Blinkit 2024 papers
Previous year Blinkit placement papers with questions and solutions
This page is a working set of Blinkit placement papers from 2025: from student reports questions, the 2025 online assessment pattern, and step-by-step solutions. Use it to see what Blinkit 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 Blinkit drive.
| Section | Questions | Time | Difficulty |
|---|---|---|---|
| Coding Problems | 2-3 | 90 min | Medium-Hard |
This section contains practice questions styled on Blinkit 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: 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: 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: 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 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)
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 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: 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)
Solution:
OS uses demand paging (and sometimes segmentation) to implement virtual memory.
Answer: Demand paging
Solution:
1NF: atomic values. 2NF: no partial dependency. 3NF: no transitive dependency.
Answer: 3NF
Solution:
Unbalanced partitions (already sorted with bad pivot) → O(n²).
Answer: O(n²)
Solution:
HAVING filters aggregates; WHERE filters rows before grouping.
Answer: HAVING
Based on recent candidate experiences from 2025 Blinkit interviews:
2025 Interview Process:
2025 Interview Trends:
Common 2025 Interview Topics:
Success Tips:
For detailed interview experiences from 2025, visit Blinkit Interview Experience page.
Blinkit 2024 papers
Previous year Blinkit placement papers with questions and solutions
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Complete collection of Blinkit coding problems with solutions
Blinkit interview experience
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Blinkit preparation guide
Comprehensive preparation strategy for Blinkit placement
Blinkit main page
Complete Blinkit placement guide with eligibility, process, and salary
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Practice 2025 papers to stay updated with latest patterns and prepare effectively!