Netflix Python Developer Interview Questions & Answers

I would begin with one lock and keep every critical section very small. If profiling proves that lock waiting is a bottleneck, I would partition the data and use one lock per partition. This lock striping approach lets operations on unrelated keys use different locks. For data with many reads and rare updates, I would consider immutable snapshots. For CPU bound work or stronger isolation, I would consider processes that own separate state and communicate through messages.

I would start with one lock because it is the easiest design to make correct. The lock must cover the complete logical operation, such as reading a value and then updating it. Slow calculation, logging, file access, and network access should stay outside the critical section.

1. Should I focus on Python language behavior, or also explain the runtime and standard library?
2. Which Python version and execution environment should I assume?
3. Would you like a small code example together with production tradeoffs and edge cases?

If profiling shows significant waiting, I would use lock striping. I would split the data into several dictionaries. Each dictionary has its own lock, and every key always maps to the same stripe. Operations on different stripes avoid waiting for the same application lock. Operations involving several keys must acquire all required stripe locks in a consistent order to prevent deadlock.

A read write lock may help when reads are long and greatly outnumber writes, but it often adds little value for very short dictionary reads. The threading module does not provide one, so another implementation adds overhead and may introduce fairness or starvation concerns.

Immutable snapshots fit data that is read often and changed rarely. A writer copies the current dictionary, applies changes, and publishes the new reference under a short lock. Readers obtain the reference under that lock, then read without holding it. Copying costs linear time and temporary linear memory.

Processes can remove shared thread state, but serialization and communication add cost.

A common mistake is assuming that the Global Interpreter Lock makes a compound action such as check then update atomic. Another mistake is protecting only one step of a logical operation. Holding a lock during slow calculation, logging, file access, or network access creates unnecessary contention. In a striped design, changing the stripe calculation or using different locks for the same key breaks correctness. Acquiring several stripe locks in inconsistent orders can deadlock. Too many stripes increase lock objects, dictionaries, memory use, and maintenance complexity. Snapshot readers must never mutate a published snapshot, and writers must not modify the old dictionary in place. A process manager dictionary should not be assumed to remove contention because proxy calls still require communication and coordination.

Start with correctness and the simplest design. Explain one short lock first. Then say that profiling may justify lock striping, a read write lock, immutable snapshots, or process ownership. Mention compound operations, stable lock ordering, snapshot copying cost, and why the Global Interpreter Lock is not a complete synchronization strategy.

I would keep timestamped latency samples in a deque and protect the deque with a threading lock. Every record and query removes expired samples and enforces a maximum count. A percentile query copies the current values while holding the lock, then releases the lock before sorting. This gives the query a consistent snapshot while allowing new records to continue during the expensive sort.

I would store each sample as a monotonic timestamp and a latency value inside a deque. A threading.Lock protects every operation that reads or changes this shared deque. The Global Interpreter Lock is not enough because pruning, appending, checking length, and copying form a multi step operation.

1. Should I focus on Python language behavior, or also explain the runtime and standard library?
2. Which Python version and execution environment should I assume?
3. Would you like a small code example together with production tradeoffs and edge cases?

The record method validates the value, acquires the lock, records time with time.monotonic, appends the sample, removes expired entries, and removes the oldest entries when the count limit is exceeded. A monotonic clock is suitable because changes to the system clock do not affect elapsed time.

The percentile method validates a value from zero through one hundred. While holding the lock, it removes expired entries and copies retained latency values. It then releases the lock, sorts the private copy, and uses linear interpolation between neighboring values. With samples 10, 20, 30, and 40, P50 is 25 and P95 is 38.5.

An empty tracker raises ValueError. Recording is amortized constant time, although one call may remove several old entries. A query takes linear time to copy, n log n time to sort, and linear extra memory. This design suits bounded local metrics. A histogram is better for much larger workloads.

LatencyTracker stores samples in recording order inside a deque. Each sample contains a time.monotonic timestamp and a latency value in milliseconds. max_age_seconds limits how long a sample is retained, while max_samples provides a second strict memory bound. The private cleanup method runs only while the lock is held. The percentile method removes expired data and copies the values under the lock, then sorts the private snapshot after releasing the lock. It maps the requested percentile to a position from zero through the final sorted index and uses linear interpolation when the position falls between two values. For the example values 10, 20, 30, and 40, the code returns 25 for P50 and 38.5 for P95.

A common mistake is assuming the Global Interpreter Lock makes the tracker thread safe. It does not protect the complete sequence of pruning, appending, checking length, and copying. Another mistake is sorting shared data while holding the lock, which blocks writers for the full sorting time. Developers may also forget a count limit, use time.time for elapsed retention, accept NaN or infinity, calculate percentiles with an inconsistent formula, or return values from expired samples. Another error is claiming every record call is strictly constant time even though one call may remove many old entries.

Start with the concurrency rule. Explain that the lock protects the shared deque, but sorting uses a private snapshot outside the lock. Then describe the age limit, count limit, interpolation rule, empty state behavior, and exact time and memory costs.

I would store the counter value in a class and protect every read and increment with the same threading.Lock. The lock lets only one participating thread enter the protected section at a time. I would not rely on the Global Interpreter Lock because the complete read, add, and write sequence is not a safe application level synchronization contract.

I would implement the counter with one integer and one threading.Lock. The increment method acquires the lock, changes the value, and returns the new value. The read method acquires the same lock before returning the current value. This makes every completed increment and read behave as one protected operation for threads using that counter.

1. Should I focus on Python language behavior, or also explain the runtime and standard library?
2. Which Python version and execution environment should I assume?
3. Would you like a small code example together with production tradeoffs and edge cases?

A Python lock provides mutual exclusion. While one thread owns the lock, another thread that tries to acquire it must wait. Releasing the lock allows a waiting thread to continue. The with statement is useful because it releases the lock even if an exception leaves the protected block.

I would not depend on the Global Interpreter Lock. A counter update includes reading the current value, calculating a result, and storing that result. Explicit locking makes the required guarantee clear and remains correct across Python runtime changes.

Each read and increment uses constant work and no growing data structure. The counter object uses constant memory. Under heavy contention, waiting for the lock can reduce throughput. This design protects threads in one process only. Separate processes or servers need a process safe or external atomic counter.

The AtomicCounter class stores one integer and one threading.Lock. Both increment and read acquire the same lock with a with statement. Increment changes the protected value and returns the new value before the lock is released. Read returns the value while holding that same lock. Each operation uses constant work and the object uses constant memory. The example starts four threads, lets each thread perform ten thousand increments, waits for all threads to finish, and safely reads the final value of forty thousand.

A common mistake is using value += 1 without a lock and assuming the Global Interpreter Lock protects the complete update. Another mistake is locking increment while reading the value without the same lock. Creating a new lock inside each method call is also wrong because the calls would not share one synchronization object. Developers should not hold the lock during slow file access, network access, sleeping, or callbacks because this increases waiting and can create deadlock risks. A threading.Lock also does not coordinate separate processes or servers.

Start with the design: one value, one shared lock, and the same lock around every read and increment. Then explain that the lock provides mutual exclusion, while the Global Interpreter Lock does not replace explicit synchronization for the complete counter operation. Finish by stating the one process limitation and the contention cost.

I would replace recursion with an explicit Python list used as a stack. For subtree information, each stack item stores a node and a visited flag. On the first visit, I schedule the parent for later and then schedule its children. On the second visit, every child result is ready, so I compute the parent result. This avoids recursion depth failures, visits each tree node once for calculation, and uses linear total memory because the code stores results and validation state.

I would use a Python list as an explicit stack instead of making recursive calls. Each stack item contains a node and a visited flag. When a node is removed with visited set to false, I place it back with visited set to true, then place its children on the stack. This creates postorder traversal, which means children finish before their parent.

1. Should I focus on Python language behavior, or also explain the runtime and standard library?
2. Which Python version and execution environment should I assume?
3. Would you like a small code example together with production tradeoffs and edge cases?

When the node is removed again with visited set to true, every child result is available. The example computes subtree sizes, so the result is one plus the sizes of all direct child subtrees.

Python lists are suitable here because append and pop at the end are normally constant time. The traversal takes linear time for a valid tree because every node and parent to child connection is handled a constant number of times.

The explicit stack avoids Python recursion depth failures, but it does not remove memory costs. A deep or wide tree can make the stack large. The result dictionary and validation set also grow with the number of nodes. The code treats a missing dictionary entry as a leaf and rejects cycles or shared child nodes because the input must be a true tree.

The tree is stored as a mapping from each node to its direct children. The stack stores a node and a visited flag. A false flag means the algorithm must schedule the children first. A true flag means all child results are complete and the node can be calculated. The validation set rejects cycles and shared child nodes, so the structure must be a true tree. A node with no mapping entry is treated as a leaf. For the sample tree, A has subtree size 5, B has subtree size 3, and C, D, and E each have subtree size 1.

A common mistake is calculating a parent during the first visit, before its child results exist. Another mistake is storing only nodes in the stack and losing the information that tells the algorithm whether a node is entering or leaving. Candidates may also use pop at the front of a Python list, which moves remaining elements and is slower. Other mistakes include forgetting that a wide tree can make the explicit stack large, claiming the iterative version uses constant memory, ignoring malformed input, and using this exact tree validation when shared nodes are valid in the real data model.

Start by saying that you replace Python call frames with a list based stack. Then explain the visited flag, why it creates postorder traversal, when the parent is calculated, and why total memory is still linear when results and validation state are stored.

I would model the tasks as a directed acyclic graph. I store outgoing edges in an adjacency list and count each task’s prerequisites with an indegree map. Then I use Kahn’s topological sort. For every task, I track its earliest start and finish time. A task starts after its latest prerequisite finishes. Independent source tasks can start together at time zero. The answer is the largest finish time. This takes O(V + E) time and O(V + E) auxiliary space.

The problem asks for the shortest total time needed to finish every task. Each task has a duration. A directed edge means one task must finish before another task can start. Because independent tasks may run at the same time, we do not add every duration. We find the longest required dependency chain by processing the graph in topological order.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input has two parts.

The first part is a dictionary of task durations. In the example, the durations are A: 3, B: 2, C: 4, D: 3, and E: 1.

The second part is a list of directed dependencies. A pair such as A to C means A must finish before C can start.

The dependencies are A to C, B to C, B to D, C to E, and D to E.

The output is one integer. It is the minimum time needed to finish all tasks when independent tasks can run in parallel. For this example, the answer is 8.

I represent the dependencies with an adjacency list. For each task, the adjacency list stores the tasks that directly depend on it.

I also store an indegree count. The indegree of a task is the number of unfinished prerequisites it has.

I use Kahn’s algorithm for topological sorting. It starts with every task whose indegree is zero. These tasks have no prerequisites, so they can start at time zero.

For each task, I track two values. earliest_start is the first time the task may begin. earliest_finish is earliest_start plus the task duration.

The central invariant is this: when a task leaves the queue, all of its prerequisites have already been processed. Therefore, its earliest start and finish times are final.

The adjacency list is:

A to [C] B to [C, D] C to [E] D to [E] E to []

The indegree values are:

A: 0 B: 0 C: 2 D: 1 E: 2

A and B have indegree zero. They enter the queue first.

Both start at time zero. A finishes at time 3. B finishes at time 2.

The initial queue is [A, B]. The initial earliest_start value is zero for every task.

First, remove A from the queue.

A finishes at time 3. C depends on A, so update C’s earliest start from 0 to max(0, 3), which is 3. C’s indegree changes from 2 to 1. C still has one unfinished prerequisite, so it does not enter the queue.

The queue is now [B].

Next, remove B.

B finishes at time 2. C also depends on B. Update C’s earliest start to max(3, 2), which remains 3. C’s indegree changes from 1 to 0. All prerequisites of C are now complete. C finishes at 3 + 4, which is 7, and enters the queue.

D also depends on B. Update D’s earliest start to max(0, 2), which is 2. D’s indegree changes from 1 to 0. D finishes at 2 + 3, which is 5, and enters the queue.

The queue is now [C, D].

Next, remove C.

C finishes at time 7. E depends on C. Update E’s earliest start to max(0, 7), which is 7. E’s indegree changes from 2 to 1. E still waits for D.

The queue is now [D].

Next, remove D.

D finishes at time 5. E also depends on D. Update E’s earliest start to max(7, 5), which remains 7. E’s indegree changes from 1 to 0. E finishes at 7 + 1, which is 8, and enters the queue.

The queue is now [E].

Finally, remove E. It has no outgoing edges. The queue becomes empty.

The finish times are A: 3, B: 2, C: 7, D: 5, and E: 8. The largest finish time is 8.

Topological order processes every prerequisite before its dependent tasks.

For each edge from u to v, the algorithm gives v the value max(earliest_start[v], earliest_finish[u]). This means v waits for its latest-finishing prerequisite.

A dependent task enters the queue only when its indegree becomes zero. At that moment, every prerequisite has contributed its finish time. Therefore, the task’s earliest start and finish values are correct.

The maximum finish time is the minimum total project time because independent tasks run together whenever their dependencies allow it. The critical path is A to C to E. Its duration is 3 + 4 + 1, which is 8. The path B to D to E takes 2 + 3 + 1, which is 6, so it does not determine the total time.

The code first creates an empty adjacency list and an indegree value for every task.

It then reads each dependency. It adds the dependent task to the prerequisite’s adjacency list and increases the dependent task’s indegree.

Next, it creates the earliest_start map, the earliest_finish map, and a deque. Every zero-indegree task enters the deque and gets a finish time equal to its duration.

The main loop removes one task at a time. It updates the answer with that task’s finish time. It then updates every dependent task. When a dependent task’s indegree becomes zero, the code calculates its finish time and adds it to the queue.

The processed counter detects a cycle. If fewer tasks are processed than exist in the input, the dependency graph is not a DAG, so no valid schedule exists.

Let V be the number of tasks and E be the number of dependencies.

Each task enters and leaves the queue once. Each dependency edge is processed once. The time complexity is O(V + E).

The graph, indegree map, timing maps, and queue may all grow with the input. The auxiliary space complexity is O(V + E).

Important cases include multiple source tasks, disconnected groups of tasks, one task with many prerequisites, a single task, and a cycle. Multiple source tasks can start together. Disconnected groups are processed independently. A cycle is rejected because no topological schedule exists.

The key idea is to combine Kahn’s topological sort with dynamic programming on a directed acyclic graph. The adjacency list stores each task’s direct dependents. The indegree map stores how many prerequisites each task still has. For every task v, earliest_start[v] is the largest finish time seen among its prerequisites. The invariant is that when v enters the queue, all of its prerequisites have been processed, so earliest_start[v] is final. We then calculate earliest_finish[v] as earliest_start[v] plus duration[v]. The largest finish time is the total completion time because independent tasks already overlap whenever possible.

Let V be the number of tasks and E be the number of directed dependencies. Building the graph takes O(V + E) time. During topological sorting, each task is removed from the queue once, and each dependency edge is examined once. Therefore, the total time is O(V + E). The adjacency list uses O(V + E) memory. The indegree map, timing maps, and queue use O(V) more memory. The total auxiliary space is O(V + E).

A common mistake is adding every task duration, which ignores parallel work. Another mistake is using the first prerequisite finish time instead of the maximum finish time across all prerequisites. Some candidates enqueue a task before its indegree reaches zero, even though another prerequisite is still unfinished. Others reverse the dependency edge and build the adjacency list in the wrong direction. It is also easy to forget disconnected source tasks or omit cycle detection. Finally, claiming O(V) time is incorrect because every dependency edge must also be processed.

State the invariant before writing code: when a task enters the queue, all of its prerequisites are complete, so its earliest start time is final. This makes the update rule and correctness argument much easier to explain.

I would use post-order depth-first search. For each node, I start with its own value. I recursively compute each child’s subtree sum and add the returned value to the current total. After every child is finished, I store the total for the node and return it to the parent. This works because each parent uses complete child results. The time complexity is O(n). The recursion stack uses O(h) auxiliary space, where h is the tree height.

The input contains a root node, a children adjacency list, and one integer value for every node. We must return a dictionary that maps each node to the sum of its own value and every descendant value. Post-order DFS fits this problem because a parent’s result depends on the completed results of its children.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The root tells us where traversal begins.

The children dictionary stores each node and its direct children. The edges are directed from parent to child.

The values dictionary stores the integer value of each node.

The output is a dictionary that maps each node to its complete subtree sum.

In the example, the edges are 1 to 2, 1 to 3, 2 to 4, 2 to 5, and 3 to 6. The node values are 1 to 5, 2 to 2, 3 to -3, 4 to 4, 5 to 1, and 6 to 6.

I use recursive depth-first search.

Each call solves one node’s complete subtree. The call returns one integer, which is that subtree’s sum.

The central invariant is: when dfs(node) returns, it has computed and stored the correct sum for node and every descendant in that node’s subtree.

The children must finish before their parent. This processing order is called post-order traversal.

I create an empty dictionary named sums. It stores node to completed subtree sum.

Inside dfs(node), I set total to values[node]. This first includes the current node’s own value.

Then I visit every child. Each child returns its complete subtree sum, which I add to total.

A leaf has no children. Its loop does not run, so it stores and returns its own value.

Start with dfs(1). Node 1 begins with total 5 and visits node 2.

Node 2 begins with total 2 and visits node 4.

Node 4 is a leaf. It stores sums[4] = 4 and returns 4.

Back at node 2, total changes from 2 to 2 + 4 = 6. Node 2 then visits node 5.

Node 5 is a leaf. It stores sums[5] = 1 and returns 1.

Node 2 now computes 6 + 1 = 7. It stores sums[2] = 7 and returns 7.

Back at node 1, total changes from 5 to 5 + 7 = 12. Node 1 then visits node 3.

Node 3 begins with total -3 and visits node 6.

Node 6 is a leaf. It stores sums[6] = 6 and returns 6.

Node 3 computes -3 + 6 = 3. It stores sums[3] = 3 and returns 3.

Finally, node 1 computes 12 + 3 = 15. It stores sums[1] = 15 and returns 15.

The post-order return order is 4, 5, 2, 6, 3, 1.

The final mapping is {1: 15, 2: 7, 3: 3, 4: 4, 5: 1, 6: 6}.

Every child call returns the exact sum of that child’s subtree.

The parent starts with its own value and adds the returned sum from every child.

Therefore, after all children finish, total contains the parent’s value and every descendant value exactly once.

Storing sums[node] after the children finish gives the correct answer for that node.

The outer function creates the sums dictionary.

The inner dfs function computes one subtree. It starts total with the current node’s value.

It reads the current node’s children with children.get(node, []). This also works when a leaf is missing from the dictionary.

For each child, it calls dfs(child) and adds the returned subtree sum to total.

After the loop, the subtree is complete. The function stores sums[node] = total and returns total to the parent.

The outer function calls dfs(root) and returns the completed sums dictionary.

Let n be the number of nodes and h be the tree height.

The time complexity is O(n). Each node is visited once, and each parent-to-child edge is processed once.

The recursion stack uses O(h) auxiliary space. A balanced tree has a smaller height. A skewed tree can have h equal to n.

The output dictionary uses O(n) space because it stores one result for every node.

Important edge cases are a leaf node, a single-node tree, negative or zero values, and a skewed tree with deep recursion.

The key idea is to compute child subtree sums before computing the parent’s sum. Each recursive DFS call returns one integer: the complete subtree sum for that node. The invariant is that when dfs(node) returns, sums[node] has already been stored and equals values[node] plus the completed sums returned by all children. Post-order DFS is suitable because the parent cannot be finished until all child results are available.

Let n be the number of nodes and h be the height of the tree. The time complexity is O(n) because every node is visited once and every parent-to-child edge is processed once. The recursion stack uses O(h) auxiliary space because the active calls form one root-to-leaf path. In a skewed tree, h can equal n. The output dictionary uses O(n) space because it stores one subtree sum for each node.

A common mistake is storing sums[node] before visiting the children. That stores an incomplete total. Another mistake is forgetting to return total to the parent. Candidates may forget that a leaf still stores and returns its own value. They may incorrectly use one shared total across all calls instead of one local total per call. They may also claim O(1) auxiliary space and forget the O(h) recursion stack.

State the invariant before writing code: when dfs(node) returns, it has stored and returned the correct sum for that complete subtree. This explains both the recursion order and why sums[node] is stored after the child loop.

I would convert each string into a set of its unique characters. Then I would compare every unordered pair once by using indices i and j, with j greater than i. For each pair, I call isdisjoint on the two cached sets. If it returns true, I add the original strings to the result. This works because isdisjoint is true exactly when the strings share no characters. The expected time is O(T + sum of min(u_i, u_j)), with O(sum of u_i) auxiliary space.

The problem asks us to return every pair of strings that does not share any character. A useful representation is a set of characters for each string. A set keeps each character only once. Python's isdisjoint method then gives us a direct way to test whether two strings share a character. We build the sets once and reuse them while checking every unordered pair.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is a list of strings. The output is a list of pairs containing the original string values.

For the example, the input is ["ab", "cd", "ac", "ef"].

The returned result is [("ab", "cd"), ("ab", "ef"), ("cd", "ef"), ("ac", "ef")].

A pair is valid only when the two strings have no common character. We return string pairs, not index pairs and not character sets.

For each string, I create a set containing its unique characters.

The cached sets are:

Index 0: "ab" becomes {'a', 'b'}.

Index 1: "cd" becomes {'c', 'd'}.

Index 2: "ac" becomes {'a', 'c'}.

Index 3: "ef" becomes {'e', 'f'}.

Then I use set.isdisjoint. It returns true when the two sets have no shared element.

The central invariant is: after each processed pair, result contains exactly the disjoint pairs among all pairs checked so far.

I build char_sets with one cached set for each input string.

I also create an empty result list.

The outer loop selects index i. The inner loop starts at i + 1, so j is always greater than i.

This skips self-pairs such as comparing "ab" with itself. It also prevents reverse duplicates such as returning both ("ab", "cd") and ("cd", "ab").

Step 1 checks indices (0, 1), which contain "ab" and "cd".

The result is initially empty. {'a', 'b'}.isdisjoint({'c', 'd'}) is true. We append ("ab", "cd"). The result becomes [("ab", "cd")].

Step 2 checks indices (0, 2), which contain "ab" and "ac".

The state before the check is [("ab", "cd")]. Both sets contain 'a', so isdisjoint returns false. We skip this pair. The result stays [("ab", "cd")].

Step 3 checks indices (0, 3), which contain "ab" and "ef".

The sets have no common character, so isdisjoint returns true. We append ("ab", "ef"). The result becomes [("ab", "cd"), ("ab", "ef")].

Step 4 checks indices (1, 2), which contain "cd" and "ac".

Both sets contain 'c', so isdisjoint returns false. We skip the pair. The result remains [("ab", "cd"), ("ab", "ef")].

Step 5 checks indices (1, 3), which contain "cd" and "ef".

The sets are disjoint, so we append ("cd", "ef"). The result becomes [("ab", "cd"), ("ab", "ef"), ("cd", "ef")].

Step 6 checks indices (2, 3), which contain "ac" and "ef".

The sets are disjoint, so we append ("ac", "ef"). The final result is [("ab", "cd"), ("ab", "ef"), ("cd", "ef"), ("ac", "ef")].

All six unordered pairs have now been checked, so the function returns the result.

Every unordered pair is checked exactly once because the inner loop uses j greater than i.

A pair is added if and only if isdisjoint returns true.

Therefore, every returned pair shares no character. Every omitted pair shares at least one character. This means the result contains exactly all valid disjoint pairs.

The list comprehension builds one character set for every input string.

The nested loops generate every unordered pair in this order: (0, 1), (0, 2), (0, 3), (1, 2), (1, 3), and (2, 3).

The condition char_sets[i].isdisjoint(char_sets[j]) checks the cached sets.

When the condition is true, the code appends the original strings. It does not append the sets or the indices.

After every pair has been processed, the function returns the complete result list.

Let T be the total number of characters across all strings. Building all sets costs O(T) time.

Let u_i be the number of unique characters in string i. For one pair, isdisjoint takes expected O(min(u_i, u_j)) time because Python sets use hashing and membership checks are O(1) on average.

The full expected time is O(T + sum over all i < j of min(u_i, u_j)). If each string has at most k unique characters, this can be written as O(T + n^2 * k).

The cached sets use O(sum of u_i) auxiliary space. The returned output uses O(p) space for p valid pairs.

An empty collection or a collection with one string returns no pairs. An empty string is disjoint with every other string. Repeated characters inside one string do not change its set. Duplicate strings at different indices are still treated as separate input elements.

The key idea is to separate character extraction from pair comparison. We convert every string once into a cached set of unique characters. Then we compare each unordered pair with set.isdisjoint. The invariant is that after each processed pair, result contains exactly the disjoint pairs among all pairs checked so far. Using j > i makes each unordered pair appear once and avoids self-pairs. This approach is easier to read and avoids rebuilding character collections during every comparison.

Let T be the total number of characters in all strings. Building the character sets takes O(T) time. Let u_i be the number of unique characters in string i. Python sets use hashing, so membership checks are O(1) on average. The expected cost of isdisjoint for one pair is O(min(u_i, u_j)). Therefore, the total expected time is O(T + sum over i < j of min(u_i, u_j)). If every string has at most k unique characters, this is O(T + n^2 * k). The cached sets use O(sum of u_i) auxiliary space. The output uses O(p) space for p returned pairs.

A common mistake is checking both (i, j) and (j, i), which returns duplicate pair orders. Another mistake is starting j at i, which compares a string with itself. Some candidates rebuild sets inside the nested loops and repeat unnecessary work. Another error is appending sets or indices when the required output contains the original string values. Candidates may also claim that Python set operations are guaranteed O(1), even though that is average behavior. Repeated characters inside one string should not be counted more than once because the representation is a set.

State the invariant before coding: after each checked pair, the result contains exactly the valid disjoint pairs seen so far. Then explain why j starts at i + 1. This makes both the traversal and the correctness argument easy to defend.

I use a sliding window and a hash map. The window runs from left to right, and the map stores each value’s latest index. When the current value already appears inside the window, I move left to one position after its previous index. After each step, I compare the valid window with the best one found so far. This returns a longest distinct contiguous subarray in O(n) expected time with O(n) auxiliary space.

The input is an integer array. We must return one longest contiguous part of the array that contains no repeated value. A sliding window fits this problem because it keeps one valid range while moving from left to right. A hash map lets us move the start of the window directly past an earlier duplicate.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The input is an integer array called nums. The output is the values from one longest contiguous subarray with no duplicates.

Contiguous means the values must come from neighboring positions in the original array. We cannot skip elements.

For nums = [5, 1, 3, 5, 2, 3, 4, 1], the returned subarray is [5, 2, 3, 4, 1]. It comes from indices 3 through 7.

Use a sliding window. The current window is nums[left:right + 1]. The left variable marks its first index. The right variable marks its last index.

Use a hash map called last_seen. It stores each value and the latest processed index where that value appeared. For example, last_seen[5] = 3 means the latest processed 5 was at index 3.

The central invariant is that, after any required movement of left, nums[left:right + 1] contains no repeated values.

Set left to 0. Start with an empty last_seen map.

Set best_start to 0 and best_len to 0. These variables describe the longest valid window found so far.

Then process the array from left to right using right and value.

At right = 0, value = 5. It has not appeared inside the current window. Store last_seen[5] = 0. The window is [5], so the best becomes [5].

At right = 1, value = 1. It is new in the current window. Store last_seen[1] = 1. The window becomes [5, 1], so the best becomes [5, 1].

At right = 2, value = 3. It is new in the current window. Store last_seen[3] = 2. The window becomes [5, 1, 3], so the best becomes [5, 1, 3].

At right = 3, value = 5. Its previous index is 0, and index 0 is still inside the current window because 0 >= left. Move left from 0 to 1. Then store last_seen[5] = 3. The valid window is now [1, 3, 5]. The best remains [5, 1, 3].

At right = 4, value = 2. It is new in the current window. Store last_seen[2] = 4. The window becomes [1, 3, 5, 2]. Its length is 4, so it becomes the new best.

At right = 5, value = 3. Its previous index is 2, and index 2 is inside the current window because 2 >= left. Move left from 1 to 3. Then store last_seen[3] = 5. The valid window becomes [5, 2, 3]. The best remains [1, 3, 5, 2].

At right = 6, value = 4. It is new in the current window. Store last_seen[4] = 6. The window becomes [5, 2, 3, 4]. Its length ties the best length. The code updates only for a strictly longer window, so it keeps [1, 3, 5, 2].

At right = 7, value = 1. Its previous index is 1, but left is already 3. The old 1 is outside the current window because 1 < left, so left stays 3. Store last_seen[1] = 7. The window becomes [5, 2, 3, 4, 1]. Its length is 5, so it becomes the final best window.

The function returns nums[3:8], which is [5, 2, 3, 4, 1].

After duplicate handling, the current window contains only distinct values. If a duplicate lies inside the window, moving left to one position after its earlier index removes the earlier copy and keeps the range contiguous.

For each right index, the algorithm keeps the longest distinct window that can end at that index. It compares that valid window with the best one found earlier. Therefore, after the final index is processed, best_start and best_len describe a longest distinct contiguous subarray.

The for loop gives the current index and value. The duplicate condition checks two facts: the value appeared before, and its previous index is greater than or equal to left. Only then is the earlier copy still inside the current window.

After moving left when needed, the code records the current index in last_seen. It calculates the current length as right - left + 1. It updates best_start and best_len only when the current window is strictly longer.

Finally, it returns nums[best_start:best_start + best_len].

The expected running time is O(n). The right index processes every array position once. The left index only moves forward. Python dictionary lookup and insertion are O(1) on average.

The auxiliary space is O(n) because last_seen may store one entry for every distinct value. The returned slice uses O(k) output space, where k is the returned subarray length.

For an empty input, the function returns an empty list. For one element, it returns that one-element subarray. If all values are distinct, it returns the whole array. If all values are equal, any one-element subarray is longest, and this implementation keeps the earliest one. Negative values and zero work without special handling.

The key idea is to keep one valid sliding window instead of checking every possible subarray. The window is nums[left:right + 1]. The hash map last_seen stores each processed value and its latest index. The invariant is that, after any required left adjustment, the current window contains no repeated values. When a repeated value has a previous index inside the window, left jumps to last_seen[value] + 1. This removes the earlier copy without moving left one position at a time. For every right index, the resulting window is the longest distinct window ending there. The algorithm records it when it is strictly longer than the current best.

The expected time is O(n), where n is the number of values in nums. The right index visits each array position once. The left index only moves forward. Python dictionary lookup and insertion take O(1) time on average, so the full algorithm takes O(n) expected time. The auxiliary space is O(n) because last_seen may contain one entry for each distinct value. The returned list also uses O(k) output space, where k is the length of the returned subarray.

A common mistake is moving left whenever a value appeared before. Left should move only when the previous index is still inside the current window. Another mistake is moving left backward when an older duplicate is already outside the window. Candidates may measure or save the window before removing an active duplicate, which can record an invalid range. Some people confuse a contiguous subarray with a subsequence and skip values. Another mistake is updating the best window on equal length even though this implementation keeps the earliest tied result. It is also incorrect to describe Python dictionary operations as guaranteed O(1); they are O(1) on average.

State the invariant before writing code: after duplicate handling, nums[left:right + 1] contains only distinct values. Then explain the condition last_seen[value] >= left. This shows exactly when left must move and prevents the main duplicate-handling bug.

I would store the latest live entry for each key in a hash map and keep expiration records in a min-heap. Every successful positive-TTL put gets a globally increasing version. Cleanup pops records whose expiration time is at or before now. It removes a map entry only when the key, version, and expiration time still match, so stale records cannot delete newer values. Put costs O(log h) plus amortized cleanup, get has expected O(1) lookup plus amortized cleanup, and compacted space is O(n).

The cache must return a value only while that value is live. It must also remove expired and stale data so memory does not grow forever. The solution uses a hash map for the newest live entry and a min-heap for expiration records. A globally increasing version separates the current entry from older heap records for the same key.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The cache supports three operations:

• put(key, value, ttl, now) stores or replaces a value.
• get(key, now) returns a value only when it has not expired.
• cleanup(now) removes records whose expiration time has arrived.

The hash map stores:

key -> (value, expires_at, version)

The min-heap stores:

(expires_at, key, version)

An entry is expired when expires_at <= now. A missing or expired key returns None.

The map contains only the newest live entry for each key. The heap may still contain records created by older writes.

Every successful put with ttl > 0 receives a new globally increasing version. Versions are never reused.

Cleanup deletes a key only when the popped record has the same key, version, and expiration time as the current map entry. A record that does not match is stale, so cleanup skips it.

This prevents an old heap record from deleting a newer value or a later reinsertion of the same key.

Both put and get call cleanup(now) first.

Cleanup repeatedly examines the smallest expiration time in the min-heap. While that time is less than or equal to now, cleanup pops the record.

If the key is no longer in the map, the record is stale and can be ignored.

If the current map entry has the same version and expiration time, cleanup deletes it. Otherwise, the popped record belongs to an older write and is skipped.

For ttl <= 0, put removes the current key and returns without creating a live entry.

For ttl > 0, put increments the version counter, calculates expires_at = now + ttl, updates the map, and pushes the new expiration record.

After cleanup, get returns None when the key is missing. Otherwise, it returns the stored value.

The consistent example uses these operations:

1. put("A", 100, ttl=5, now=0)
2. put("B", 200, ttl=3, now=1)
3. put("A", 101, ttl=4, now=3)
4. get("B", now=4)
5. cleanup(now=5)
6. get("A", now=6)
7. cleanup(now=8)
8. get("A", now=8)

The initial state is:

store = {} expiry_heap = [] next_version = 0

Step 1 stores A.

expires_at = 0 + 5 = 5 version = 1

The state becomes:

store = {A: (100, 5, 1)} heap = [(5, A, 1)]

Step 2 stores B.

expires_at = 1 + 3 = 4 version = 2

The state becomes:

store = {A: (100, 5, 1), B: (200, 4, 2)} heap = [(4, B, 2), (5, A, 1)]

Step 3 updates A.

expires_at = 3 + 4 = 7 version = 3

The map replaces A v1 with A v3:

store = {A: (101, 7, 3), B: (200, 4, 2)} heap = [(4, B, 2), (5, A, 1), (7, A, 3)]

The old record (5, A, 1) remains in the heap, but it is now stale.

Step 4 calls get("B", now=4).

Cleanup pops (4, B, 2). It matches the current B entry, so B is deleted. The later lookup misses, so get returns None.

The state is:

store = {A: (101, 7, 3)} heap = [(5, A, 1), (7, A, 3)]

Step 5 calls cleanup(now=5).

Cleanup pops (5, A, 1). The current A entry is (101, 7, 3). Its version and expiration time do not match the popped record. The record is stale, so cleanup skips it.

The state is:

store = {A: (101, 7, 3)} heap = [(7, A, 3)]

Step 6 calls get("A", now=6).

The earliest expiration time is 7, which is later than 6. A is still live, so get returns 101.

Step 7 calls cleanup(now=8).

Cleanup pops (7, A, 3). It matches the current A entry, so A is deleted.

The state becomes:

store = {} heap = []

Step 8 calls get("A", now=8).

The key is missing, so get returns None.

The three get results are None, 101, and None.

Repeated updates can leave stale records inside the heap. Those records are safe because of the version check, but they still use memory.

The implementation rebuilds the heap when:

len(expiry_heap) > 2 * len(store) + 32

The rebuild creates one heap record for each current map entry and then calls heapq.heapify.

This prevents stale records from accumulating indefinitely. The threshold avoids rebuilding after every update.

The map always stores the newest live entry for each key.

The min-heap exposes the earliest record that may have expired.

A popped record can delete a map entry only when its key, version, and expiration time all match the current entry.

Globally increasing versions are never reused. Therefore, a stale record cannot match a later reinsertion of the same key.

Because get runs cleanup first, it does not return an entry whose expiration time is at or before now.

Let h be the heap size, n be the number of live entries, and k be the number of records popped during one cleanup call.

A positive-TTL put performs one heap push. This costs O(log h), plus cleanup work.

A get performs an expected O(1) Python dictionary lookup, plus cleanup work.

Cleanup costs O(k log h) when it pops k expired or stale records.

A heap rebuild costs O(n). This work is amortized because rebuilding happens only after the heap becomes much larger than the live map.

The map uses O(n) space. The compacted heap also uses O(n) space, with a fixed extra allowance from the +32 threshold.

Important edge cases include updating a key before its old TTL expires, ttl <= 0, a missing key, a request exactly at the expiration time, reinserting a removed key, and repeatedly rewriting one key.

Use two data structures with different jobs. The hash map stores the latest live entry as key -> (value, expires_at, version). The min-heap stores expiration candidates as (expires_at, key, version), so its top record is the next one that may need removal. The central invariant is that a heap record may delete an entry only when its key, version, and expiration time all match the current map entry. Every successful positive-TTL put receives a globally increasing version that is never reused. This makes lazy deletion safe. Periodic heap rebuilding removes stale records that have not yet reached the top.

Let h be the heap size, n be the number of live keys, and k be the number of records removed during cleanup. A positive-TTL put pushes one heap record, so that part costs O(log h), plus amortized cleanup work. A get uses an expected O(1) Python dictionary lookup, plus amortized cleanup work. Cleanup costs O(k log h) when it pops k records. Rebuilding the heap costs O(n), but it happens only when the heap is much larger than the live map. The map and compacted heap use O(n) auxiliary space, with a fixed extra allowance from the +32 threshold.

A common mistake is deleting a key whenever any heap record for that key expires. An older record could then delete a newer value. Another mistake is deriving a version only from the current map entry. Removing and later reinserting a key could reuse an old version. Candidates may also forget to call cleanup before get, use expires_at < now instead of expires_at <= now, keep a live entry when ttl <= 0, omit heap compaction, or claim that every get is strictly O(1) even when it performs cleanup.

State the invariant before coding: the map owns the current value, and a heap record may delete it only when the key, version, and expiration time all match.

I would use a dictionary for the current record of each key and a min-heap for expiration records. Each heap item stores the expiration time, version, and key. Before every put or get, I pop records whose expiration time is less than or equal to now. A version check prevents an old heap record from deleting a newer value. Dictionary operations are O(1) on average, heap operations are O(log h), and auxiliary space is O(k + h).

The cache stores values in memory and makes each value unavailable after a fixed TTL. Scanning every key on each operation would be wasteful. Instead, the solution uses a dictionary for direct key lookup and a min-heap for expiration order. A version number makes overwrites safe when an older expiration record is still inside the heap.

1. What input sizes, value ranges, and edge cases should the solution handle?
2. What output should be returned for empty, invalid, or duplicate input?
3. Should I prioritize execution time or memory use, and may I use the standard library?

The constructor receives ttl_seconds.

put(key, value, now) stores the value under the key. Its expiration time is now + ttl_seconds.

get(key, now) returns the stored value when the current record has not expired. It returns None when the key is missing or expired.

A record is expired when expires_at <= now.

The dictionary maps each key to:

(value, expires_at, version)

It stores the newest record for that key.

The min-heap stores:

(expires_at, version, key)

Python heapq is a min-heap. Therefore, the record with the smallest expiration time is always at the top.

The version number identifies one specific write. It prevents an old heap record from deleting a newer value written under the same key.

The dictionary starts empty. The heap starts empty. next_version starts at zero.

Before every put and get, the cache runs _evict_expired(now).

After this cleanup, every record remaining in the dictionary has expires_at greater than now.

The heap may still contain stale records. A popped record may delete a dictionary entry only when both its expiration time and version match the current dictionary record.

The TTL is 5 seconds.

At t=0, put A=100.

The cache creates version 1 and calculates expires_at = 0 + 5 = 5. The dictionary becomes A -> (100, 5, 1). The heap becomes [(5, 1, A)].

At t=1, put B=200.

The cache creates version 2 and calculates expires_at = 1 + 5 = 6. The dictionary contains A -> (100, 5, 1) and B -> (200, 6, 2). The heap contains (5, 1, A) and (6, 2, B).

At t=3, put A=150.

No record has expired. The cache creates version 3 and calculates expires_at = 3 + 5 = 8. The dictionary replaces A with A -> (150, 8, 3). The heap receives (8, 3, A). The older record (5, 1, A) remains in the heap, but it is stale.

At t=5, get A.

Cleanup pops (5, 1, A). The current dictionary record for A is (150, 8, 3). The expiration time and version do not match, so the popped record is stale and cannot delete A. The cache returns 150. The dictionary remains unchanged. The heap contains (6, 2, B) and (8, 3, A).

At t=6, get B.

Cleanup pops (6, 2, B). It matches the current dictionary record for B, so B is deleted. The following lookup does not find B, so get returns None. The dictionary contains only A -> (150, 8, 3). The heap contains (8, 3, A).

At t=9, get A.

Cleanup pops (8, 3, A). It matches the current dictionary record for A, so A is deleted. The following lookup does not find A, so get returns None.

The returned values are [150, None, None]. The final dictionary is empty, and the final heap is empty.

The heap always exposes the next possible expiration in time order.

Cleanup removes every heap record whose expiration time is less than or equal to now.

A matching expiration time and version proves that the popped record still represents the current dictionary record. It is therefore safe to delete that key.

A mismatch proves that the heap record belongs to an older write. Ignoring it prevents a stale record from deleting a newer value.

After cleanup, every dictionary record is valid at the supplied time. Therefore, get returns a value only when that value has not expired.

The constructor validates the TTL and creates the dictionary, heap, and version counter.

_evict_expired repeatedly examines the heap top. It pops every record with expires_at <= now. It deletes a key only when the popped expiration time and version match the current dictionary record.

put runs cleanup first. It creates a new version, calculates the expiration time, updates the dictionary, and pushes one heap record.

get also runs cleanup first. It then returns the value from the current dictionary record or None when the key is absent.

Let h be the number of heap records before cleanup. Let r be the number of expired or stale records popped during one operation.

A put takes O(r log h + log h) time for cleanup and the new heap push, plus O(1) average dictionary work.

A get takes O(r log h) cleanup time plus O(1) average dictionary lookup.

Each heap record is pushed once and popped at most once. Cleanup work is therefore amortized across the sequence of operations.

Auxiliary space is O(k + h), where k is the number of current dictionary records. Repeated overwrites can make h larger than k because stale heap records remain until they reach the top.

A negative TTL is rejected. A TTL of zero makes an entry expire at the same timestamp. The next operation at that time or later removes it.

The key insight is to separate direct lookup from expiration order. The dictionary provides O(1) average access to the newest record for a key. The min-heap exposes the next possible expiration without scanning all dictionary entries. The central invariant is that after _evict_expired(now), every dictionary record has expires_at greater than now. Version numbers make lazy cleanup safe. A heap record can delete a key only when its expires_at and version still match the current dictionary record. Any mismatch means the heap record belongs to an older overwrite and must be ignored.

Let h be the number of heap records before cleanup, and let r be the number of expired or stale records popped during the operation. A put takes O(r log h + log h) time for cleanup and the new heap push, plus O(1) average dictionary work. A get takes O(r log h) time for cleanup, plus O(1) average dictionary lookup. Each heap record is pushed once and popped at most once, so cleanup work is amortized across many operations. Auxiliary space is O(k + h), where k is the number of current dictionary records. Repeated overwrites can make h larger than k.

One mistake is deleting a key whenever any old heap record expires. That can remove a newer value written under the same key. Compare both expires_at and version before deleting. Another mistake is running cleanup only during get. put must also clean expired records before writing. A third mistake is using expires_at < now instead of expires_at <= now. In this design, an entry is expired exactly at its expiration timestamp. Candidates may also claim every operation is O(1), even though heap pushes and pops cost O(log h). Finally, repeated overwrites can leave stale records in the heap, so space can grow beyond the number of current keys.

Say the invariant before coding: after cleanup at time now, every dictionary record expires after now, and a popped heap record may delete a key only when both its expiration time and version match the current record.