Most common Adobe coding Interview questions

Securing a role at Adobe requires more than technical skills—it requires demonstrating your ability to innovate, collaborate, and contribute to a creative environment. As a leader in digital media and design software, Adobe values candidates who are technically proficient and passionate about bringing creative ideas to life.

In this blog, we’ll explore the common types of interview questions Adobe asks, from coding challenges to System Design problems and behavioral interviews, while providing helpful insights and preparation strategies to set you up for success.

Adobe coding interview questions

Adobe’s technical interviews are designed to evaluate your core programming knowledge, algorithmic thinking, and problem-solving abilities. These rounds typically involve challenges related to data structures, algorithms, and sometimes real-world scenarios that test your ability to write clean, efficient code. To prepare, it’s helpful to practice various coding problems—especially those that highlight common patterns like sliding windows, dynamic programming, and graph traversal.

The “Grokking the Coding Interview Patterns” course is a great resource that systematically teaches these problem-solving strategies.

Let’s dive into a few examples of questions typically seen in Adobe interviews.

1. Count pairs of similar strings

Problem statement: You’re given a 0-indexed array of strings, words. Two strings are considered similar if they contain the same set of characters, regardless of their order or frequency. Return the number of index pairs (i, j) such that 0 ≤ i < j < words.length and the two strings words[i] and words[j] are similar.

Solution: We first transform each word into a unique key by creating a sorted string of its distinct characters. This key allows us to identify words with the same set of characters. We then use a dictionary to count how many times each signature appears. For each new word, if its signature has already been seen, we add the current count of that signature to the total number of similar pairs, and then increment the signature’s count. This approach ensures we count all valid pairs without comparing every word to every other word.

Let’s look at the code for the algorithm we just discussed.

from collections import defaultdict

def similar_pairs(words):
    # Dictionary to count how many times each unique character set appears
    freq = defaultdict(int)
    
    # Variable to count the number of index pairs
    count = 0

for word in words:
        # Convert the word to a sorted set of characters
        word_set = set(word)
        
        # Sort the set and covert it to a tuple
        key = tuple(sorted(word_set))
        
        # For every previous word with the same character set, we can form a pair
        count += freq[key]
        
        # Update the frequency for this key
        freq[key] += 1

# Return count as the number of valid pairs
    return count
    
# Driver code
def main():
    words_arr = [
        ["abba", "aba", "cab", "acba"],
        ["stop", "tops", "post", "spot", "pots"],
        ["zyx", "wvu", "tsr"],
        ["a", "b", "a", "c", "b"],
        ["aabbcc", "cab", "bca", "ccccaaaabbbb"]
    ]

for i, words in enumerate(words_arr):
        print(f"{i + 1}.\twords: {words}\n")
        print(f"\tNumber of index pairs: {similar_pairs(words)}")
        print('-' * 100)

if __name__ == "__main__":
    main()

Count Pairs of Similar Strings

Code explanation:

Line 10: We iterate through each word in the words array:
- Line 12: We convert it to a set to remove duplicate characters.
- Line 15: Then, we sort the set and convert it to a tuple to be used as a dictionary key.
- Line 18: We add the current frequency of the key to the count (as all previous words with this same key can form valid pairs).
- Line 21: We increment the frequency of this key.

Time complexity: The time complexity is \( O(n) \), where \( n \) is the length of the words array. This is because we traverse the array once and perform constant-time lookups in the hash map.

Space complexity: The space complexity is \( O(n) \), as we may store up to \( n \) tuples in the hash map.

2. Top K frequent elements

Problem statement: Given an array of integers and an integer, k, return the k most frequent elements. You can return the answer in any order.

Solution: We maintain the top k elements in a min heap based on their frequency. Initially, a frequency map is created to count the occurrences of each element. Then, we use a min heap, which keeps elements ordered based on frequency. At first, up to k, elements are added to the min heap. When the size exceeds k, and the current element has a higher frequency than the top of the heap, we replace the least common element from the heap (top of the min heap) with the current element. This ensures that only the most frequent k elements remain in the min heap. By the end of the process, the min heap contains exactly the k most frequent elements.

Let’s look at the code for the algorithm we just discussed.

from collections import defaultdict
from heapq import heappush, heappop

def top_k_frequent(nums, k):
    # Dictionary to count how many times each number appears
    freq_map = defaultdict(int)
    
    # Iterate through the list and increment the count for each number
    for num in nums:
        freq_map[num] += 1

# Initialize an empty list to act as a min-heap
    min_heap = []
    
    # Iterate through each (num, freq) pair in the frequency map
    for num, freq in freq_map.items():
        # Push (freq, num) into the heap
        heappush(min_heap, (freq, num))
        
        # If the heap size exceeds k, remove the element with the lowest frequency
        if len(min_heap) > k:
            heappop(min_heap)

# Create an empty list result
    result = []
    
    # Iterate through each (freq, num) in the heap and append num to the result
    for freq, num in min_heap:
        result.append(num)

# Return the result list containing the k most frequent elements
    return result
    
# Driver code
def main():
    nums_arr = [[1, 3, 5, 12, 11, 12, 11, 12, 5],
                [1, 3, 5, 14, 18, 14, 5],
                [9, 8, 7, 6, 6, 5, 4, 3, 2, 1],
                [2, 4, 3, 2, 3, 4, 5, 4, 4, 4],
                [2, 3]]
                
    k = [3, 2, 1, 3, 2]

for i in range(len(k)):
        print(i+1, ".\tnums: ", nums_arr[i])
        print("\tk: ", k[i])
        print("\n\tTop", k[i], "frequent Elements: ",
              top_k_frequent(nums_arr[i], k[i]))
        print("-" * 100)

if __name__ == '__main__':
    main()

Top K Frequent Elements

Code explanation:

Lines 9–10: We iterate through nums and increment the count for each number.
Line 16: We iterate through each (num, freq) pair in freq:
- Line 18: We push (freq, num) into the heap.
- Lines 21–22: If the heap size exceeds k, we remove the top element with the lowest frequency.
Lines 28–29: We iterate through each (freq, num) in the heap and append num to the result.

Time complexity: The time complexity of this solution is \( O(n \log k) \), where \( n \) is the number of elements in the input array, and k is the number of top frequent elements to return. For each unique element, we perform a heap operation that takes \( O(\log k) \), and as there can be up to unique \( n \) elements, this part contributes \( O(n \log k) \) in total.

Space complexity: The space complexity is \( O(n) \) because we store the frequency of each number in a hash map, which may have up to \( n \) unique entries.

3. Minimum window substring

Problem statement: Given two strings, s and t, return the minimum window substring of s that contains all characters from t, including duplicates. If no such substring exists, return an empty string.

Solution: We use a sliding window to efficiently track the smallest possible substring in s that contains all characters from t. We first count the frequency of each character needed from t, and as we slide a window across s, we keep track of how many required characters we have in the current window. Once the window contains all characters in the right amounts, we try to shrink it from the left to find the smallest valid window. By adjusting the window size dynamically, we process each character only a few times while ensuring the minimum window is found.

Let’s look at the code for the algorithm we just discussed.

def min_window(s: str, t: str) -> str:
    # If `t` is empty, return an empty string as no window is possible
    if not t:
        return ""

# Dictionaries to store the required character counts and the current window's character counts
    req_count = {}
    window = {}

# Populate `req_count` with the character frequencies of `t`
    for char in t:
        req_count[char] = req_count.get(char, 0) + 1

# Variables to track the number of characters that match the required frequencies
    current = 0  # Count of characters in the current window that meet the required frequency
    required = len(req_count)  # Total number of unique characters in `t`

# Result variables to track the best window
    res = [-1, -1]  # Stores the start and end indices of the minimum window
    res_len = float("inf")  # Length of the minimum window

# Left pointer of the window
    left = 0  
    for right in range(len(s)):
        char = s[right]

# If `char` is in `t`, update the window count
        if char in req_count:
            window[char] = window.get(char, 0) + 1
            
            # If the frequency of `char` in the window matches the required frequency, update `current`
            if window[char] == req_count[char]:
                current += 1

# Try to contract the window while all required characters are present
        while current == required:
            # Update the result if the current window is smaller than the previous best
            if (right - left + 1) < res_len:
                res = [left, right]
                res_len = (right - left + 1)

left_char = s[left]
            
            if left_char in req_count:
                # Decrement the count of `left_char` in the window
                window[left_char] -= 1
                
                # If the frequency of `left_char` in the window is less than required, update `current`
                if window[left_char] < req_count[left_char]:
                    current -= 1
            
            left += 1  # Move the left pointer to shrink the window

# Return the minimum window if found, otherwise return an empty string
    return s[res[0]:res[1] + 1] if res_len != float("inf") else ""
    
# Driver code
def main():
    s_arr = ["PATTERN", "LIFE", "ABRACADABRA", "STRIKER", "DFFDFDFVD"]
    t_arr = ["TN", "I", "ABC", "RK", "VDD"]

for i in range(len(s_arr)):
        print(f"{i + 1}.\ts: {s_arr[i]}\n\tt: {t_arr[i]}\n\n\tThe minimum substring containing {t_arr[i]} is: {min_window(s_arr[i], t_arr[i])}")
        print("-" * 100)

if __name__ == '__main__':
    main()

Minimum Window Substring

Code explanation:

Lines 11–12: Populate req_count with the character frequencies of t.
Line 23: We set the left pointer of the window to 0.
Line 24: We also set the right pointer to 0, which iterates till the end of s.
- Lines 28–29: If the character at right is in t, update the current window’s character counts in window.
- Lines 32–33: If the frequency of char in the window matches the required frequency, update the current.
Lines 36: We try to contract the window while all required characters are present.
Lines 38–40: We update the result variables if the current window is smaller than the previous best.
Lines 44: If the character at left is in req_count:
- Lines 46: We decrement the count of that character in the window.
- Lines 49–50: If the frequency of that character in the window is less than required, update the current.
Lines 52: We move the left pointer to shrink the window.

Time complexity: The time complexity of the algorithm is \( O(n+m) \), where \( n \) is the length of the string s and \( m \) is the length of the string t. This complexity arises because the algorithm processes each character of s exactly once during the sliding window traversal, and the preprocessing step to build the hash map req_count for string t takes \( O(m) \).

Space complexity: As the characters in t are limited to uppercase and lowercase English letters, there is a maximum of \( 52 \) possible characters. Therefore, the size of the req_count and window hash maps will be at most \( 52 \), regardless of the length of t. Therefore, the space complexity of this solution is \( O(1) \).

System Design interview questions at Adobe

Adobe’s System Design interviews focus on your ability to architect systems that meet real-world demands. You’ll work through designing platforms that can grow with users, remain reliable under pressure, and deliver a seamless experience tailored to creative and collaborative work. The interview emphasizes clear reasoning about your choices and how you handle challenges such as performance limits and system failures.

The “Grokking the Modern System Design Interview” course is an excellent resource to help sharpen these skills.

1. Design a scalable cloud storage system

Problem statement: Design a system like Google Drive or Dropbox that allows users to upload, store, retrieve, and share files over the cloud. The system should be scalable enough to support millions of users, have high availability, and have effective data management.

Key features:

Allow users to upload and download files of varying types and sizes.
Support file sharing with other users via public/private links or access permissions.
Enable folder organization, versioning, and metadata handling.
Optional: Support collaborative features like simultaneous file editing.

Design highlights:

Use object storage solutions like Amazon S3 or Google Cloud Storage for scalable and cost-effective file storage.
Use a metadata service to track file ownership, access permissions, version history, and folder structures.
Implement load balancers and CDNs (content delivery networks) to handle high request volumes and reduce latency.
Shard metadata and user data across multiple servers or databases for better scalability and reduced bottlenecks.

2. Design a real-time collaboration tool

Problem statement: Design a real-time collaborative design platform (similar to Adobe XD or Figma) where multiple users can edit the same project simultaneously. The system should support live syncing of changes, version control, and real-time communication between users.

Key features:

Enable multiple users to view and edit designs in real time.
Maintain consistency and synchronization across all clients during collaboration.
Provide version control and undo/redo capabilities.
Allow commenting and annotations within projects.
Optional: Integrate voice/video chat, asset libraries, and plug-ins.

Design highlights:

Store design documents as structured data (e.g., JSON trees) in a distributed database like MongoDB or Firebase.
Use a central document server to track all changes and broadcast real-time updates to connected clients.
Persist snapshots of the document periodically and store deltas (incremental changes) for optimized version history.
Enable access control and user roles (e.g., viewer, editor) using an authentication system like OAuth and permission management.

3. Design an API rate limiter

Problem statement: Build a system that limits the number of requests a user can make in a given period.

Key features:

Prevent abuse by limiting requests based on time windows (e.g., 100 requests per minute).
Provide immediate feedback when limits are exceeded.
Optional: Different limits for user tiers (e.g., free vs. premium users).

Design highlights:

Use algorithms like the token bucket or leaky bucket.
Store counters in a distributed cache (e.g., Redis).
Distribute rate-limiting state across multiple servers using a shared cache like Redis.

Additional System Design questions

Here are some additional System Design questions that assess your ability to architect scalable and efficient systems:

Design an online code editor (like Adobe Brackets or CodePen): Create a collaborative coding environment that supports real-time updates, file management, and version control across users and sessions.
Design a search engine for Adobe’s asset libraries: Create a scalable and fast search system that indexes design assets (icons, fonts, templates) and provides real-time suggestions and filters based on user queries, categories, and metadata.
Design an image processing pipeline (relevant to Adobe’s creative tools): Build a backend service to handle media uploads and asynchronous image processing tasks such as resizing, compression, and format conversion.
Design a web caching system: Implement a system that accelerates data retrieval using cache layers, LRU eviction policies, and integration with content delivery networks (CDNs).
Design a URL shortener: Build a service that converts long URLs into short links with features like redirection, analytics tracking, and scalability using hashing and persistent storage.

Behavioral interview questions

At Adobe, behavioral interviews explore how you navigate challenges, collaborate with others, and help build a culture that thrives on creativity and innovation. Interviewers will focus on real experiences from your past, looking at moments where you resolved conflicts, made tough decisions, took initiative, or adjusted to new situations. As Adobe strongly emphasizes innovation and working across teams, your responses should highlight how you embody these values. Structuring your answers with the STAR method (situation, task, action, result) helps ensure clarity and impact. Below are some example questions to help you practice effectively.

1. Collaborating across diverse teams

Question: Tell me about a time when you had to work with cross-functional teams to achieve a shared goal.

STAR answer:

Situation: I was tasked with leading a project that required collaboration between several departments to deliver a product on time.
Task: My goal was to align all teams, gather feedback, and ensure the product was ready for launch with input from all relevant stakeholders.
Action: I facilitated regular meetings, clarified roles, and ensured open communication to address all teams’ concerns.
Result: The project was delivered successfully, meeting the target deadline and receiving positive feedback from internal and external stakeholders.

2. Managing multiple priorities

Question: Describe a time when you had to manage multiple competing priorities and how you handled it.

STAR answer:

Situation: I had to juggle multiple projects, each with varying deadlines and levels of urgency.
Task: I needed to determine which tasks were most critical and prioritize them without sacrificing the quality of my work.
Action: I assessed the urgency of each task, communicated with stakeholders, and delegated tasks to ensure effective handling.
Result: I successfully met all deadlines while maintaining the quality of work, and all stakeholders were satisfied with the outcomes.

3. Overcoming a technical challenge

Question: Can you describe a time when you faced a significant technical challenge and how you overcame it?

STAR answer:

Situation: I encountered a significant performance issue in a system affecting user experience.
Task: I was responsible for resolving the issue without affecting the system’s core functionality.
Action: I analyzed the problem, collaborated with the technical team, and implemented an optimized solution to improve system performance.
Result: The performance issue was resolved, improving user satisfaction and overall system efficiency.

Additional behavioral questions

Influencing stakeholders: Describe when you convinced a stakeholder to support your idea or project. How did you approach it, and what was the outcome?
Handling ambiguity: Share an example of when you had to work in a situation with limited or unclear instructions. How did you navigate through it, and what was the result?
Driving innovation: Tell me about when you introduced an innovative solution to a problem. What was the challenge, and how did your solution make an impact?
Learning from mistakes: Describe a situation where you made a mistake in your work. How did you handle it, and what did you learn from it?
Achieving results under pressure: Tell me about when you had to deliver a project under tight deadlines. How did you manage your time and resources to meet the goal?

Check out this blog on behavioral interviews and tips on preparing and excelling.

Level up your skills with AI-driven mock interviews

As you strengthen your problem-solving and coding skills, preparing for the real-world pressure of interviews is equally important. One effective way to bridge the gap between practice and performance is through an AI-powered mock interview. These realistic simulations mirror actual interview conditions and offer immediate feedback on both technical and behavioral responses. With tailored insights and performance tips, you can pinpoint areas for improvement, boost your confidence, and walk into your interviews better prepared than ever.

Summary

Cracking Adobe’s interviews takes more than just solving coding challenges. It requires a well-rounded strategy that includes mastering core problem-solving patterns, gaining a solid grasp of System Design concepts, and crafting thoughtful, experience-backed responses to behavioral questions. With focused preparation and a structured approach, you’ll be well-equipped to make a strong impression and move closer to landing your role at Adobe.

Boost your preparation with these hand-picked resources:

Company Interview Questions

Most common Adobe coding Interview questions

Adobe coding interview questions

1. Count pairs of similar strings

2. Top K frequent elements

3. Minimum window substring

More Adobe coding interview questions

System Design interview questions at Adobe

1. Design a scalable cloud storage system

2. Design a real-time collaboration tool

3. Design an API rate limiter

Additional System Design questions

Behavioral interview questions

1. Collaborating across diverse teams

2. Managing multiple priorities

3. Overcoming a technical challenge

Additional behavioral questions

Level up your skills with AI-driven mock interviews

Summary

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