance at the same time and then write it back after adding money, they could overwrite each other’s updates. This can result in the loss of money due to the race condition. A simplified scenario of a race condition can be:

balance = 100  # Initial balance
# Thread 1
balance = balance + 50  # Adding 50
# Thread 2
balance = balance + 50  # Adding 50

The result would be 150, not 200, if both threads operate without proper synchronization. This example shows how two threads can manipulate the shared balance variable simultaneously, leading to an incorrect final value due to a race condition.

9. What is Semaphore?

A semaphore is a synchronization early used to control access to a shared resource in a concurrent system, like in a multi-threaded or multi-process environment. It’s often used to avoid race conditions and ensure that only a certain number of processes or threads can access a resource at the same time. A semaphore operates by maintaining an internal counter that indicates the number of resources available or the number of processes allowed to enter a critical section.

For example, imagine a printing system with only two printers available. If more than two users try to print documents simultaneously, the semaphore will limit the access to the printers. The semaphore can be initialized to 2 (for two printers). When a user requests access, the semaphore is decremented; once a printer is done, the semaphore is incremented, making the resource available again.

import threading
semaphore = threading.Semaphore(2)  # Two resources available

def print_document():
    semaphore.acquire()  # Acquire the resource
    # Printing logic here
    semaphore.release()  # Release the resource

In this code, the semaphore is initialized with a count of 2, representing two available resources (e.g., printers). Each thread must acquire the semaphore before printing and release it when done, ensuring that only two threads can access the printer simultaneously, preventing resource contention.

10. What is Mutex?

A mutex (short for mutual exclusion) is a synchronization object used to protect shared resources from being simultaneously accessed by multiple threads in a concurrent system. Unlike semaphores, which allow multiple threads to access a resource, a mutex ensures that only one thread can access a resource at any given time, preventing race conditions and ensuring data integrity.

For example, in a banking system, a mutex can be used to ensure that only one thread can access the account balance at a time, preventing multiple threads from modifying it simultaneously and causing inconsistent results. Here’s a simple mutex example in Python:

import threading
mutex = threading.Lock()

def update_balance():
    with mutex:  # Lock the mutex
        # Critical section: updating the balance
        pass  # Balance update logic here

The mutex in the above example ensures that only one thread can execute the code inside the with mutex block at a time. By acquiring and releasing the mutex lock, we prevent race conditions when multiple threads attempt to update the shared account balance concurrently.

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11. What is Priority Inversion?

Priority inversion occurs when a higher-priority task is indirectly preempted by a lower-priority task, which causes the higher-priority task to wait for the lower-priority task to finish. This situation can be particularly problematic in real-time systems, where tasks with higher priority should ideally preempt tasks with lower priority. The inversion happens when a low-priority task holds a resource that a high-priority task needs, and the low-priority task is preempted by a medium-priority task, causing the high-priority task to wait. This can lead to performance degradation and system instability.

A simple example can be seen in a real-time operating system where a high-priority task is waiting for a resource held by a low-priority task:

# High-priority task waiting for a resource locked by low-priority task
mutex.acquire()  # Low-priority task acquires the mutex
# High-priority task now waits, resulting in priority inversion

To prevent priority inversion, a system can implement priority inheritance, where the lower-priority task temporarily inherits the priority of the higher-priority task until the resource is released.

12. What is Memory management?

Memory management refers to the process of efficiently allocating, managing, and freeing memory within a computer system or application. It involves tracking every byte in a computer’s memory and ensuring that memory is used optimally by different applications and processes. In a system, memory is allocated to processes as needed, and when processes finish their work, the memory is freed for reuse. Effective memory management is crucial to ensure that programs run smoothly and don’t encounter errors such as out-of-memory exceptions or performance slowdowns. There are several techniques for memory management, such as paging, segmentation, and garbage collection, each serving to manage different types of memory access.

For example, in a heap memory model, a program allocates and deallocates memory dynamically during its execution, and a garbage collector periodically cleans up unused memory.

import gc
gc.collect()  # Manually invoking garbage collection to free memory

In the code above, the gc.collect() method is used to trigger garbage collection, which helps in freeing up memory by reclaiming memory that is no longer in use, preventing memory leaks.

13. What is Deadlock?

A deadlock is a situation in which two or more processes are unable to proceed because each is waiting for the other to release a resource. In a multi-threaded or multi-process system, deadlock occurs when processes are blocked, waiting indefinitely for resources that are held by each other. There are four necessary conditions for a deadlock to occur: mutual exclusion, hold and wait, no preemption, and circular wait. For example, consider two threads where thread A holds resource X and is waiting for resource Y, while thread B holds resource Y and is waiting for resource X. This circular waiting leads to a deadlock, where neither thread can proceed.

Here’s an example scenario:

# Thread A locks resource X and waits for resource Y
mutex_X.acquire()  
mutex_Y.acquire()  # This will never be reached if Thread B locks Y first

# Thread B locks resource Y and waits for resource X
mutex_Y.acquire()
mutex_X.acquire()  # This will never be reached if Thread A locks X first

In this situation, both threads are waiting for each other to release the resources, and neither can proceed, causing a deadlock.

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14. What is Mem leaks?

Memory leaks occur when a program allocates memory dynamically but fails to release it when it is no longer needed. Over time, if memory leaks are not fixed, the application will use more memory than required, eventually leading to out-of-memory errors or performance degradation. Memory leaks are a common issue in low-level programming languages like C or C++, where the programmer must explicitly manage memory allocation and deallocation. In languages like Python or Java, memory leaks can still occur due to improper handling of references or cyclic references between objects. For example, in C++:

int* ptr = new int(10);  // Memory allocated dynamically
// Memory is not freed, causing a memory leak

In this case, the memory allocated by new int(10) is never freed, causing a memory leak. In managed languages, memory leaks typically occur when objects are still referenced, even though they are no longer needed, preventing the garbage collector from reclaiming the memory.

15. What do you understand by Data Redundancy?

Data redundancy occurs when the same piece of data is stored multiple times in a database or data storage system, leading to inefficiencies and potential data inconsistency. In a relational database, redundancy can happen when the same data is repeated across multiple tables or rows, violating the normalization principles. This redundancy can lead to wasted storage, slower performance, and increased complexity when performing updates, since changes must be made in multiple places to ensure data consistency.

For example, if an address is stored separately for each customer in a customer table, multiple customers with the same address may lead to redundancy. Here’s an example of data redundancy in a customer database:

-- Redundant storage of address information for each customer
INSERT INTO customer (name, address) VALUES ('Alice', '123 Main St');
INSERT INTO customer (name, address) VALUES ('Bob', '123 Main St');

To avoid data redundancy, databases often use normalization, a process that organizes data into separate tables and minimizes redundancy by ensuring that each piece of information is stored only once. This improves efficiency, consistency, and ease of maintenance.

16. Explain Normalization and De-Normalization.

In my experience, normalization is a database design technique that organizes data to reduce redundancy and dependency. The main goal is to divide large tables into smaller, manageable ones while ensuring that relationships among the data are logically structured. By doing this, we minimize duplicate data and improve data integrity, ensuring that updates, insertions, and deletions can be done efficiently. Normalization typically involves creating multiple tables to store related data and using foreign keys to link them together. A common example would be splitting a customer and address table, so customer information isn’t duplicated.

On the other hand, de-normalization involves merging normalized tables to improve read performance. It is typically used in scenarios where query performance is critical, and the database must handle large amounts of data. De-normalization sacrifices some of the benefits of normalization to reduce the number of joins needed to fetch data. For example, in reporting applications, instead of joining multiple normalized tables to fetch customer details and orders, it might be beneficial to store them together in one table for faster access.

Here’s an example of denormalization:

-- De-normalized table with customer and orders combined
CREATE TABLE customer_orders (
    customer_id INT,
    customer_name VARCHAR(100),
    order_id INT,
    order_date DATE
);

The above example combines both customer and order information, improving performance but at the cost of data redundancy.

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17. Find the length of the longest substring without repeated characters given a string s.

In my experience, the problem of finding the length of the longest substring without repeating characters can be efficiently solved using the sliding window technique. I would use two pointers to traverse the string, one representing the start of the window and the other representing the end. As I go through the string, I move the end pointer to explore new characters and keep track of them. If I encounter a repeated character, I shift the start pointer to the right until the duplicate is removed from the window. The key is to keep updating the maximum length of the substring without repeating characters as I move through the string.

Here’s a Python code snippet that implements this sliding window approach:

def longest_substring(s):
    start, max_len = 0, 0
    seen = {}
    for end in range(len(s)):
        if s[end] in seen and seen[s[end]] >= start:
            start = seen[s[end]] + 1
        seen[s[end]] = end
        max_len = max(max_len, end - start + 1)
    return max_len

This code keeps track of the last seen index of each character using a dictionary seen. When a duplicate character is encountered, the start pointer is moved to the right of the previous occurrence of that character. The result is the longest substring without repeated characters.

18. Get the longest palindromic substring contained in s from a given string s.

When I think about finding the longest palindromic substring, I think about two primary approaches: expand around center and dynamic programming. In my experience, the “expand around center” technique is both time-efficient and easy to implement. The idea is that each character in the string can be the center of a potential palindrome. I expand outward from this center, checking if the substring is a palindrome, and keep track of the longest one found. This approach only requires a single pass over the string and has a time complexity of O(n^2).

Here’s an implementation in Python using the expand around center approach:

def longest_palindrome(s):
    def expand_center(left, right):
        while left >= 0 and right < len(s) and s[left] == s[right]:
            left -= 1
            right += 1
        return s[left+1:right]
    
    longest = ""
    for i in range(len(s)):
        palindrome1 = expand_center(i, i)  # Odd length palindrome
        palindrome2 = expand_center(i, i+1)  # Even length palindrome
        longest = max(longest, palindrome1, palindrome2, key=len)
    return longest

This function uses a helper function expand_center to find the longest palindrome centered at each position in the string. It checks both odd-length and even-length palindromes, updating the longest palindrome found. This approach efficiently finds the longest palindromic substring in the string by expanding from the center for each possible center in the string.

19. You are provided an n-length height integer array. There are n vertical lines made using (i, height[i]) as the two ends of the ith line. Find two lines that, when joined with the x-axis, produce a container with the most water inside.

This problem can be efficiently solved using the two-pointer technique. I start by placing one pointer at the beginning of the array (left) and the other at the end of the array (right). I calculate the area formed between the lines at these two pointers, and then I try to maximize the area. The key observation here is that the area between two lines is determined by the shorter of the two lines and the distance between them. Therefore, to maximize the area, I move the pointer pointing to the shorter line towards the other pointer, hoping to find a taller line that could potentially increase the area.

Here is a Python code snippet that implements this approach:

def max_area(height):
    left, right = 0, len(height) - 1
    max_area = 0
    while left < right:
        area = min(height[left], height[right]) * (right - left)
        max_area = max(max_area, area)
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    return max_area

In this code, left and right are the pointers that initially point to the first and last indices of the array, respectively. The function calculates the area between the lines at these positions and updates the maximum area found so far. Depending on which line is shorter, it moves the corresponding pointer inward, aiming to find a taller line to increase the area.

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20. Find three integers in the integer array nums of length n such that the total is as near to goal as possible. Give back the three integers’ sum. You may suppose that there is only one possible solution for each input.

For this problem, I would first sort the array to facilitate easier handling of the sum. After sorting, I use a combination of two-pointer and greedy approaches. The idea is to fix one element (say, at index i) and then use two pointers (left and right) to try to find two other numbers whose sum, along with the fixed element, is as close to the goal as possible. I move the left pointer to the right and the right pointer to the left to get closer to the goal, adjusting the pointers based on the current sum.

Here’s a Python implementation of this approach:

def threeSumClosest(nums, target):
    nums.sort()
    closest_sum = float('inf')
    for i in range(len(nums) - 2):
        left, right = i + 1, len(nums) - 1
        while left < right:
            total = nums[i] + nums[left] + nums[right]
            if abs(total - target) < abs(closest_sum - target):
                closest_sum = total
            if total < target:
                left += 1
            elif total > target:
                right -= 1
            else:
                return total
    return closest_sum

In this code, the array is sorted to make the process more efficient. The two-pointer approach checks all possible pairs that, combined with the fixed element, get closest to the target sum. The function returns the sum that is closest to the goal.

21. Return an array of all the distinct quadruplets [nums[a], nums[b], nums[c], nums[d]] from an array nums of n integers in the following way: A, B, C, and D are distinct if 0 = a, b, c, and d. target = nums[a] + nums[b] + nums[c] + nums[d].

For this problem, I would first sort the array to allow efficient checking of possible quadruplets using the two-pointer technique. Then, I iterate through each element, fixing it as one of the quadruplet elements, and use two more pointers to find the remaining three elements that sum up to the target. This reduces the problem from a four-pointer problem to a more manageable two-pointer problem.

Here’s the Python implementation:

def fourSum(nums, target):
    nums.sort()
    result = []
    for i in range(len(nums) - 3):
        if i > 0 and nums[i] == nums[i-1]:
            continue
        for j in range(i + 1, len(nums) - 2):
            if j > i + 1 and nums[j] == nums[j-1]:
                continue
            left, right = j + 1, len(nums) - 1
            while left < right:
                total = nums[i] + nums[j] + nums[left] + nums[right]
                if total == target:
                    result.append([nums[i], nums[j], nums[left], nums[right]])
                    while left < right and nums[left] == nums[left+1]:
                        left += 1
                    while left < right and nums[right] == nums[right-1]:
                        right -= 1
                    left += 1
                    right -= 1
                elif total < target:
                    left += 1
                else:
                    right -= 1
    return result

This implementation uses the sorted array to efficiently find the quadruplets. It also handles duplicates by skipping over repeated values of i and j, ensuring that only unique quadruplets are added to the result.

22. Return all potential subsets for an integer array with nums unique items (the power set). There cannot be any duplicate subsets in the solution set. You can return the answers in any order.

The problem of generating all possible subsets can be solved using a backtracking approach. I would start with an empty subset and recursively add each element from the array to it. After each addition, I would backtrack by removing the last added element and continue the process with the next elements. This ensures that all possible subsets are generated. Since the input has unique elements, there won’t be duplicate subsets.

Here’s the Python implementation of the backtracking approach:

def subsets(nums):
    result = []
    def backtrack(start, current_subset):
        result.append(list(current_subset))
        for i in range(start, len(nums)):
            current_subset.append(nums[i])
            backtrack(i + 1, current_subset)
            current_subset.pop()
    
    backtrack(0, [])
    return result

In this code, the backtrack function recursively explores all subsets by either including or excluding each element. The result is a list of all subsets, which includes the empty set and all non-empty combinations of the elements in nums.

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23. Return the number of structurally unique binary search trees (BSTs) that have exactly n nodes and unique values ranging from 1 to n, given an integer n.

This problem can be solved using dynamic programming and is based on the idea of Catalan numbers. The number of unique BSTs that can be formed with n nodes is given by the Catalan number formula:

Where C_0 is initialized as 1 (since there’s one empty tree). Each number from 1 to n can be the root of the tree, and the left and right subtrees must also be unique BSTs.
Here’s the Python code to compute the number of unique BSTs using dynamic programming:

def numTrees(n):
    dp = [0] * (n + 1)
    dp[0], dp[1] = 1, 1
    for i in range(2, n + 1):
        for j in range(1, i + 1):
            dp[i] += dp[j - 1] * dp[i - j]
    return dp[n]

In this code, the dp array stores the number of unique BSTs that can be formed with i nodes. For each i, the function calculates the number of unique trees by considering each number as the root and multiplying the number of trees formed by the left and right subtrees. The result is the number of structurally unique BSTs that can be formed with n nodes.

24. You are given two non-empty linked lists that each represent a non-negative integer. The digits are kept in reverse order, with each node containing only one digit. Add the two numbers together and return the total as a linked list.

This problem is essentially about adding two numbers represented by linked lists, where each node contains a digit and the digits are stored in reverse order. I would iterate through both linked lists, adding corresponding digits and taking care of any carry-over that results from a sum greater than 9. If there is a carry after the final addition, I create a new node for the carry.
Here’s how I would approach this problem:

• Start from the head of both lists.
• Add the digits from the corresponding nodes, and add any carry from the previous sum.
• If the sum is 10 or greater, set the carry to 1, and store the remainder in the current node.
• Continue this process until both linked lists have been fully processed. Here’s the Python implementation:

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def addTwoNumbers(l1, l2):
    dummy_head = ListNode()
    current = dummy_head
    carry = 0
    
    while l1 or l2 or carry:
        val1 = l1.val if l1 else 0
        val2 = l2.val if l2 else 0
        
        total = val1 + val2 + carry
        carry = total // 10
        current.next = ListNode(total % 10)
        
        current = current.next
        if l1: l1 = l1.next
        if l2: l2 = l2.next
    
    return dummy_head.next

# Example Input
l1 = ListNode(2, ListNode(4, ListNode(3)))  # Represents 342
l2 = ListNode(5, ListNode(6, ListNode(4)))  # Represents 465

result = addTwoNumbers(l1, l2)

# Output the result
while result:
    print(result.val, end=" ")
    result = result.next

Output:

7 0 8

Explanation

• I start by initializing a dummy node and a carry variable.
• I loop through both linked lists, adding corresponding digits and carry.
• After adding the digits, if the sum is greater than 9, I set the carry for the next addition.
• I return the resultant linked list after performing all additions.
• The linked list 342 + 465 results in 807 which is printed as 7 0 8.

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25. Reverse the nodes of the list from position left to position right, and then return the reversal list, given the head of a singly linked list and two numbers left and right, where left = right.

This problem involves reversing a section of a singly linked list between positions left and right. Since left and right are the same, it means we’re essentially reversing a single node, so no changes would be made.
For cases where left != right, here’s how I would solve it:

• Traverse the list until reaching the left node.
• Use a helper function to reverse the portion of the list between left and right.
• After the reversal, connect the modified section back to the original list.
  Here’s the Python code for this scenario:

def reverseBetween(head, left, right):
    if not head or left == right:
        return head
    
    dummy = ListNode(0)
    dummy.next = head
    prev = dummy
    
    for _ in range(left - 1):
        prev = prev.next
    
    current = prev.next
    for _ in range(right - left):
        temp = current.next
        current.next = temp.next
        temp.next = prev.next
        prev.next = temp
    
    return dummy.next

# Example Input
head = ListNode(1, ListNode(2, ListNode(3, ListNode(4, ListNode(5)))))
left, right = 2, 4

result = reverseBetween(head, left, right)

# Output the result
while result:
    print(result.val, end=" ")
    result = result.next

1 4 3 2 5

Explanation

• The input list is 1 -> 2 -> 3 -> 4 -> 5.
• I use a dummy node and a prev pointer to manage the reversal.
• I move prev to the node before left.
• Then, I reverse the section from left to right.
• Finally, the list becomes 1 -> 4 -> 3 -> 2 -> 5, as expected after the reversal.

26. Given an integer targetSum and a binary tree representing a root-to-leaf path that adds up all the values along the path, return true if and only if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

To solve this problem, I would use a depth-first search (DFS) approach. Starting from the root of the tree, I recursively subtract the node’s value from targetSum. If I reach a leaf node (a node with no children), I check if the value of the leaf node equals the remaining target sum. If it does, I return True. If no path is found that satisfies this condition, I return False.
Here’s the Python code for this solution:

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def hasPathSum(root, targetSum):
    if not root:
        return False
    if not root.left and not root.right:
        return root.val == targetSum
    return (hasPathSum(root.left, targetSum - root.val) or
            hasPathSum(root.right, targetSum - root.val))

# Example Input
root = TreeNode(5)
root.left = TreeNode(4)
root.right = TreeNode(8)
root.left.left = TreeNode(11)
root.left.left.left = TreeNode(7)
root.left.left.right = TreeNode(2)
root.right.left = TreeNode(13)
root.right.right = TreeNode(4)
root.right.right.right = TreeNode(1)

targetSum = 22

# Output the result
print(hasPathSum(root, targetSum))

True

Explanation

• I use a depth-first search (DFS) approach.
• Starting from the root, I subtract the node value from targetSum.
• If I reach a leaf node and targetSum equals the node’s value, I return True.
• The path 5 -> 4 -> 11 -> 2 gives the sum 22, so the output is True.

See also: Tech Mahindra FullStack Developer Interview Questions

27. To compute the product of two non-negative integers num1 and num2, string each of them.

To compute the product of two non-negative integers represented as strings, I would convert both strings to integers, perform the multiplication, and then return the result as a string. Since the numbers are represented as strings, it’s important to handle them properly and convert them back to string format after multiplication.

Here’s the Python code for this problem:

def multiply(num1, num2):
    # Convert the string numbers to integers
    num1 = int(num1)
    num2 = int(num2)
    
    # Multiply the numbers
    product = num1 * num2
    
    # Convert the result back to string and return it
    return str(product)

# Example Input
num1 = "123"
num2 = "456"

# Output the result
print(multiply(num1, num2))

Output:

56088

Explanation

• I first convert the string inputs num1 and num2 to integers.
• Then, I multiply the two integers together.
• Finally, I convert the result back to a string before returning it.
• The product of 123 and 456 is 56088, which is returned as a string.

28. To reverse only the vowels in a string s, let s be given. Only the vowels 'a', 'e', 'i', 'o', 'u' are reversed, and they may appear in any order.

Code

def reverseVowels(s):
    vowels = set("aeiouAEIOU")
    s = list(s)
    left, right = 0, len(s) - 1
    
    while left < right:
        while left < right and s[left] not in vowels:
            left += 1
        while left < right and s[right] not in vowels:
            right -= 1
        s[left], s[right] = s[right], s[left]
        left += 1
        right -= 1
    
    return "".join(s)

# Example Input
s = "hello"

# Output the result
print(reverseVowels(s))

Output:

holle

Explanation

• Convert the string into a list for easy swapping.
• Use two pointers (left and right) to find the vowels from both ends.
• Swap the vowels and move the pointers toward the center.
• Continue until all vowels are reversed while keeping non-vowels in place.
• For "hello", the vowels 'e' and 'o' are swapped, resulting in "holle".

29. Find the largest contiguous subarray (containing at least one number) of an integer array nums and return its value. An array subrange is a contiguous portion of an array.

Code

def maxSubArray(nums):
    max_sum = nums[0]
    current_sum = nums[0]
    
    for i in range(1, len(nums)):
        current_sum = max(nums[i], current_sum + nums[i])
        max_sum = max(max_sum, current_sum)
    
    return max_sum

# Example Input
nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4]

# Output the result
print(maxSubArray(nums))

Output:

6

Explanation

• Use Kadane’s Algorithm to track the maximum sum.
• Start with the first element as both current_sum and max_sum.
• Iterate through the array and decide whether to add the current number to current_sum or start fresh.
• Update max_sum whenever current_sum exceeds it.
• The subarray [4, -1, 2, 1] has the largest sum 6, so the output is 6.

See also: Top 50 Hibernate interview questions

30. A given number array nums must have unique elements in order to obtain all possible subsets (the power set). The solution set must not have duplicate subsets. Return the solution in any order.

Code

def subsets(nums):
    result = []
    
    def backtrack(start, path):
        result.append(path[:])
        for i in range(start, len(nums)):
            path.append(nums[i])
            backtrack(i + 1, path)
            path.pop()
    
    backtrack(0, [])
    return result

# Example Input
nums = [1, 2, 3]

# Output the result
print(subsets(nums))

Output:

[[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]]

Explanation

• Use backtracking to explore all possible subsets.
• Start with an empty list and recursively build subsets by adding elements.
• Use a loop to include each element and move forward.
• Remove the last added element (pop()) after recursion to explore other combinations.
• The subsets of [1,2,3] include all possible selections: {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}.

Qualcomm Interview Preparation

Preparing for a Qualcomm interview requires a strong grasp of technical concepts, problem-solving skills, and domain expertise. Qualcomm primarily focuses on embedded systems, software engineering, VLSI, and wireless technologies. Here’s a structured guide to help you prepare efficiently.

1. Understanding the Interview Process

The Qualcomm hiring process typically consists of the following rounds:

• Online Assessment (OA) – Covers DSA, C/C++, Python, and aptitude questions.
• Technical Interviews (2-3 Rounds) – Focuses on Data Structures, Algorithms, OS, Networks, C/C++, Embedded Systems, or VLSI (for hardware roles).
• System Design & Problem-Solving – For senior roles, involving designing scalable systems.
• HR/Behavioral Round – Questions on projects, leadership skills, and cultural fit.

2. Key Topics to Prepare

A. Data Structures & Algorithms (DSA)

• Arrays, Strings, Linked Lists
• Stack, Queue, Heap, Hashing
• Trees (BST, AVL), Graphs (BFS, DFS)
• Sorting (QuickSort, MergeSort), Searching Algorithms
• Dynamic Programming, Bit Manipulation

Example Question:
Find the longest palindromic substring in a given string.

def longestPalindrome(s):
    if not s:
        return ""
    start, max_len = 0, 1
    for i in range(len(s)):
        for j in range(i, len(s)):
            if s[i:j+1] == s[i:j+1][::-1] and (j-i+1) > max_len:
                start, max_len = i, j-i+1
    return s[start:start+max_len]

# Example Usage
print(longestPalindrome("babad"))

Output: "bab" (or "aba")
Explanation: We check each substring to find the longest palindrome.

See also: Infosys FullStack Interview Questions

B. Operating Systems (OS)

• Processes & Threads, IPC Mechanisms
• Memory Management, Paging, Segmentation
• Deadlocks, Synchronization (Mutex, Semaphore)
• File Systems & Scheduling Algorithms

Example Question:
What is a Mutex and how is it used?

• A Mutex (Mutual Exclusion) is used to prevent race conditions in multi-threaded programs.
• It allows only one thread to access the critical section at a time.
• Example using Python’s threading module:

import threading

lock = threading.Lock()
counter = 0

def increment():
    global counter
    lock.acquire()
    counter += 1
    lock.release()

t1 = threading.Thread(target=increment)
t2 = threading.Thread(target=increment)
t1.start()
t2.start()
t1.join()
t2.join()

Explanation:

• lock.acquire() ensures that only one thread can increment counter at a time.
• lock.release() allows the next thread to proceed.

C. Computer Networks

• TCP/IP, UDP, HTTP/HTTPS, DNS
• OSI & TCP/IP Model, Socket Programming
• Firewalls, NAT, VLANs, Load Balancing
• Wi-Fi, 5G, Bluetooth, Qualcomm Snapdragon Wireless Technologies

Example Question:
How does a Firewall work?

• A firewall filters incoming and outgoing traffic based on security rules.
• It can be packet-filtering, stateful, or proxy-based.
• It helps in preventing unauthorized access and attacks.

D. Embedded Systems (For Embedded Engineer Roles)

• Microcontrollers, ARM Architecture
• Memory Management (RAM, Flash, Cache)
• RTOS (Real-Time Operating Systems), Interrupts
• Serial Communication (UART, SPI, I2C)

Example Question:
What is an Interrupt in Embedded Systems?

• Interrupts allow the CPU to respond to events asynchronously.
• It suspends the current task, executes an Interrupt Service Routine (ISR), and resumes execution.
• Example of an interrupt in C (for microcontrollers):

#include <avr/interrupt.h>

ISR(TIMER1_COMPA_vect) {
    // Timer Interrupt Code
}

void setup() {
    cli(); // Disable global interrupts
    sei(); // Enable global interrupts
}

Explanation:

• ISR(TIMER1_COMPA_vect): Defines the Interrupt Service Routine for Timer1.
• cli() / sei(): Disable and enable interrupts globally.

See also: Microsoft Software Engineer Interview Questions

3. Behavioral & HR Questions

Apart from technical rounds, Qualcomm assesses communication, problem-solving, and leadership skills.

Common HR Questions:

• Why do you want to work at Qualcomm?
• Tell me about a challenging project you worked on.
• How do you handle tight deadlines?

4. Tips for Cracking the Qualcomm Interview

✅ Understand the job role – Whether Software, Embedded, VLSI, or Hardware, prepare accordingly.
✅ Solve coding problems daily – Use platforms like LeetCode, GeeksforGeeks, CodeChef.
✅ Practice system design – For senior roles, focus on scalability, load balancing, caching.
✅ Revise OS, Networks, DBMS – Qualcomm asks conceptual low-level programming questions.
✅ Work on Embedded projects – If applying for Embedded/VLSI, build projects using Microcontrollers, FPGA, ARM.
✅ Mock interviews – Practice DSA & problem-solving in a timed environment.

See also: Python Full-Stack Developer Interview Questions

Summing Up

Cracking a Qualcomm interview demands expertise in data structures, algorithms, embedded systems, and wireless communication. Strong coding skills in C, C++, or Python, along with problem-solving and system design knowledge, are essential. Hands-on experience and deep technical understanding set candidates apart in this competitive process.

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