Adobe India Placement Papers 2026

Meta Description: Adobe India placement papers 2026 with latest exam pattern, coding questions, interview tips, and preparation strategy. Crack Adobe SDE roles with PapersAdda!

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Introduction

Adobe India is a premier destination for software engineers, especially those interested in creative technology, cloud services, and digital experience platforms. With major development centers in Noida and Bangalore, Adobe offers exciting opportunities in products like Photoshop, Acrobat, Experience Cloud, and Creative Cloud. This guide provides complete information about Adobe India's 2026 placement process, including exam patterns, coding questions, and preparation strategies.

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Adobe India Exam & Interview Pattern 2026

Round	Type	Duration	Topics Covered
Round 1	Online Assessment	90-120 mins	Aptitude, Coding, Technical MCQs
Round 2	Technical Interview 1	60 mins	Coding, Data Structures, Algorithms
Round 3	Technical Interview 2	60 mins	Advanced coding, System design basics
Round 4	Technical/Managerial	45-60 mins	Project discussion, Problem-solving
Round 5	HR Interview	30-45 mins	Behavioral, Culture fit, Expectations

Key Highlights:

• Online test includes aptitude, coding (2-3 questions), and technical MCQs
• Heavy focus on C++ and Java programming
• Questions often related to graphics, image processing, and algorithms
• Puzzle-solving questions are common

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Practice Questions with Detailed Answers

Question 1: Matrix Spiral Order

Problem: Given an m x n matrix, return all elements in spiral order.

Solution:

def spiralOrder(matrix):
    if not matrix:
        return []
    
    result = []
    top, bottom = 0, len(matrix) - 1
    left, right = 0, len(matrix[0]) - 1
    
    while top <= bottom and left <= right:
        # Traverse right
        for i in range(left, right + 1):
            result.append(matrix[top][i])
        top += 1
        
        # Traverse down
        for i in range(top, bottom + 1):
            result.append(matrix[i][right])
        right -= 1
        
        # Traverse left
        if top <= bottom:
            for i in range(right, left - 1, -1):
                result.append(matrix[bottom][i])
            bottom -= 1
        
        # Traverse up
        if left <= right:
            for i in range(bottom, top - 1, -1):
                result.append(matrix[i][left])
            left += 1
    
    return result

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Question 2: Set Matrix Zeroes

Problem: Given an m x n matrix, if an element is 0, set its entire row and column to 0.

Solution:

def setZeroes(matrix):
    if not matrix:
        return
    
    rows, cols = len(matrix), len(matrix[0])
    first_row_zero = any(matrix[0][j] == 0 for j in range(cols))
    first_col_zero = any(matrix[i][0] == 0 for i in range(rows))
    
    # Use first row and column as markers
    for i in range(1, rows):
        for j in range(1, cols):
            if matrix[i][j] == 0:
                matrix[i][0] = 0
                matrix[0][j] = 0
    
    # Set zeroes based on markers
    for i in range(1, rows):
        for j in range(1, cols):
            if matrix[i][0] == 0 or matrix[0][j] == 0:
                matrix[i][j] = 0
    
    if first_row_zero:
        for j in range(cols):
            matrix[0][j] = 0
    
    if first_col_zero:
        for i in range(rows):
            matrix[i][0] = 0

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Question 3: Rotate Image

Problem: You are given an n x n 2D matrix representing an image, rotate it by 90 degrees clockwise.

Solution:

def rotate(matrix):
    n = len(matrix)
    
    # Transpose
    for i in range(n):
        for j in range(i, n):
            matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
    
    # Reverse each row
    for i in range(n):
        matrix[i].reverse()

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Question 4: Word Search

Problem: Given an m x n grid of characters and a word, find if the word exists in the grid.

Solution:

def exist(board, word):
    rows, cols = len(board), len(board[0])
    
    def dfs(r, c, index):
        if index == len(word):
            return True
        if r < 0 or c < 0 or r >= rows or c >= cols or board[r][c] != word[index]:
            return False
        
        temp = board[r][c]
        board[r][c] = '#'
        
        found = (dfs(r+1, c, index+1) or
                 dfs(r-1, c, index+1) or
                 dfs(r, c+1, index+1) or
                 dfs(r, c-1, index+1))
        
        board[r][c] = temp
        return found
    
    for r in range(rows):
        for c in range(cols):
            if dfs(r, c, 0):
                return True
    return False

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Question 5: Longest Palindromic Substring

Problem: Given a string s, return the longest palindromic substring in s.

Solution:

def longestPalindrome(s):
    if not s:
        return ""
    
    start, max_len = 0, 1
    
    def expandAroundCenter(left, right):
        while left >= 0 and right < len(s) and s[left] == s[right]:
            left -= 1
            right += 1
        return right - left - 1
    
    for i in range(len(s)):
        len1 = expandAroundCenter(i, i)
        len2 = expandAroundCenter(i, i + 1)
        length = max(len1, len2)
        
        if length > max_len:
            max_len = length
            start = i - (length - 1) // 2
    
    return s[start:start + max_len]

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Question 6: Regular Expression Matching

Problem: Implement regular expression matching with support for '.' and '*'.

Solution:

def isMatch(s, p):
    m, n = len(s), len(p)
    dp = [[False] * (n + 1) for _ in range(m + 1)]
    dp[0][0] = True
    
    for j in range(2, n + 1):
        if p[j-1] == '*':
            dp[0][j] = dp[0][j-2]
    
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if p[j-1] == '.' or p[j-1] == s[i-1]:
                dp[i][j] = dp[i-1][j-1]
            elif p[j-1] == '*':
                dp[i][j] = dp[i][j-2]
                if p[j-2] == '.' or p[j-2] == s[i-1]:
                    dp[i][j] = dp[i][j] or dp[i-1][j]
    
    return dp[m][n]

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Question 7: Container With Most Water

Problem: Given n non-negative integers representing heights, find two lines that together with the x-axis form a container that holds the most water.

Solution:

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

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Question 8: Jump Game

Problem: Given an array of non-negative integers, you are initially at the first index. Determine if you can reach the last index.

Solution:

def canJump(nums):
    max_reach = 0
    
    for i in range(len(nums)):
        if i > max_reach:
            return False
        max_reach = max(max_reach, i + nums[i])
    
    return True

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Question 9: Subsets

Problem: Given an integer array of unique elements, return all possible subsets.

Solution:

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

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Question 10: LRU Cache

Problem: Design a data structure that follows the constraints of a Least Recently Used (LRU) cache.

Solution:

class DLinkedNode:
    def __init__(self, key=0, value=0):
        self.key = key
        self.value = value
        self.prev = None
        self.next = None

class LRUCache:
    def __init__(self, capacity):
        self.cache = {}
        self.capacity = capacity
        self.size = 0
        self.head = DLinkedNode()
        self.tail = DLinkedNode()
        self.head.next = self.tail
        self.tail.prev = self.head
    
    def _add_node(self, node):
        node.prev = self.head
        node.next = self.head.next
        self.head.next.prev = node
        self.head.next = node
    
    def _remove_node(self, node):
        prev = node.prev
        new = node.next
        prev.next = new
        new.prev = prev
    
    def _move_to_head(self, node):
        self._remove_node(node)
        self._add_node(node)
    
    def _pop_tail(self):
        res = self.tail.prev
        self._remove_node(res)
        return res
    
    def get(self, key):
        node = self.cache.get(key)
        if not node:
            return -1
        self._move_to_head(node)
        return node.value
    
    def put(self, key, value):
        node = self.cache.get(key)
        if not node:
            new_node = DLinkedNode(key, value)
            self.cache[key] = new_node
            self._add_node(new_node)
            self.size += 1
            
            if self.size > self.capacity:
                tail = self._pop_tail()
                del self.cache[tail.key]
                self.size -= 1
        else:
            node.value = value
            self._move_to_head(node)

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Question 11: Find Peak Element

Problem: A peak element is greater than its neighbors. Find any peak element's index in O(log n) time.

Solution:

def findPeakElement(nums):
    left, right = 0, len(nums) - 1
    
    while left < right:
        mid = (left + right) // 2
        if nums[mid] > nums[mid + 1]:
            right = mid
        else:
            left = mid + 1
    
    return left

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Question 12: Kth Largest Element in Array

Problem: Find the kth largest element in an unsorted array.

Solution:

import heapq

def findKthLargest(nums, k):
    return heapq.nlargest(k, nums)[-1]

# Quickselect approach
import random

def findKthLargestQuick(nums, k):
    def partition(left, right, pivot_idx):
        pivot = nums[pivot_idx]
        nums[pivot_idx], nums[right] = nums[right], nums[pivot_idx]
        store_idx = left
        
        for i in range(left, right):
            if nums[i] < pivot:
                nums[store_idx], nums[i] = nums[i], nums[store_idx]
                store_idx += 1
        
        nums[right], nums[store_idx] = nums[store_idx], nums[right]
        return store_idx
    
    def select(left, right, k_smallest):
        if left == right:
            return nums[left]
        
        pivot_idx = random.randint(left, right)
        pivot_idx = partition(left, right, pivot_idx)
        
        if k_smallest == pivot_idx:
            return nums[k_smallest]
        elif k_smallest < pivot_idx:
            return select(left, pivot_idx - 1, k_smallest)
        else:
            return select(pivot_idx + 1, right, k_smallest)
    
    return select(0, len(nums) - 1, len(nums) - k)

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Preparation Strategy

Step 1: Aptitude & Technical MCQs (Week 1)

• Practice quantitative aptitude (percentages, ratios, time-speed-distance)
• Review C++ and Java fundamentals
• Study OS, DBMS, and networking concepts

Step 2: Core Programming (Weeks 2-4)

• Master pointers and memory management (C++)
• Practice object-oriented programming concepts
• Solve array and string manipulation problems

Step 3: Data Structures (Weeks 5-7)

• Arrays, Matrices, and 2D operations
• Linked Lists, Stacks, Queues
• Trees and Graphs
• Hash Tables and Heaps

Step 4: Algorithms (Weeks 8-10)

• Sorting and Searching
• Dynamic Programming
• Backtracking
• Greedy Algorithms

Step 5: Mock Tests & Revision (Weeks 11-12)

• Take timed practice tests
• Solve previous year Adobe questions
• Focus on puzzle-solving skills

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Resources & Books to Follow

Coding Platforms

• LeetCode - Practice medium-hard problems
• GeeksforGeeks - Adobe-specific questions
• HackerRank - Aptitude and coding practice

Recommended Books

• "Cracking the Coding Interview" by Gayle Laakmann McDowell
• "Data Structures and Algorithms Made Easy" by Narasimha Karumanchi
• "Let Us C++" by Yashavant Kanetkar
• "Head First Design Patterns"

Adobe-Specific Preparation

• Study Adobe's product portfolio
• Learn basics of image processing
• Understand PDF and document management concepts

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Adobe India Salary & Package Details 2026

Role	Level	Experience	CTC (LPA)
Member of Technical Staff I	MTS 1	0-2 years	₹15-22 LPA
Member of Technical Staff II	MTS 2	2-5 years	₹25-40 LPA
Computer Scientist	CS	5-8 years	₹45-65 LPA
Senior Computer Scientist	Sr. CS	8+ years	₹70-100 LPA

Benefits:

• Adobe Stock (RSUs)
• Health and wellness programs
• Creative Cloud subscription
• Learning and development budget
• Flexible work arrangements

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Conclusion

Adobe India offers a unique blend of creative and technical challenges. The interview process tests both your coding skills and your ability to think creatively. Focus on matrix operations, string algorithms, and image-related problems. Prepare well for the aptitude section and showcase your passion for building products that empower creativity.

Visit PapersAdda.com for more placement papers, interview experiences, and job preparation resources!

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Keywords: Adobe placement papers 2026, Adobe interview questions, Adobe coding test, Adobe recruitment process, Adobe India careers