10 Amazon Technical Interview Questions You Should Be Ready to Solve

Amazon technical interviews are not just about getting the code to work.

You can know Python, Java, C++, data structures and algorithms inside out and still struggle in an Amazon interview if you cannot explain why your solution works, identify edge cases, improve a brute-force approach, or respond when the interviewer changes one condition.

That is what makes Amazon-style technical interviews challenging. The good news is that you do not need to memorize hundreds of unrelated problems.

A relatively small collection of algorithmic patterns appears again and again across candidate reports: hash maps, arrays, strings, sliding windows, stacks, trees, graphs, BFS/DFS, heaps, sorting, greedy algorithms and data-structure design.

The research behind this guide includes candidate-reported Amazon questions from sources such as Glassdoor, LeetCode Discuss and Reddit, with Two Sum, Valid Parentheses, Maximum Subarray, Number of Islands, LRU Cache, Word Ladder and other problems appearing repeatedly in the collected reports.

So instead of giving you a giant list of 100 or 200 questions, this guide focuses on 10 REAL Amazon interview problems and, more importantly, how to think about them in an interview. These are the kinds of questions where knowing the final code is only half the battle. Also, in-depth walkthrough of Amazon interview questions with solutions coming soon!

Before the Questions: What Amazon Is Actually Testing

Suppose an interviewer gives you a problem involving an array.

They are rarely testing only whether you know a particular trick.

They are also evaluating:

• How quickly you understand the requirements
• Whether you ask useful clarifying questions
• How you approach a brute-force solution
• Whether you can identify its bottleneck
• Whether you recognize the appropriate data structure
• Whether you can explain your reasoning
• Whether you consider edge cases
• Whether your complexity analysis is correct
• Whether you write clean, maintainable code
• Whether you test your own solution
• How you respond to hints
• Whether you can adapt when the problem changes

A strong candidate therefore does not immediately start typing.

A much better pattern is:

1. Clarify the problem

Ask questions that actually affect the solution.

For example:

“Can the array contain duplicate values?”

“Can the input be empty?”

“Are negative numbers possible?”

“Do we need to return the indices or the values?”

“Can I modify the input?”

These questions are not filler.

They determine the algorithm.

2. State the brute-force approach

Even if you know an optimized solution immediately, explain the obvious approach first.

For example:

“The straightforward approach would be to compare every pair. That gives us O(n²) time.”

Then identify the bottleneck.

3. Improve it

Now explain the data structure or algorithm that removes that bottleneck.

“We repeatedly need to determine whether a complementary value exists. A hash table gives us expected O(1) lookup, so we can reduce the solution to O(n).”

4. State complexity

Always explicitly state:

• Time complexity
• Space complexity

5. Test the solution

Do not stop after writing code.

Walk through:

• Normal input
• Empty input
• One element
• Duplicate values
• Negative values
• Boundary cases
• Maximum/minimum values where relevant

This habit alone can dramatically improve interview performance.

1. Two Sum

Difficulty: Easy
Primary concepts: Arrays, Hash Maps
Typical interview time: Around 30 minutes
Common level: SDE I and coding/OA-style rounds

The problem:

Given an array of integers nums and an integer target, return the indices of the two numbers whose sum equals target.

For example:

nums = [2, 7, 11, 15]
target = 9
Output:
[0, 1]

Because:

nums[0] + nums[1]
= 2 + 7
= 9

Two Sum is deceptively simple.

The real interview question is usually:

Can you improve the obvious O(n²) solution?

The brute-force solution

The first idea is to examine every pair.

def twoSum(nums, target):
    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            if nums[i] + nums[j] == target:
                return [i, j]
    return []

This works.

But there are potentially n² pairs.

Complexity

Time:  O(n²)
Space: O(1)

If the array contains one million elements, this approach becomes impractical.

The key observation

Suppose:

target = 9
current = 2

We do not need to search for every possible number.

We specifically need:

9 - 2 = 7

So while scanning the array, ask:

“Have I already seen the number I need?”

A hash map gives us approximately O(1) average lookup.

Optimized solution

def twoSum(nums, target):
    seen = {}
    for i, x in enumerate(nums):
        complement = target - x
        if complement in seen:
            return [seen[complement], i]
        seen[x] = i
    return []

Walkthrough

For:

nums = [2, 7, 11, 15]
target = 9

Start:

seen = {}

Read 2.

We need:

9 - 2 = 7

7 isn’t in the map.

Store:

seen = {2: 0}

Read 7.

We need:

9 - 7 = 2

2 is already present.

Therefore:

[0, 1]

Complexity

Time:  O(n)
Space: O(n)

We traded memory for speed.

That is an extremely common interview trade-off.

Amazon-style follow-ups

An interviewer may immediately change the problem.

What if the array is sorted?

Then you can use two pointers:

left  = 0
right = n - 1

If:

nums[left] + nums[right] < target

move left.

If:

nums[left] + nums[right] > target

move right.

That gives:

O(n)

time and:

O(1)

extra space.

What if there are multiple valid pairs?

Now the problem becomes different.

You need to clarify whether:

• duplicate pairs matter
• indices or values are required
• each element can be used once
• all possible pairs must be returned

That is exactly why clarifying the requirements before coding matters.

What the interviewer is looking for

Do not simply say:

“I’ll use a dictionary.”

Explain why:

“The brute-force solution repeatedly searches for a complement. Since hash-table lookup is expected O(1), I can store previously seen values and find the complement in one pass.”

That explanation demonstrates algorithmic thinking rather than memorization.

2. Maximum Subarray

Difficulty: Medium
Primary concepts: Dynamic Programming, Kadane’s Algorithm
Typical interview time: Around 45 minutes

The problem:

Given an integer array, find the contiguous subarray having the largest sum.

Example:

[-2,1,-3,4,-1,2,1,-5,4]

The answer is:

[4,-1,2,1]

with sum:

6

This is a classic example of how Amazon-style interviews can test whether you recognize an algorithmic pattern.

Brute force

Generate every possible subarray and calculate its sum.

That quickly becomes expensive.

Depending on implementation, you can reach:

O(n²)

or even:

O(n³)

The key idea

At every element, ask:

Is it better to extend the current subarray, or start a new subarray here?

Suppose:

current = -2
x = 5

Keeping -2 gives:

-2 + 5 = 3

Starting fresh gives:

5

Clearly, starting fresh is better.

So:

current = max(x, current + x)

This is Kadane’s algorithm.

Solution

def maxSubArray(nums):
    current = nums[0]
    best = nums[0]
    for x in nums[1:]:
        current = max(x, current + x)
        best = max(best, current)
    return best

Complexity

Time:  O(n)
Space: O(1)

Important edge case: all negative numbers

This is where many candidates make a mistake.

Consider:

[-8, -3, -5]

The answer is:

-3

not:

0

because the problem asks for a non-empty subarray.

Initializing:

current = 0
best = 0

can therefore produce an incorrect result.

Correct initialization uses the first element:

current = best = nums[0]

Follow-up

An interviewer may ask:

“What if I want the actual subarray, not just its sum?”

Now you need to track the starting index.

That transforms a simple Kadane implementation into a slightly richer state-tracking problem.

This is a common interview pattern:

Solve the original problem first, then modify your state to return additional information.

3. Group Anagrams

Difficulty: Medium
Primary concepts: Hash Maps, Strings, Sorting

The problem:

Given an array of strings, group strings that are anagrams of each other.

Example:

["eat","tea","tan","ate","nat","bat"]

Possible result:

[
    ["eat","tea","ate"],
    ["tan","nat"],
    ["bat"]
]

The important question is:

How do we identify that two strings belong to the same group?

The key concept: canonical representation

Anagrams contain the same characters with the same frequencies.

For example:

eat
tea
ate

Sort each string:

eat -> aet
tea -> aet
ate -> aet

Now they all have the same key.

Solution

from collections import defaultdict
def groupAnagrams(strs):
    groups = defaultdict(list)
    for s in strs:
        key = ''.join(sorted(s))
        groups[key].append(s)
    return list(groups.values())

Complexity

If:

• n = number of strings
• k = maximum string length

Sorting each string costs:

O(k log k)

Therefore:

O(n * k log k)

approximately.

A stronger solution

Instead of sorting, you can count characters.

For lowercase English letters, construct:

(a_count, b_count, c_count, ..., z_count)

as the hash key.

That can reduce the per-string processing to:

O(k)

giving approximately:

O(nk)

time.

This is exactly the sort of optimization an interviewer may ask about after you produce the first correct solution.

Follow-up questions you should expect

An interviewer could ask:

• What if strings contain Unicode?
• What if capitalization matters?
• What if spaces should be ignored?
• What if input is extremely large?
• Can you avoid sorting?
• Can you stream the input?
• What if you need only the largest anagram group?

The important lesson is not memorizing two implementations.

It is recognizing the pattern:

When objects need to be grouped by an invariant, construct a canonical key.

4. Number of Islands

Difficulty: Medium
Primary concepts: Graphs, DFS, BFS, Matrix Traversal

This is one of the most important graph problems to understand.

The problem:

Given a grid containing land (1) and water (0), count the number of connected islands.

Example:

1 1 0 0 0
1 1 0 0 0
0 0 1 0 0
0 0 0 1 1

There are:

3

islands.

The important insight

A grid is essentially a graph.

Each cell can be considered a node.

Adjacent land cells are connected.

So the problem becomes:

Count connected components.

That immediately suggests:

• DFS
• BFS

DFS approach

When we encounter land:

1. Increment the island count.
2. Traverse every connected land cell.
3. Mark them visited.
4. Continue scanning.

def numIslands(grid):
    if not grid:
        return 0
    rows = len(grid)
    cols = len(grid[0])
    def dfs(r, c):
        if (
            r < 0 or
            c < 0 or
            r >= rows or
            c >= cols or
            grid[r][c] != '1'
        ):
            return
        grid[r][c] = '0'
        dfs(r + 1, c)
        dfs(r - 1, c)
        dfs(r, c + 1)
        dfs(r, c - 1)
    count = 0
    for r in range(rows):
        for c in range(cols):
            if grid[r][c] == '1':
                count += 1
                dfs(r, c)
    return count

The collected interview research specifically identifies Number of Islands as a frequently reported Amazon graph problem.

Complexity

Every cell is visited at most once.

Therefore:

Time: O(rows × columns)

Recursive DFS can use:

O(rows × columns)

stack space in the worst case.

Important interview issue: recursion depth

In Python, very large grids can cause recursion-depth problems.

An interviewer may ask:

“Can you implement this iteratively?”

Then use a queue or explicit stack.

This is a useful lesson:

Know the algorithm, not just the implementation.

DFS does not have to mean recursive code.

Follow-ups

Possible variations include:

• Count the largest island.
• Return the area of every island.
• Find the perimeter.
• Allow diagonal connections.
• Find the shortest path between two cells.
• Find the number of connected components in a general graph.

Once you understand the underlying connected-component idea, these variations become much easier.

5. LRU Cache

Difficulty: Medium
Primary concepts: Hash Maps, Doubly Linked Lists, Data Structure Design

This is one of the most important data-structure design questions in the set.

The problem is to design an LRU cache supporting:

get(key)
put(key, value)

with:

O(1)

average time.

The supplied research identifies LRU Cache as a reported Amazon SDE-II/L5-style question.

What does LRU mean?

LRU stands for:

Least Recently Used.

Suppose the cache capacity is:

2

You insert:

A
B

Then access:

A

Now:

A = recently used
B = least recently used

If you insert:

C

you should evict:

B

Why isn’t a normal dictionary enough?

A hash map gives:

O(1)

lookup.

But it does not naturally give you:

Which item was used least recently?

We therefore need two structures.

Hash map

Maps:

key -> node

for O(1) lookup.

Doubly linked list

Maintains:

least recently used <----> most recently used

This gives O(1) insertion/removal when we already have the node.

Conceptual structure

LRU                         MRU
 |                           |
 v                           v
[Node A] <-> [Node B] <-> [Node C]

When B is accessed:

A <-> C <-> B

When capacity is exceeded, remove the leftmost node.

Python implementation

Python provides OrderedDict, which already provides the required ordering operations.

from collections import OrderedDict
class LRUCache:
    def __init__(self, capacity):
        self.capacity = capacity
        self.cache = OrderedDict()
    def get(self, key):
        if key not in self.cache:
            return -1
        self.cache.move_to_end(key)
        return self.cache[key]
    def put(self, key, value):
        if key in self.cache:
            self.cache.move_to_end(key)
        self.cache[key] = value
        if len(self.cache) > self.capacity:
            self.cache.popitem(last=False)

But be careful

Some interviewers may explicitly say:

“Do not use a built-in ordered dictionary.”

Now you need to implement:

• Hash map
• Doubly linked list
• Sentinel head
• Sentinel tail

This is much more representative of the actual data-structure knowledge being tested.

What interviewers may ask next

Why a doubly linked list?

Because removing an arbitrary node from a singly linked list requires finding its predecessor.

That would be:

O(n)

A doubly linked list allows direct removal:

previous <-> node <-> next

in:

O(1)

6. Word Ladder

Difficulty: Hard
Primary concepts: Graphs, BFS, Shortest Path

The problem:

Given a beginning word, ending word and dictionary, find the shortest transformation sequence where one character changes at a time.

Example:

begin = hit
end   = cog
dictionary =
hot
dot
dog
lot
log
cog

One shortest path is:

hit
hot
dot
dog
cog

Length:

5

The collected research identifies Word Ladder as a repeatedly reported Amazon SDE-II graph/BFS problem.

The critical observation

The word transformations form a graph.

For example:

hit
 |
hot
 / \
dot lot
 |   |
dog log
 \   /
  cog

We want the shortest path.

Whenever the question asks for an unweighted shortest path, BFS should immediately come to mind.

Why not DFS?

DFS can find a path.

But it does not naturally guarantee the shortest path without potentially exploring many alternatives.

BFS explores level by level:

distance 1
distance 2
distance 3
...

Therefore the first time we reach the destination, we have found the shortest transformation.

The tricky part

How do we find neighboring words efficiently?

For:

hot

generate patterns:

*ot
h*t
ho*

Then:

hit -> h*t
hot -> h*t

so they are neighbors.

Solution

from collections import defaultdict, deque
def ladderLength(beginWord, endWord, wordList):
    if endWord not in wordList:
        return 0
    patterns = defaultdict(list)
    L = len(beginWord)
    for word in wordList:
        for i in range(L):
            pattern = word[:i] + '*' + word[i + 1:]
            patterns[pattern].append(word)
    queue = deque([(beginWord, 1)])
    visited = {beginWord}
    while queue:
        word, distance = queue.popleft()
        for i in range(L):
            pattern = word[:i] + '*' + word[i + 1:]
            for next_word in patterns[pattern]:
                if next_word == endWord:
                    return distance + 1
                if next_word not in visited:
                    visited.add(next_word)
                    queue.append((next_word, distance + 1))
            patterns[pattern] = []
    return 0

Why clear the pattern?

This line:

patterns[pattern] = []

prevents repeatedly scanning the same neighbors.

That is an optimization worth explaining if asked.

Follow-ups

Expect questions such as:

• What if we need the actual sequence?
• What if the graph is bidirectional?
• Can you use bidirectional BFS?
• What is the complexity?
• What if words have different lengths?
• What if substitutions have different costs?

The last question can fundamentally change the algorithm.

For example, if edges have different costs, BFS may no longer be sufficient and Dijkstra’s algorithm could become relevant.

7. Reorganize String

Difficulty: Medium, sometimes challenging
Primary concepts: Greedy Algorithms, Heaps, Frequency Counting

The problem:

Rearrange a string so that no two identical characters are adjacent.

For example:

aab

can become:

aba

But:

aaab

cannot be reorganized.

The first thing to determine

Suppose:

n = 7

The most frequent character cannot appear more than:

ceil(7 / 2)

which is:

4

If a character occurs five times:

aaaaab

there is no way to separate all five as.

So the feasibility condition is critical.

Why use a heap?

We repeatedly want the most frequent characters.

A max heap allows us to efficiently retrieve them.

The general strategy is:

1. Count characters.
2. Put frequencies into a max heap.
3. Pick the two most frequent different characters.
4. Append them.
5. Decrease their counts.
6. Put them back if they still have remaining occurrences.

Solution

import heapq
from collections import Counter
def reorganizeString(s):
    counts = Counter(s)
    heap = [
        (-freq, char)
        for char, freq in counts.items()
    ]
    heapq.heapify(heap)
    result = []
    while len(heap) > 1:
        freq1, char1 = heapq.heappop(heap)
        freq2, char2 = heapq.heappop(heap)
        result.extend([char1, char2])
        if freq1 + 1 < 0:
            heapq.heappush(
                heap,
                (freq1 + 1, char1)
            )
        if freq2 + 1 < 0:
            heapq.heappush(
                heap,
                (freq2 + 1, char2)
            )
    if heap:
        freq, char = heapq.heappop(heap)
        if freq < -1:
            return ""
        result.append(char)
    return "".join(result)

The collected candidate-question research specifically includes Reorganize String among the reported Amazon SDE-II-style questions.

What makes this problem interesting?

It tests whether you recognize a greedy strategy.

You want to avoid using the same high-frequency character twice in a row.

Therefore, after using one character, you deliberately choose a different character with the next highest frequency.

This is a recurring algorithmic pattern:

Use the most constrained resource first while maintaining feasibility.

8. Find All Possible Recipes from Given Supplies

Difficulty: Medium
Primary concepts: Graphs, Topological Sort, BFS

This is a particularly useful problem because it looks like a recipe problem but is actually a dependency-resolution problem.

The problem gives:

• Recipes
• Ingredients required by each recipe
• Initial supplies

You need to determine which recipes can be produced.

Suppose:

Recipe A requires:
flour, water
Recipe B requires:
A, sugar
Supplies:
flour, water, sugar

Then:

A

can be produced.

Once A exists:

B

can also be produced.

Convert the problem into a graph

Think:

ingredient -> recipe

For example:

flour ----\
           -> A ----\
water ----/          \
                     -> B
sugar ---------------/

A recipe becomes available when all of its dependencies are satisfied.

That is exactly what topological sorting is designed for.

Track indegree

For every recipe, calculate:

number of unavailable ingredients

If that becomes:

0

the recipe can be created.

Solution

from collections import defaultdict, deque
def findAllRecipes(recipes, ingredients, supplies):
    graph = defaultdict(list)
    indegree = {recipe: 0 for recipe in recipes}
    supply_set = set(supplies)
    for recipe, required in zip(recipes, ingredients):
        for ingredient in required:
            if ingredient not in supply_set:
                indegree[recipe] += 1
                graph[ingredient].append(recipe)
    queue = deque(
        recipe
        for recipe in recipes
        if indegree[recipe] == 0
    )
    result = []
    while queue:
        recipe = queue.popleft()
        result.append(recipe)
        for next_recipe in graph[recipe]:
            indegree[next_recipe] -= 1
            if indegree[next_recipe] == 0:
                queue.append(next_recipe)
    return result

The supplied interview research describes this as a candidate-reported Amazon graph/topological-sort question.

Why this matters for interviews

You may be given a completely different story:

• Build software modules
• Resolve package dependencies
• Process jobs
• Install packages
• Schedule tasks
• Manufacture products

The underlying problem can still be:

Topological sorting.

Learning patterns is far more powerful than memorizing problem statements.

9. Merge Overlapping Intervals with Priority

Difficulty: Medium/Hard
Primary concepts: Sorting, Greedy Algorithms, Intervals

Suppose you are given:

[start, end, priority]

For example:

[1, 5, 2]
[3, 8, 5]
[10, 12, 1]

The first two intervals overlap.

They should become:

[1, 8, 5]

because the combined interval has the maximum priority.

The collected candidate research describes this as a reported L5/bar-raiser-style problem.

The key observation

Sort intervals by their starting position.

Then process them from left to right.

For each interval:

If it starts after the current merged interval ends:

start > current_end

there is no overlap.

Otherwise, merge them.

Solution

def mergeIntervals(intervals):
    intervals.sort(key=lambda x: x[0])
    merged = []
    for start, end, priority in intervals:
        if not merged or start > merged[-1][1]:
            merged.append([
                start,
                end,
                priority
            ])
        else:
            merged[-1][1] = max(
                merged[-1][1],
                end
            )
            merged[-1][2] = max(
                merged[-1][2],
                priority
            )
    return merged

Complexity

Sorting dominates:

Time: O(n log n)
Space: O(n)

depending on whether the output storage is counted.

The important follow-up

An interviewer may change:

“Instead of maximum priority, what if the interval with the highest priority determines which portion survives?”

That is no longer a simple merge.

You may need to split intervals.

This is an important interview lesson:

Do not assume that a small wording change leaves the algorithm unchanged.

Understand the exact semantics.

10. Valid Parentheses

Difficulty: Easy
Primary concepts: Stack, Strings

This is probably the simplest question in the list, but simple does not mean unimportant.

The problem:

Determine whether a string containing (), {}, and [] is correctly balanced.

Examples:

"()[]{}"   -> True
"([{}])"   -> True
"(]"        -> False
"([)]"      -> False

The key observation

Opening brackets must be closed in reverse order.

That is exactly Last In, First Out.

And the data structure for LIFO behavior is:

Stack

Solution

def isValid(s):
    stack = []
    pairs = {
        ')': '(',
        '}': '{',
        ']': '['
    }
    for char in s:
        if char in '([{':
            stack.append(char)
        elif char in pairs:
            if not stack:
                return False
            if stack.pop() != pairs[char]:
                return False
        else:
            return False
    return not stack

Walkthrough

Consider:

([{}])

Process:

(
[
{

Stack:

( [ {

Then:

}

matches:

{

Then:

]

matches:

[

Finally:

)

matches:

(

Stack is empty.

Therefore:

True

Complexity

Each character is processed once.

Time:  O(n)
Space: O(n)

The collected interview research lists Valid Parentheses among the commonly reported entry-level Amazon questions.

The 10 Problems Are Really Testing These 10 Patterns

This is the part many students miss.

You should not memorize these ten solutions independently.

Look at what they are actually teaching:

Once you recognize these patterns, dozens of seemingly unrelated questions become easier.

What Happens After You Solve the Problem?

This is where an Amazon interview can become significantly harder.

Imagine you solve Two Sum in 12 minutes.

You might think:

“Great. I’m done.”

The interviewer may instead ask:

“Can you do it without extra space?”

Or:

“What if the array is sorted?”

Or:

“What if we need all unique pairs?”

Or:

“What if the data arrives as a stream?”

Your original solution may no longer be sufficient.

That is intentional.

The Follow-Up Trap

A common mistake is preparing only the exact LeetCode problem.

Instead, prepare the neighborhood around the problem.

For every problem you practice, ask yourself:

What happens if the input is huge?

This forces you to think about:

• Memory
• Streaming
• External storage
• Distributed processing

What happens if the input is sorted?

This may introduce:

• Binary search
• Two pointers
• Greedy techniques

What happens if duplicates are allowed?

You need to reason about:

• Sets
• Frequency maps
• Duplicate handling

What happens if the output must contain the actual objects?

You may need:

• Parent pointers
• Index tracking
• Backtracking
• Reconstruction

What happens if operations happen concurrently?

Now you may need to consider:

• Locks
• Atomic operations
• Race conditions
• Thread safety

What happens if the data cannot fit in memory?

Now the problem is no longer just an algorithm question.

You might have to discuss:

• External sorting
• Streaming algorithms
• Databases
• Distributed processing

That is how a basic coding problem can turn into a senior-level discussion.

What You Should Say While Coding

One of the worst interview habits is silently typing for 20 minutes.

The interviewer cannot evaluate your reasoning if you do not communicate it.

Instead, narrate your important decisions.

For example:

“The brute-force approach would compare every pair, which is O(n²). The repeated operation is looking for the complement, so I’ll use a hash map to make that lookup O(1) on average.”

Then:

“I’ll scan the array once. Before inserting the current value, I’ll check whether its complement has already appeared. This also naturally handles duplicates.”

Then:

“The time complexity is O(n) and the additional space is O(n).”

That is concise.

You are not talking constantly.

You are exposing the reasoning behind the implementation.

What If You Get Stuck?

Getting stuck does not automatically mean failing.

A much better response than sitting silently is:

“I’m considering two approaches. The first is a straightforward brute-force solution, but its complexity is too high. I’m trying to find a way to avoid repeatedly scanning the input.”

Now the interviewer knows where you are.

You can also ask:

“Would it be reasonable to assume the input is sorted?”

or:

“Can I use additional memory?”

or:

“Do I need to preserve the input array?”

These questions can materially change the solution.

The Edge Cases You Should Automatically Check

Before declaring your solution complete, run through a mental checklist.

Empty input

[]

What should happen?

Single element

[5]

Does your algorithm crash?

Duplicate values

[2, 2, 2]

Does your hash-map or set logic still work?

Negative values

[-5, -2, -10]

Does your initialization assume positive numbers?

Very large input

Does your algorithm accidentally become:

O(n²)

or worse?

Already sorted input

Does your solution still behave correctly?

Reverse-sorted input

Same question.

Maximum/minimum values

Could arithmetic overflow occur?

This matters particularly in languages such as Java and C++.

Amazon Technical Interviews Are Not Only Coding

Another major mistake is assuming:

“Amazon interview = LeetCode.”

The supplied interview research also identifies system-design questions and behavioral/Leadership Principles discussions as important parts of the interview process, especially as seniority increases.

Representative design prompts in the research include:

• URL shortening service
• Warehouse fulfillment system
• Inventory management
• Shopping cart/checkout
• Amazon Locker
• Notification service
• Prime Video personalization
• Scaling Amazon.com to handle 10× traffic

So your preparation should eventually expand beyond algorithms.

The System Design Questions You Should Eventually Practice

If you are targeting SDE II or higher, start becoming comfortable with questions such as:

Design a URL Shortener

You should be prepared to discuss:

Client
   |
Load Balancer
   |
URL Service
   |
Database
   |
Cache

Then go deeper:

• How are IDs generated?
• How do you avoid collisions?
• How do you scale reads?
• What happens when a URL becomes extremely popular?
• How do you handle expiration?
• How do you rate-limit abuse?

The supplied research specifically highlights key-generation strategies such as Base62 encoding and the need to consider collision handling and caching.

Design an Inventory System

Now the critical problem becomes consistency.

Imagine:

Inventory = 1

Two customers attempt to purchase it simultaneously.

Both requests cannot be allowed to successfully reserve the same item.

You need to discuss:

• Atomic updates
• Transactions
• Locks
• Reservations
• Consistency
• Failure handling
• Idempotency

The supplied research specifically highlights preventing overselling and using ACID-style transactions for inventory correctness.

Design a Notification System

Suppose millions of users need delivery notifications.

A naive architecture:

Order Service
     |
     v
Email Service
     |
     v
User

can become fragile.

A better architecture introduces asynchronous processing:

Order Service
     |
     v
Message Queue
     |
     +------> Email Workers
     |
     +------> SMS Workers
     |
     +------> Push Workers

Now you can discuss:

• Queues
• Retries
• Dead-letter queues
• Idempotency
• Backpressure
• Fanout
• Delivery guarantees
• User preferences

The research likewise highlights queues, retry logic, fanout and eventual consistency for notification systems.

And Then There Are Amazon’s Leadership Principles

Technical preparation alone is not enough.

Amazon interviews can also include behavioral questions connected to its Leadership Principles.

Examples include questions around:

• Delivering under a tight deadline
• Taking ownership
• Raising the quality bar
• Taking on work outside your formal responsibility
• Handling failure
• Disagreeing with a decision
• Customer impact
• Making difficult decisions

The supplied candidate research notes that behavioral questions can be interleaved with technical rounds rather than being completely isolated from them.

A strong approach is to prepare stories using the STAR structure:

Situation
Task
Action
Result

But don’t memorize a robotic script.

Know the story.

Know the decisions you made.

Know the measurable result.

And be ready for:

“What would you do differently?”

That follow-up is often more difficult than the original question.

A Practical Amazon Interview Preparation Strategy

If you have four weeks, don’t try to solve 300 random problems.

Build pattern recognition.

Week 1: Arrays, Strings and Hashing

Focus on:

• Two Sum
• Group Anagrams
• Maximum Subarray
• Frequency counting
• Two pointers
• Sliding windows
• Prefix sums

Goal:

Recognize when hashing eliminates repeated work.

Week 2: Trees and Graphs

Practice:

• DFS
• BFS
• Number of Islands
• Tree traversal
• Binary search trees
• Shortest paths
• Connected components
• Topological sorting

Goal:

Learn to convert a problem statement into a graph.

Week 3: Advanced Patterns

Focus on:

• Heaps
• Greedy algorithms
• Dynamic programming
• Backtracking
• Intervals
• Linked lists
• LRU Cache

Goal:

Become comfortable choosing the right data structure rather than forcing one technique onto every problem.

Week 4: Interview Simulation

Now stop simply solving problems.

Start simulating interviews.

Give yourself:

5 minutes

to understand the problem.

Then:

5-10 minutes

to discuss approaches.

Then:

15-20 minutes

to implement.

Finally:

5 minutes

to test and analyze complexity.

Do this repeatedly.

The research provided for this guide similarly recommends structured preparation, progressively covering arrays, trees, graphs, dynamic programming, system design, mocks and behavioral preparation.

Your Interview-Day Checklist

Before the interview:

• Review common patterns, not entire solutions.
• Know your preferred programming language.
• Review complexity analysis.
• Practice explaining code verbally.
• Review common data structures.
• Practice at least a few timed problems.
• Prepare behavioral stories.
• Sleep properly.

During the interview:

Don’t rush into code.

First understand the problem.

Don’t hide your reasoning.

Explain important decisions.

Don’t ignore brute force.

Use it to demonstrate how you arrived at the optimized solution.

Don’t forget edge cases.

Test your own implementation.

Don’t panic when the interviewer asks a follow-up.

The follow-up is often designed to see how you adapt.

The Bigger Lesson

The biggest mistake students make while preparing for Amazon is treating interview preparation as a memorization exercise.

They collect lists such as:

Top 100 Amazon Questions
Top 200 Amazon Questions
Top 500 LeetCode Questions

and start memorizing solutions.

That strategy eventually breaks.

An interviewer can change:

"return the pair"

to:

"return all unique pairs"

or:

"the input is sorted"

or:

"you cannot use additional memory"

or:

"the data arrives continuously"

or:

"multiple threads access this structure"

Suddenly the memorized answer is useless.

Pattern recognition survives those changes.

That’s why these 10 questions matter.

Not because these are the only 10 questions Amazon can ask.

They are useful because they expose a collection of patterns that can transfer to many other problems.

The 10 Questions You Should Know Cold

Before your interview, make sure you can solve these without looking at the answer:

1. Two Sum

Pattern: Hash map

2. Maximum Subarray

Pattern: Kadane / Dynamic Programming

3. Group Anagrams

Pattern: Canonical key + hashing

4. Number of Islands

Pattern: DFS/BFS

5. LRU Cache

Pattern: Hash map + doubly linked list

6. Word Ladder

Pattern: BFS / shortest path

7. Reorganize String

Pattern: Greedy + heap

8. Find All Possible Recipes

Pattern: Topological sorting

9. Merge Intervals with Priority

Pattern: Sorting + greedy

10. Valid Parentheses

Pattern: Stack

And don’t just memorize the final code.

For each one, you should be able to answer:

What is the brute-force solution?

Why is it inefficient?

What is the optimized approach?

Why does the optimization work?

What is the time complexity?

What is the space complexity?

What are the edge cases?

What happens if the interviewer changes the constraints?

Can you implement it without built-in helpers?

Can you explain the solution line by line?

If you can comfortably answer all of those, you are much closer to being interview-ready than someone who has simply solved hundreds of problems without understanding the underlying patterns.

More Amazon Interview Prep Is Coming

I’m putting together a video series focused specifically on technical interview preparation, where we will go much deeper into the things that are difficult to learn from a question list alone.

The goal is to break down the actual thinking process:

Problem
   ↓
Clarify Requirements
   ↓
Brute Force
   ↓
Identify Bottleneck
   ↓
Choose Data Structure
   ↓
Optimize
   ↓
Code
   ↓
Test
   ↓
Handle Follow-ups

We’ll go beyond simply showing the final solution.

The idea is to make you understand how to arrive at the solution yourself.

The video series is coming soon.

But you don’t need to become a member to get value from this article.

Start with these 10 problems.

Solve them yourself.

Then solve them again without looking at the solution.

Then explain them out loud as if you were sitting across from an Amazon interviewer.

That final step is where many candidates discover the difference between:

“I solved this problem.”

and:

“I can solve this problem in an interview.”

And that difference matters.

Final Takeaway

Amazon interview preparation isn’t about finding a magical list of questions that guarantees an offer.

There is no such list.

The better strategy is to understand the patterns behind the questions.

A hash map can turn a quadratic search into a linear scan.

A stack can turn nested-order validation into a straightforward state machine.

BFS can turn a shortest-transformation problem into a level-order traversal.

A heap can make greedy selection efficient.

Topological sorting can turn a dependency problem into a graph problem.

A hash map combined with a doubly linked list can turn an LRU cache into an O(1) data structure.

Once you start seeing those patterns, interview questions stop looking like hundreds of unrelated puzzles.

They start looking like variations of problems you already understand.

That’s the skill you actually want to build before your Amazon interview.