Adobe Coding Round Questions 2026: Patterns + Solutions

Adobe's coding rounds reward correct, well-explained solutions and trip up candidates on output-prediction traps. This guide gives you the repeating patterns, worked solutions, and the snippet-reading skills the OA demands.

Adobe Coding Round Topics

Compiled from candidate reports across 2023 to 2025; relative frequency, not an official syllabus:

Relative frequencies from candidate aggregates (2023 to 2025), labelled approximate. Adobe pairs DSA with CS-fundamentals snippet reading.

Worked Solution 1: Linked Lists

Problem: Reverse a singly linked list.

Approach: Iterative pointer reversal. Walk the list, redirecting each node's next pointer to its predecessor.

def reverse_list(head):
    prev = None
    curr = head
    while curr:
        nxt = curr.next
        curr.next = prev
        prev = curr
        curr = nxt
    return prev

Complexity: O(n) time, O(1) space. Adobe often follows up with "reverse in groups of k" or "do it recursively", so know both variants.

Worked Solution 2: Linked List Cycle

Problem: Detect whether a linked list has a cycle.

Approach: Floyd's tortoise and hare. Advance one pointer by one and another by two; if they meet, there is a cycle.

def has_cycle(head):
    slow = fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
        if slow is fast:
            return True
    return False

Complexity: O(n) time, O(1) space. The follow-up "find the cycle's start node" uses the same two pointers with a second phase from the head.

Worked Solution 3: Trees (Recursion)

Problem: Given a binary tree, return its maximum depth.

Approach: Recursive depth-first. A node's depth is one plus the larger of its children's depths.

def max_depth(root):
    if not root:
        return 0
    return 1 + max(max_depth(root.left), max_depth(root.right))

Complexity: O(n) time, O(h) recursion stack where h is the height. Adobe likes recursion, so be ready to convert this to an iterative BFS or DFS if asked about deep trees and stack overflow.

Worked Solution 4: Dynamic Programming

Problem: Given two strings, find the length of their longest common subsequence.

Approach: Classic 2D DP. dp[i][j] is the LCS length of the first i characters of one string and the first j of the other.

def lcs(a, b):
    m, n = len(a), len(b)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if a[i - 1] == b[j - 1]:
                dp[i][j] = dp[i - 1][j - 1] + 1
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
    return dp[m][n]

Complexity: O(m times n) time and space. The space can be reduced to O(n) with two rows, a common Adobe optimisation follow-up.

Worked Solution 5: Strings

Problem: Check whether a string is a valid palindrome, considering only alphanumeric characters and ignoring case.

Approach: Two pointers from both ends, skipping non-alphanumeric characters and comparing case-insensitively.

def is_palindrome(s):
    left, right = 0, len(s) - 1
    while left < right:
        while left < right and not s[left].isalnum():
            left += 1
        while left < right and not s[right].isalnum():
            right -= 1
        if s[left].lower() != s[right].lower():
            return False
        left += 1
        right -= 1
    return True

Complexity: O(n) time, O(1) space. The two-pointer-with-skip pattern recurs across Adobe string problems.

Output-Prediction Traps in the OA

Adobe's OA includes "what does this print" questions designed to catch shallow understanding. Common traps:

• Operator precedence and associativity: mixing increment, assignment, and arithmetic.
• Pointer vs value semantics: passing by reference vs by value in C/C++/Java.
• Integer overflow and division: integer division truncation, overflow wraparound.
• Static vs instance behaviour: static variables shared across calls.
• Short-circuit evaluation: side effects that do or do not run in logical expressions.

Practise by predicting output for tricky snippets before running them, then verifying. The discipline of reading code precisely is exactly what these questions reward.

How to Prepare for Adobe Coding Rounds

• Drill arrays, strings, linked lists, trees, recursion, and DP until automatic.
• Practise output-prediction questions in your OA language until the traps are familiar.
• Be ready to explain every solution clearly; Adobe weights explanation, not just passing tests.
• Combine coding prep with CS-fundamentals revision, since Adobe interleaves them.

See Adobe interview process 2026 for the full loop.

More Worked Solutions

Worked Solution 6: Merge Two Sorted Linked Lists

Splice the smaller head each step using a dummy node.

def merge(l1, l2):
    dummy = tail = ListNode(0)
    while l1 and l2:
        if l1.val <= l2.val:
            tail.next, l1 = l1, l1.next
        else:
            tail.next, l2 = l2, l2.next
        tail = tail.next
    tail.next = l1 or l2
    return dummy.next

Time O(n + m), space O(1). A common Adobe linked-list question; the dummy head avoids special-casing the first node.

Worked Solution 7: Kadane's Maximum Subarray

Track the best sum ending at each index, resetting when it turns negative.

def max_subarray(nums):
    best = cur = nums[0]
    for x in nums[1:]:
        cur = max(x, cur + x)
        best = max(best, cur)
    return best

Time O(n), space O(1). Be ready to return the subarray itself or handle all-negative input.

Worked Solution 8: Detect and Remove Loop in a Linked List

Use Floyd's cycle detection, then find the loop start by resetting one pointer to the head.

def detect_cycle(head):
    slow = fast = head
    while fast and fast.next:
        slow, fast = slow.next, fast.next.next
        if slow is fast:
            ptr = head
            while ptr is not slow:
                ptr, slow = ptr.next, slow.next
            return ptr
    return None

Time O(n), space O(1). The two-phase trick (meet, then walk from head) finds the entry node.

Worked Solution 9: Climbing Stairs (DP)

Count ways to climb n stairs by 1 or 2 steps; this is Fibonacci.

def climb(n):
    a, b = 1, 1
    for _ in range(n):
        a, b = b, a + b
    return a

Time O(n), space O(1).

Pattern and Complexity Reference

A useful way to prepare for Adobe's coding rounds is to internalise the small set of patterns that recur and the complexity each implies, so that when you see a problem you immediately reach for the right tool. For arrays and strings, the two-pointer and sliding-window patterns solve most subarray and substring problems in linear time, replacing a naive O(n squared) scan. For lookups, frequency counts, and deduplication, a hash map gives O(1) average access and turns many quadratic problems linear at the cost of O(n) space. For linked lists, the core skills are careful pointer manipulation (reversal), the fast-and-slow two-pointer idiom (cycle detection and finding the middle), and merging.

For trees, recursion is the default tool: most problems reduce to "solve for the children, then combine", which is O(n) time and O(h) stack space where h is the height. Breadth-first search with a queue handles level-order problems and shortest paths in unweighted graphs. For optimisation and counting problems with overlapping subproblems, dynamic programming applies; recognise it by the presence of a choice at each step and a result that depends on smaller subproblems, and remember that many one-dimensional DPs can be reduced from O(n) space to O(1) with rolling variables, an optimisation Adobe interviewers like to ask for.

Knowing the expected complexity also lets you read constraints as hints. If the input size can be up to roughly ten to the fifth, an O(n squared) solution will likely time out and you should aim for O(n log n) or better. If it is small, say up to a few hundred, a quadratic or even exponential-with-pruning approach may be acceptable. Training yourself to glance at the constraints and infer the target complexity before you start coding is one of the highest-leverage habits for any coding round, and it is especially useful at Adobe where clean, appropriately efficient solutions are valued over brute force.

How to Read Output-Prediction Questions Like Adobe Wants

Output-prediction questions are where Adobe separates candidates who have truly understood a language from those who have only used it, and they are worth dedicated practice because they are quick marks if you are disciplined and quick losses if you are not. The skill is not memorising outputs; it is tracing execution precisely, statement by statement, the way a compiler would.

The traps cluster into a few families. Operator precedence and associativity questions mix increment, assignment, and arithmetic, and the answer often hinges on whether an increment is pre or post and in what order sub-expressions evaluate. In C and C++ specifically, when multiple side effects happen between sequence points, the behaviour can be unspecified or undefined, and the correct answer is to say so rather than to guess a number. Integer questions exploit truncating division, overflow wraparound, and signed-versus-unsigned comparison surprises. Object-oriented snippets test static versus instance members (a static variable shared across calls is a favourite), constructor and destructor order, and virtual dispatch. Short-circuit evaluation questions check whether a side effect inside a logical expression actually runs.

The practical method is to slow down and annotate. For each line, write the current value of every variable involved, evaluate sub-expressions in the correct order, and only then compute the result. When you see something that looks like undefined behaviour, name it as such; Adobe values the candidate who recognises the trap over the one who confidently produces a wrong, non-portable number. Build this habit by predicting the output of ten tricky snippets a day and then running them to check, and within a couple of weeks the common traps become instantly recognisable.

Why Candidates Lose Marks in the Adobe Coding Round

Candidate reports point to recurring reasons strong candidates underperform:

• Failing output-prediction traps. Mispredicting pointer, precedence, or overflow behaviour costs easy marks; practise reading code precisely.
• Not explaining the solution. Adobe weights clear explanation; producing code silently loses signal.
• Weak linked-list pointer handling. Off-by-one and lost-pointer bugs in reversal and cycle problems are common.
• Skipping edge cases. Empty input, single node, and duplicates are frequently the hidden tests.
• Treating coding as separate from CS fundamentals. Adobe interleaves them; expect a fundamentals question attached to a coding problem.

Preparation Timeline (6 to 8 Weeks)

• Weeks 1 to 2: Foundations. Arrays, strings, and linked lists, plus daily output-prediction practice in your OA language.
• Weeks 3 to 4: Core DSA. Trees, recursion, and basic DP, with attention to clean implementation and explanation.
• Weeks 5 to 6: Depth. More medium problems, the linked-list variants (reverse in k, cycle start), and CS-fundamentals revision alongside.
• Weeks 7 to 8: Mocks. Timed mock coding sessions narrating out loud, plus targeted output-prediction drills.

Related Resources

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• Adobe placement papers 2026, solved past-drive papers
• Adobe India placement papers 2026, India-specific solved papers
• 7-day coding round crash plan 2026, last-week prep
• System design interview questions freshers 2026, design fundamentals
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