Adobe 2025 paper 1
Latest Adobe placement paper with coding problems and solutions
This page is a working set of Adobe placement papers from 2025: from student reports questions, the 2025 online assessment pattern, and step-by-step solutions. Use it to see what Adobe actually asked in the latest cycle, how hard the rounds were, and which themes (DSA, system design, aptitude, or role-specific topics) mattered most. Practice the problems below under timed conditions, then cross-check with the interview and preparation guides if you are targeting an upcoming Adobe drive.
Adobe 2025 paper 1
Latest Adobe placement paper with coding problems and solutions
Adobe 2025 paper 2
Additional 2025 Adobe paper with detailed solutions
Adobe 2025 paper 3
Another 2025 Adobe paper with comprehensive solutions
The 2025 exam pattern remains similar to 2024. For detailed exam pattern, see 2024 Papers.
Note: The pattern may have minor variations. Check the latest updates from the company.
This section contains practice questions styled on Adobe placement papers 2025 (recent-cycle pattern), with worked solutions. Use them as timed sectional drills - from student reports drives vary by college and role, so treat this as a high-signal practice bank, not an official paper dump.
Problem Statement: Find the longest common prefix string amongst an array of strings.
Example:
Input: ["flower","flow","flight"]Output: "fl"Solution (Java):
public String longestCommonPrefix(String[] strs) { if (strs.length == 0) return ""; String pref = strs[0]; for (int i = 1; i < strs.length; i++) { while (!strs[i].startsWith(pref)) { pref = pref.substring(0, pref.length() - 1); if (pref.isEmpty()) return ""; } } return pref;}Time Complexity: O(S)
Space Complexity: O(1)
Problem Statement: Merge two sorted linked lists and return a new sorted list.
Example:
Input: 1→2→4 , 1→3→4Output: 1→1→2→3→4→4Solution (Java):
public ListNode mergeTwoLists(ListNode a, ListNode b) { ListNode dummy = new ListNode(0), cur = dummy; while (a != null && b != null) { if (a.val <= b.val) { cur.next = a; a = a.next; } else { cur.next = b; b = b.next; } cur = cur.next; } cur.next = (a != null) ? a : b; return dummy.next;}Time Complexity: O(n + m)
Space Complexity: O(1)
Problem Statement: Return the level-order traversal of a binary tree.
Example:
Input: [3,9,20,null,null,15,7]Output: [[3],[9,20],[15,7]]Solution (Java):
public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if (root == null) return res; Queue<TreeNode> q = new ArrayDeque<>(); q.add(root); while (!q.isEmpty()) { int sz = q.size(); List<Integer> level = new ArrayList<>(); for (int i = 0; i < sz; i++) { TreeNode n = q.poll(); level.add(n.val); if (n.left != null) q.add(n.left); if (n.right != null) q.add(n.right); } res.add(level); } return res;}Time Complexity: O(n)
Space Complexity: O(n)
Problem Statement: Find the contiguous subarray with the largest sum.
Example:
Input: [-2, 1, -3, 4, -1, 2, 1, -5, 4]Output: 6 // [4, -1, 2, 1]Solution (Java):
public int maxSubArray(int[] nums) { int best = nums[0], cur = nums[0]; for (int i = 1; i < nums.length; i++) { cur = Math.max(nums[i], cur + nums[i]); best = Math.max(best, cur); } return best;}Time Complexity: O(n)
Space Complexity: O(1)
Problem Statement: Given a string, return it reversed.
Example:
Input: "placement"Output: "tnemecalp"Solution (Java):
public String reverse(String s) { return new StringBuilder(s).reverse().toString();}Time Complexity: O(n)
Space Complexity: O(n)
Problem Statement: Return the first non-repeating character in a string, or ‘_’ if none.
Example:
Input: "swiss"Output: 'w'Solution (Java):
public char firstUnique(String s) { int[] freq = new int[256]; for (char c : s.toCharArray()) freq[c]++; for (char c : s.toCharArray()) if (freq[c] == 1) return c; return '_';}Time Complexity: O(n)
Space Complexity: O(1)
Problem Statement: Given a string of brackets, determine if it is valid.
Example:
Input: "()[]{}"Output: trueSolution (Java):
public boolean isValid(String s) { Deque<Character> st = new ArrayDeque<>(); Map<Character, Character> pair = Map.of(')', '(', ']', '[', '}', '{'); for (char c : s.toCharArray()) { if (pair.containsValue(c)) st.push(c); else if (st.isEmpty() || st.pop() != pair.get(c)) return false; } return st.isEmpty();}Time Complexity: O(n)
Space Complexity: O(n)
Solution:
TCP is connection-oriented; UDP is connectionless.
Answer: TCP
Solution:
OS uses demand paging (and sometimes segmentation) to implement virtual memory.
Answer: Demand paging
Solution:
Each step halves the search space → O(log n).
Answer: O(log n)
Solution:
1NF: atomic values. 2NF: no partial dependency. 3NF: no transitive dependency.
Answer: 3NF
Hiring volume
2025 Data: Adobe is actively hiring 600-1200 candidates in 2025. The company is conducting placement drives at 60+ colleges across India.
Salary packages
2025 Packages: ₹35-45 LPA for freshers (updated packages)
Process updates
2025 Updates: Latest assessment tools, improved interview process
Based on recent candidate experiences from 2025 Adobe interviews:
2025 Interview Process:
2025 Interview Trends:
Common 2025 Interview Topics:
Success Tips:
For detailed interview experiences from 2025, visit Adobe Interview Experience page.
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