Blinkit 2025 papers
Latest Blinkit placement papers with current year questions
This page collects Blinkit placement papers from 2024 with previous-year questions, solutions, and the 2024 exam pattern. It is useful when you want real drive history: what the OA looked like, which question types repeated, and how solutions were approached. Work through the papers below to build speed and accuracy, then compare against newer 2025 material so your prep matches both established Blinkit patterns and the latest shifts.
| Section | Questions | Time | Difficulty |
|---|---|---|---|
| Coding Problems | 2-3 | 90 min | Medium-Hard |
This section contains practice questions styled on Blinkit placement papers 2024 (previous-year 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: 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)
Problem Statement: Return true if the string reads the same forward and backward (ignore case).
Example:
Input: "Level"Output: trueSolution (Java):
public boolean isPalindrome(String s) { s = s.toLowerCase(); int i = 0, j = s.length() - 1; while (i < j) { if (s.charAt(i++) != s.charAt(j--)) return false; } return true;}Time Complexity: O(n)
Space Complexity: O(1)
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 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 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: 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)
Solution:
OS uses demand paging (and sometimes segmentation) to implement virtual memory.
Answer: Demand paging
Solution:
FIFO = First In First Out → Queue. Stack is LIFO.
Answer: Queue
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
Based on candidate experiences from 2024 Blinkit interviews:
2024 Interview Process:
Common 2024 Interview Topics:
Success Tips:
For detailed interview experiences, visit Blinkit Interview Experience page.
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Practice 2024 papers to understand Blinkit OA pattern and prepare effectively!