Expected hiring
- Total Hires: 180+ freshers
- SDE-1: 162+ selections
- SDE-2: 18+ selections
- Locations: Bengaluru, Hyderabad, hybrid/remote
This page is a working set of Swiggy placement papers from 2025: from student reports questions, the 2025 online assessment pattern, and step-by-step solutions. Use it to see what Swiggy 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 Swiggy drive.
| Section | Questions | Time | Difficulty | Focus Areas |
|---|---|---|---|---|
| Coding Problems | 2-3 | 60-80 min | Medium-Hard | Arrays, trees, graphs, DP |
| Debugging | 1-2 | 20-30 min | Medium | Code fixes, logic errors |
Total: 3-5 problems, 90-120 minutes
Platform: HackerRank or Swiggy’s internal platform
Languages Allowed: Java, C++, Python, Go
Success Rate: ~10-15% cleared OA and advanced to interviews
This section contains practice questions styled on Swiggy 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: 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: 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: 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: 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: 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: 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)
Solution:
1NF: atomic values. 2NF: no partial dependency. 3NF: no transitive dependency.
Answer: 3NF
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:
This is the definition of Encapsulation (often paired with abstraction in interviews).
Answer: Encapsulation
Expected hiring
Salary packages
Question trends
Based on recent candidate experiences from 2025 Swiggy interviews:
2025 Interview Process:
2025 Interview Trends:
Common 2025 Interview Topics:
2025 Interview Questions Examples:
Success Tips:
For detailed interview experiences from 2025, visit Swiggy Interview Experience page.
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