A
Amazon India Interview Guide
For Freshers — India 2026
Hard Difficulty💰 ₹20–35 LPA👥 2,000+ SDE freshers/year📝 4 Questions
About the Amazon Interview Process
Amazon's SDE-1 interview is highly competitive. The process focuses on DSA problem-solving (LeetCode Medium-Hard) and Amazon's 16 Leadership Principles. The Bar Raiser ensures consistent hiring standards.
Interview Rounds
- 1Online Assessment (2 coding problems)
- 2Phone Screen
- 3Virtual Onsite (4–5 rounds)
- 4Bar Raiser
Insider Tips to Crack Amazon
- ✓Solve 150+ LeetCode problems — focus on arrays, trees, graphs, DP
- ✓Study Amazon's 16 Leadership Principles — every behavioral question maps to one
- ✓Prepare STAR (Situation, Task, Action, Result) stories for each LP
- ✓System design basics: load balancing, caching, databases — even for SDE-1
- ✓The Bar Raiser is a senior engineer — they ask the hardest questions
Amazon Interview Questions & Answers
Q1. Tell me about a time you dealt with a difficult teammate.
HR / BehavioralModel Answer:
Use STAR format. Map to 'Earn Trust' or 'Have Backbone; Disagree and Commit' LP. Structure: Situation (conflict/context) → Task (your role) → Action (how you approached it respectfully) → Result (positive outcome). Always end with what you learned.
Q2. Find the longest substring without repeating characters.
CodingModel Answer:
Sliding window with a set:
```python
def length_of_longest(s):
seen, left, res = set(), 0, 0
for right in range(len(s)):
while s[right] in seen:
seen.remove(s[left]); left += 1
seen.add(s[right])
res = max(res, right - left + 1)
return res
```
O(n) time. LeetCode #3.
Q3. Design a URL shortener (like bit.ly).
TechnicalModel Answer:
Key components: (1) Hash function — map long URL to 6-char alphanumeric (base62 encoding of auto-increment ID); (2) Database — store mapping {short_code: long_url}; (3) Redirect — 301 (permanent) or 302 (temporary); (4) Scale: cache popular URLs in Redis, sharding for DB. Handle collisions with retry.
Q4. Binary tree level order traversal.
CodingModel Answer:
Use BFS with a queue:
```python
from collections import deque
def level_order(root):
if not root: return []
res, q = [], deque([root])
while q:
level = []
for _ in range(len(q)):
node = q.popleft()
level.append(node.val)
if node.left: q.append(node.left)
if node.right: q.append(node.right)
res.append(level)
return res
```
Also Prepare For
Cracked the Interview? Get Direct HR Access
JobHuntDaily connects you directly with HRs at top companies — skip job portals that ghost you.
Get Matched — Start for ₹2