Meesho 2024 papers
Previous year Meesho placement papers with questions and solutions
This page is a working set of Meesho placement papers from 2025: from student reports questions, the 2025 online assessment pattern, and step-by-step solutions. Use it to see what Meesho 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 Meesho drive.
| Section | Questions | Time | Difficulty | Focus Areas |
|---|---|---|---|---|
| Coding Problems | 2-3 | 60-90 min | Medium-Hard | Arrays, trees, graphs, DP |
Total: 2-3 problems, 60-90 minutes
Platform: HackerRank or similar
Languages Allowed: Python, Java, JavaScript
Success Rate: ~15-20% cleared OA and advanced to interviews
This section contains practice questions styled on Meesho 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: Rotate the array to the right by k steps.
Example:
Input: [1,2,3,4,5,6,7], k = 3Output: [5,6,7,1,2,3,4]Solution (Java):
public void rotate(int[] nums, int k) { k %= nums.length; reverse(nums, 0, nums.length - 1); reverse(nums, 0, k - 1); reverse(nums, k, nums.length - 1);}void reverse(int[] a, int l, int r) { while (l < r) { int t = a[l]; a[l++] = a[r]; a[r--] = t; }}Time Complexity: O(n)
Space Complexity: O(1)
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: 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: You can climb 1 or 2 steps. How many distinct ways to climb n stairs?
Example:
Input: n = 4Output: 5Solution (Java):
public int climbStairs(int n) { if (n <= 2) return n; int a = 1, b = 2; for (int i = 3; i <= n; i++) { int c = a + b; a = b; b = c; } return b;}Time Complexity: O(n)
Space Complexity: O(1)
Problem Statement: Given an array of integers and a target, return indices of two numbers that add up to target.
Example:
Input: nums = [2, 7, 11, 15], target = 9Output: [0, 1]Solution (Java):
public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { int need = target - nums[i]; if (map.containsKey(need)) return new int[]{map.get(need), i}; map.put(nums[i], i); } return new int[]{};}Time Complexity: O(n)
Space Complexity: O(n)
Problem Statement: Return true if the linked list has a cycle.
Example:
Input: 3→2→0→-4→(back to 2)Output: trueSolution (Java):
public boolean hasCycle(ListNode head) { ListNode slow = head, fast = head; while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; if (slow == fast) return true; } return false;}Time Complexity: O(n)
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)
Solution:
This is the definition of Encapsulation (often paired with abstraction in interviews).
Answer: Encapsulation
Solution:
Unbalanced partitions (already sorted with bad pivot) → O(n²).
Answer: O(n²)
Solution:
TCP is connection-oriented; UDP is connectionless.
Answer: TCP
Solution:
1NF: atomic values. 2NF: no partial dependency. 3NF: no transitive dependency.
Answer: 3NF
Based on recent candidate experiences from 2025 Meesho interviews:
2025 Interview Process:
2025 Interview Trends:
Common 2025 Interview Topics:
Success Tips:
For detailed interview experiences from 2025, visit Meesho Interview Experience page.
Meesho 2024 papers
Previous year Meesho placement papers with questions and solutions
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Meesho preparation guide
Comprehensive preparation strategy for Meesho placement
Meesho main page
Complete Meesho placement guide with eligibility, process, and salary
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Practice 2025 papers to stay updated with latest patterns and prepare effectively!