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This page collects Netflix 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 Netflix patterns and the latest shifts.
| Section | Questions | Time | Difficulty | Focus Areas |
|---|---|---|---|---|
| Coding Problems | 2-3 | 60-80 min | Medium-Hard | Arrays, trees, graphs, DP |
| Debugging | 1 | 20-30 min | Medium | Code fixes |
Total: 3-4 problems, 90-120 minutes
Platform: HackerRank or Codility
Languages Allowed: Java, Python, Go, C++
Success Rate: ~10-15% cleared OA and advanced to interviews
This section contains practice questions styled on Netflix 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: 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: 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: 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 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: 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 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)
Solution:
Each step halves the search space → O(log n).
Answer: O(log n)
Solution:
HAVING filters aggregates; WHERE filters rows before grouping.
Answer: HAVING
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)
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Common 2024 Interview Topics:
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For detailed interview experiences, visit Netflix Interview Experience page.