Apple Python Developer Interview Questions & Answers

I would use None only for an explicit null value and use private sentinel objects for a wildcard and a missing query field. I would store rules in a nested dictionary with one level per lookup dimension. During lookup, I would explore the exact branch and the wildcard branch, count exact matches, and return the unique rule with the highest specificity. A missing query field can match only a wildcard. If several different rules have the same best specificity, I would raise an ambiguity error.

I would represent exact values, nulls, wildcards, and missing fields as separate states. None means an explicit null. A private ANY sentinel means any value. A separate MISSING sentinel means the query did not provide that field. This matters because query.get alone cannot distinguish a missing key from a key whose value is None.

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?

I would index rules in a nested dictionary with one level for each lookup dimension. A rule that omits a dimension uses ANY for that level. During lookup, an exact branch, including a None branch, adds one specificity point. An ANY branch adds no point. A missing query field can follow only ANY because there is no exact value to compare.

The lookup returns the unique rule with the highest specificity. If different rules share the best score, it raises an ambiguity error instead of depending on insertion order. Building the index costs O(n times d) time and up to O(n times d) memory, where n is the number of rules and d is the number of dimensions. Prefix sharing can reduce memory. Lookup may explore up to O(2 to the power d) paths in the worst case, so this design works best with a small, fixed number of dimensions.

The code uses ANY as the wildcard sentinel, MISSING as the absent query field sentinel, and LEAF as a private storage key for rules. None remains an ordinary exact lookup value. Rules are inserted into a nested dictionary in a fixed dimension order. An omitted rule dimension becomes ANY. Lookup explores an exact branch when the query contains the field and also explores an ANY branch when one exists. Each exact branch adds one specificity point. Missing query fields can follow only ANY. The code returns the unique rule with the highest score and raises AmbiguousMatchError when different rules share that score. Index construction uses O(n times d) time and up to O(n times d) memory. Worst case lookup explores O(2 to the power d) paths.

Common mistakes include using None for both null and wildcard, using query.get without a missing sentinel, treating a missing field as an explicit null, returning the first match found, and using insertion order to hide ambiguous rules. Another mistake is claiming lookup always costs O(d). Exact and wildcard branches may both exist at each level, so worst case lookup can explore O(2 to the power d) paths. It is also a mistake to ignore the memory used by nested dictionary nodes.

Start with the representation rule. Say that None means explicit null, ANY means wildcard, and MISSING means an absent query field. Then explain specificity scoring, ambiguity handling, and the worst case branching and memory costs.

I would parse the stream incrementally and build a compact index during the same pass. Each valid event would store its identifier, byte offset, record length, and schema version. Later, Python can open the immutable source in binary mode, seek to that offset, read one framed record, and run the same decoder and validator. I would rebuild the index whenever the source changes because saved offsets are valid only for the exact bytes that were indexed.

I would separate sequential parsing from random access. The sequential layer reads one framed record at a time and yields validated events, so Python keeps only the current payload in memory. Each record has a fixed four byte header that stores the payload length. This stable boundary tells the parser where one event ends and the next begins.

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?

For every valid event, I would store compact metadata: the event identifier, byte offset, total record length, and schema version. Invalid JSON, unsupported schemas, missing fields, and duplicate identifiers can be isolated at a complete record boundary. A truncated header or payload is different because the next boundary is unknown, so the scan must stop safely.

A later lookup opens the immutable source in binary mode, seeks to the saved byte offset, reads exactly one frame, and runs the same decoder and validator. Building the index takes one sequential pass. Each lookup reads one record, while memory use is the index plus the largest payload being processed. Production code should treat indexed files as immutable, store source identity metadata, and rebuild the index after replacement, truncation, or rewriting.

The code writes every JSON event as a four byte length header followed by one payload. The index builder opens the file in binary mode and scans it once. It validates each complete record and stores compact metadata only for valid unique event identifiers. Invalid payloads are reported while later framed records continue. Truncated framing or an unreasonable payload length stops the scan because continuing would not be safe. The code records file identity metadata from the open file before and after indexing. A lookup opens the source, verifies the identity on that same file handle, seeks to the saved byte offset, reads the exact frame, checks the stored length, and applies the same decoder and validator used during indexing.

Common mistakes include loading the full stream into a list, storing complete event objects instead of compact index metadata, recording text positions instead of byte offsets, opening the source in text mode, and trusting an unbounded length field. Another mistake is catching every exception around the whole file, which allows one bad record to stop all processing. Candidates may also continue after truncated framing even though the next boundary is unknown, validate only during the first pass, or reuse offsets after the source file has changed.

Start with the practical split: incremental parsing for normal processing and a byte offset index for later direct reads. Then explain stable framing, shared validation, record level error isolation, bounded memory use, and stale index detection.

I would define the accepted shapes as part of the function contract and reject every unexpected shape before doing the calculation. I would use clear names such as batch_size and class_count, raise ValueError with the expected and received shapes, and allow broadcasting only after validating that it is intentional. I would also check important numerical conditions and use deterministic tests for valid input, mismatched dimensions, empty batches, invalid values, and boundary cases.

I would treat each tensor shape as part of the Python function contract. For example, logits must have shape batch_size by class_count, while bias must have shape class_count. I would validate rank and dimension sizes before NumPy performs arithmetic. I would not call reshape, flatten, or squeeze merely to make an operation succeed, because that can hide an upstream error.

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?

NumPy broadcasting can copy the logical effect of a smaller array across a larger one without physically copying every value. That behavior is useful here, but only after checking that bias has exactly one value per class.

The function should raise ValueError with the parameter name, expected shape, and received shape. It should also reject non real data, non finite values, an empty class dimension, and an invalid temperature. An empty batch can be allowed because its output shape remains well defined.

Deterministic tests should use fixed arrays and expected results. Shape checks take constant time because shape metadata is already stored. Value checks and softmax each scan all elements. The implementation therefore takes linear time and linear temporary memory. Converting input to float64 may also allocate a copy when the original type differs.

The function accepts logits with shape batch_size by class_count, bias with shape class_count, and one finite positive temperature value. It first rejects unsupported data types and converts valid real numeric inputs to float64. It then checks rank, requires at least one class, and verifies the exact bias shape before using intentional NumPy broadcasting. An empty batch is accepted and returns an empty float64 array with the same two dimensional shape. For a nonempty batch, the function rejects non finite inputs and catches overflow or invalid arithmetic while adding bias and dividing by temperature. It computes stable softmax by subtracting each row maximum before exponentiation. The deterministic tests verify output shape, row sums, equal score behavior, empty batch behavior, shape failures, non finite data, and invalid temperature. For N tensor elements, runtime is O(N). Temporary arrays and a required float64 conversion can use O(N) extra memory.

Common mistakes include reshaping or squeezing data until an operation runs, trusting every shape that NumPy can broadcast, comparing only the total number of elements, and returning a vague error. Other mistakes include silently converting complex values to real values, ignoring empty dimensions, accepting a non scalar temperature, and checking only successful examples. Using assert for caller supplied input is also wrong because Python can remove assertions when optimization is enabled.

State the exact tensor contract first. Then explain which broadcasting is intentional, which failures raise exceptions, how empty batches behave, and which deterministic tests protect the contract. Finish by separating constant time shape checks from linear value scans and temporary array allocations.

I would use a running prefix sum and a hash map that stores how many times each earlier prefix sum has appeared. I start the map with 0 mapped to 1. For each number, I update the prefix sum, look for prefix_sum minus k, add that frequency to the answer, and then record the current prefix sum. This handles negative values and repeated prefix sums. The solution takes O(n) expected time and O(n) auxiliary space.

The problem asks for the number of contiguous subarrays whose values add up to k. A contiguous subarray contains neighboring elements. I use a running prefix sum and a hash map of prefix-sum frequencies. This method works with positive, negative, and zero values because it does not depend on the sum moving in only one direction.

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 nums and an integer k. The output is one integer. It is the number of contiguous subarrays whose sum equals k.

For the diagram example, nums = [1, -1, 1, 1] and k = 1. The expected answer is 5. The valid subarrays are [0..0], [0..2], [2..2], [1..3], and [3..3].

I keep a running prefix sum. A prefix sum is the total of all values processed so far.

Suppose the current prefix sum is S. If an earlier prefix sum equals S - k, then the values between that earlier position and the current position add up to k.

The hash map is named prefix_count. Each key is a prefix sum value. Each value is the number of times that prefix sum has appeared earlier.

The central invariant is that prefix_count stores frequencies of prefix sums from earlier positions only.

Set prefix_sum = 0 and count = 0.

Set prefix_count = {0: 1}. The zero represents the empty prefix before the array starts. This lets the algorithm count a valid subarray that begins at index 0.

Traversal begins at index 0.

Before processing any element, prefix_sum = 0, count = 0, and prefix_count = {0: 1}.

At index 0, num is 1. Update prefix_sum to 1. Compute needed = 1 - 1 = 0. Before this step, prefix_count is {0: 1}. The needed value appears once, so add 1 to count. Count becomes 1. Then record prefix sum 1. The map becomes {0: 1, 1: 1}. This step counts [0..0].

At index 1, num is -1. Update prefix_sum to 0. Compute needed = 0 - 1 = -1. Before this step, prefix_count is {0: 1, 1: 1}. The needed value does not appear, so count stays 1. Then increment the frequency of prefix sum 0. The map becomes {0: 2, 1: 1}.

At index 2, num is 1. Update prefix_sum to 1. Compute needed = 1 - 1 = 0. Before this step, prefix_count is {0: 2, 1: 1}. Prefix sum 0 has appeared twice, so add 2 to count. Count becomes 3. Then increment the frequency of prefix sum 1. The map becomes {0: 2, 1: 2}. This step counts [0..2] and [2..2].

At index 3, num is 1. Update prefix_sum to 2. Compute needed = 2 - 1 = 1. Before this step, prefix_count is {0: 2, 1: 2}. Prefix sum 1 has appeared twice, so add 2 to count. Count becomes 5. Then record prefix sum 2. The map becomes {0: 2, 1: 2, 2: 1}. This step counts [1..3] and [3..3].

All four elements have been processed. Return 5.

At each index, let the current prefix sum be S. Every earlier prefix sum equal to S - k defines one contiguous subarray ending at the current index whose sum is k.

The map stores the frequency of each earlier prefix sum. Therefore, prefix_count[S - k] is exactly the number of valid subarrays ending at the current index.

The algorithm adds matches before recording the current prefix sum. This means the current position cannot be treated as an earlier prefix. Each valid subarray is counted exactly once at its ending index.

The code creates prefix_count with {0: 1}. It also initializes prefix_sum and count to zero.

For each number, it first adds the number to prefix_sum. It then evaluates prefix_sum - k and uses dictionary get to find how many times that needed prefix sum appeared earlier. That frequency is added to count.

After counting matches, the code increments the frequency of the current prefix sum. When the loop ends, it returns count.

The expected time is O(n). The loop processes each array element once. Python dictionary lookup and insertion take O(1) time on average.

The auxiliary space is O(n). In the worst input shape, the dictionary may contain a different prefix sum for many positions.

An empty array returns 0. Negative values work. Repeated prefix sums correctly create multiple matches. Zero and duplicate values also work.

The key insight is that a subarray sum can be written as the difference between two prefix sums. If the current prefix sum is S, an earlier prefix sum S - k creates a subarray ending at the current index with sum k. The hash map stores prefix sum value to frequency. Its invariant is that it contains counts of earlier prefix sums only. Frequencies are required because the same prefix sum may appear more than once, and each occurrence creates a different valid subarray. This method is more suitable than a sliding window because negative values can make a window sum rise or fall unpredictably.

Let n be the number of elements in nums. The expected time is O(n) because the code processes each element once. Each Python dictionary lookup and insertion takes O(1) time on average. This is expected time rather than a guaranteed worst-case bound for hashing. The auxiliary space is O(n) because the dictionary may store many different prefix sums. The empty input still uses only the initial map entry and returns 0.

A common mistake is using a sliding window even though negative values are allowed. Another mistake is forgetting to initialize prefix_count with {0: 1}, which misses valid subarrays that start at index 0. Some candidates record the current prefix sum before counting matches. The lookup must happen first so the map represents earlier prefixes only. Another mistake is storing only whether a prefix sum exists. The map must store frequencies because repeated prefix sums can create multiple valid subarrays. It is also incorrect to claim guaranteed O(n) time because Python dictionary operations are O(1) on average.

State the invariant before writing code: prefix_count stores the frequency of every prefix sum seen before the current position. Then explain that matches are counted before the current prefix sum is recorded.

I would use vertical scanning. I take the first string as the reference and check its characters from left to right. At each index, I compare that character with the character at the same index in every other string. If a word is too short or has a different character, I return the confirmed prefix immediately. The first mismatch proves that no longer prefix can work. The time complexity is O(n × m). With the shown Python slice, auxiliary space is O(n).

The problem asks for the longest starting part shared by every string. I use the first string as the reference and verify its characters one position at a time. The prefix grows only while every string has the same character at that position. When a string ends or a mismatch appears, the algorithm returns the prefix before that position.

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 one string containing the longest prefix shared by all strings.

For the example ["flower", "flow", "flight"], the correct output is "fl".

An empty list returns "". A list with one string returns that string. If any string is empty, the answer is "". Comparisons are case-sensitive, so uppercase and lowercase letters are different.

I use the first string as the reference. Each character position in that string is a possible position in the common prefix.

At index i, I compare first[i] with the character at index i in every other string.

The invariant is that before checking index i, first[:i] is the longest prefix confirmed to match every string so far.

I first check whether the list is empty. If it is empty, I return "".

I then store the first string in first. If there is only one string, I return it because the whole string is its own longest common prefix.

The character scan begins at index 0 of first.

The input is ["flower", "flow", "flight"]. The reference string is "flower".

At index 0, the reference character is 'f'. flow[0] is 'f', and flight[0] is also 'f'. All strings match, so the confirmed prefix becomes "f".

At index 1, the reference character is 'l'. flow[1] and flight[1] are also 'l'. The confirmed prefix becomes "fl".

At index 2, the reference character is 'o'. flow[2] is 'o', but flight[2] is 'i'. The condition fails because the characters are different.

The function returns first[:2], which is "fl". It stops immediately and does not process later positions.

Before index 2 is checked, the prefix first[:2], which is "fl", already matches every string.

The mismatch at index 2 proves that no prefix containing that position can be shared by all strings. Therefore, "fl" is the longest possible common prefix.

The outer loop reads the first string from left to right with enumerate(first). This gives the current index i and character ch.

The inner loop processes strings[1:], which contains every string except the reference string. For each word, the code checks whether index i exists and whether word[i] equals ch.

If the word is too short or the character is different, the function returns first[:i]. If every character in the first string matches every other string, the function returns first.

Let n be the number of strings. Let m be the number of positions checked before a mismatch or the end of the shortest relevant string.

The time complexity is O(n × m). At each checked position, the algorithm may compare the character with every other string. It may stop early after the first mismatch.

The shown code uses strings[1:]. That expression creates a new list containing references to the remaining strings. Its peak auxiliary space is therefore O(n). The returned prefix slice also uses space proportional to the returned prefix, but output space is usually reported separately.

Important edge cases are an empty list, one string, an empty string, no shared first character, and case-sensitive input.

The key insight is to use the first string as the reference and check its characters vertically across the other strings. At each index, every string must contain the same character for the common prefix to continue. The invariant is that before checking index i, first[:i] matches every string. The first short string or mismatching character proves that no longer prefix can be common, so returning first[:i] is correct.

Let n be the number of strings. Let m be the number of character positions checked before the algorithm stops. For each checked position, the code may compare one character in every string, so the time complexity is O(n × m). The algorithm can stop early when it finds a mismatch. The shown code uses strings[1:], which creates a temporary list of up to n - 1 string references. Therefore, its peak auxiliary space is O(n), excluding the returned prefix.

A common mistake is reading word[i] before checking whether the word is long enough. This can cause an index error. Another mistake is returning first[:i + 1], which incorrectly includes the mismatching position. Candidates may forget the empty-list, one-string, or empty-string cases. Some continue scanning after the first mismatch even though the result is already known. Another mistake is treating uppercase and lowercase letters as equal. Candidates may also claim O(1) auxiliary space while using strings[1:], which creates a temporary list.

State the invariant before coding: before checking index i, first[:i] already matches every string. This makes the early return and correctness argument easy to explain.

I group words by their length, first letter, and last letter, because only words with the same signature can collide. Inside each group, I sort the words but keep their original indices. For each word, I compare it with its previous and next sorted neighbors using longest common prefix lengths. The needed prefix is one more than the larger LCP. If the abbreviation is not shorter, I keep the original word. The time is O(n log n · m), and the auxiliary space is O(n).

The problem asks us to create the shortest unique abbreviation for every word while preserving the original input order. An abbreviation contains a prefix, the number of omitted middle letters, and the final letter. The main challenge is handling words that would otherwise produce the same abbreviation. The diagram solves this by grouping possible collisions, sorting each group, and comparing neighboring words.

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?

Two words can produce the same abbreviation only when they have the same length, first letter, and last letter.

I use the tuple (length, first letter, last letter) as the signature.

For the example, internal and interval share (8, i, l). The words intension and intrusion share (9, i, n). Every other word is in a singleton group.

The hash map stores:

• Key: the signature.
• Value: a list of (word, original_index) pairs.

The original index is saved because each group is sorted temporarily.

I sort each signature group in lexicographic order.

After sorting, the word that shares the longest prefix with the current word must be one of its two adjacent neighbors. Therefore, I compare only the previous and next words.

This avoids comparing every word with every other word in the same group.

For each word, I calculate:

• The longest common prefix with the previous word.
• The longest common prefix with the next word.

The required prefix length is:

1 + max(left LCP, right LCP)

For internal and interval, the shared prefix is inter, which has length 5. Both words therefore need a prefix length of 6.

For intension and intrusion, the shared prefix is int, which has length 3. Both words need a prefix length of 4.

The abbreviation contains:

• The required prefix.
• The number of omitted middle letters.
• The final letter.

The omitted count is:

len(word) - prefix_len - 1

If len(word) - prefix_len <= 2, the abbreviation would not be shorter than the original word. In that case, I keep the original word.

This is why internal and interval remain unchanged. Their required prefix length is 6, so the abbreviation would not save any characters.

The pair intension and intrusion becomes inte4n and intr4n.

The input is:

["like", "god", "internal", "me", "internet", "interval", "intension", "face", "intrusion"]

The execution summary is:

• like: prefix length 1, result l2e.
• god: prefix length 1, keep god because the abbreviation is not shorter.
• internal: LCP 5 with interval, prefix length 6, keep internal.
• me: prefix length 1, keep me because it is too short.
• internet: singleton group, prefix length 1, result i6t.
• interval: LCP 5 with internal, prefix length 6, keep interval.
• intension: LCP 3 with intrusion, prefix length 4, result inte4n.
• face: prefix length 1, result f2e.
• intrusion: LCP 3 with intension, prefix length 4, result intr4n.

The final result in original order is:

["l2e", "god", "internal", "me", "i6t", "interval", "inte4n", "f2e", "intr4n"]

Only words in the same signature group can share an abbreviation.

Inside one group, sorting places words with similar prefixes next to each other. For any word, its maximum shared prefix with another word in the group is reached by its previous or next sorted neighbor. Therefore, one plus the larger neighboring LCP is the smallest prefix that separates the word from every other word in that group.

If the resulting abbreviation is not shorter, returning the original word follows the stated rule.

The code first builds the signature groups. It then sorts each group and calculates one required prefix length per word. Finally, it writes each abbreviation into the result array using the saved original index.

Let n be the number of words and m be the maximum word length. Grouping takes O(n). Sorting the groups and comparing strings takes O(n log n · m) overall. The groups, saved indices, prefix-length arrays, and result array use O(n) auxiliary space besides the returned strings.

The key insight is that only words with the same length, first letter, and last letter can collide. The algorithm groups words by that signature and stores each word with its original index. It sorts each group lexicographically. In sorted order, a word's largest shared prefix with any other word is achieved by its previous or next neighbor. The required prefix is therefore one plus the larger neighboring LCP. The invariant is that each chosen prefix is the smallest prefix that makes the word unique inside its signature group. If the abbreviation is not shorter, the original word is kept.

Let n be the number of words and m be the maximum word length. Building the signature groups takes O(n). Sorting all groups takes O(n log n) comparisons in the largest case, and each string comparison may inspect up to m characters. The neighboring LCP checks also inspect at most m characters per comparison. The overall time is O(n log n · m). The groups, saved original indices, prefix-length arrays, and result array use O(n) auxiliary space besides the returned strings.

Common mistakes are comparing every word with every other word, losing original indices after sorting, checking only one neighboring LCP, and forgetting that the final letter is not part of the omitted count. Another mistake is returning an abbreviation when it is not shorter than the original word. Candidates may also state O(n log n) time and forget that string comparisons can inspect up to m characters.

Explain the collision signature first. Then explain why sorting makes the previous and next words sufficient for the LCP check. This gives the interviewer the main correctness argument before you write the implementation.

I would use two stacks. New values go into in_stack. For dequeue, I first use out_stack if it already contains values. If it is empty, I move every value from in_stack to out_stack. This reverses the order, so the oldest value becomes the next one removed. An empty queue raises IndexError. Enqueue is O(1). Dequeue is amortized O(1), with O(n) worst-case time during a transfer. Auxiliary space is O(n).

The problem asks us to build a first-in-first-out queue using two last-in-first-out stacks. A queue removes the oldest value first, while a stack removes the newest value first. We solve this difference by using one stack for new values and another stack for values that are ready to leave. Moving values between the stacks reverses their order and restores FIFO behavior.

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?

I use in_stack and out_stack.

in_stack stores newly enqueued values. Its newest value is on top.

out_stack stores values in dequeue order. When out_stack is not empty, its top is the current front of the queue.

The stack top is the right side in list notation.

To enqueue a value, I append it to in_stack.

For example, after enqueue(10), enqueue(20), and enqueue(30), the state is:

in_stack = [10, 20, 30] out_stack = []

The value 30 is on top of in_stack because it is on the right.

For dequeue, I first check out_stack.

If out_stack already contains values, I pop its top and return it.

If out_stack is empty, I move every value from in_stack to out_stack. Each move pops from in_stack and appends to out_stack.

This reverses the order. The oldest enqueued value becomes the top of out_stack.

If out_stack is still empty after the transfer, both stacks were empty. The method raises IndexError("dequeue from empty queue").

The operations are enqueue(10), enqueue(20), enqueue(30), dequeue(), enqueue(40), dequeue().

Step 1: enqueue(10).

Before: in_stack = [], out_stack = [].

Action: Push 10 to in_stack.

After: in_stack = [10], out_stack = [].

Step 2: enqueue(20).

Before: in_stack = [10], out_stack = [].

Action: Push 20 to in_stack.

After: in_stack = [10, 20], out_stack = [].

Step 3: enqueue(30).

Before: in_stack = [10, 20], out_stack = [].

Action: Push 30 to in_stack.

After: in_stack = [10, 20, 30], out_stack = [].

Step 4: dequeue().

Before: in_stack = [10, 20, 30], out_stack = [].

out_stack is empty, so the code transfers 30, then 20, then 10 into out_stack. The transfer produces out_stack = [30, 20, 10]. The code then pops 10.

After: in_stack = [], out_stack = [30, 20].

Returned value: 10.

Step 5: enqueue(40).

Before: in_stack = [], out_stack = [30, 20].

Action: Push 40 to in_stack.

After: in_stack = [40], out_stack = [30, 20].

Step 6: dequeue().

Before: in_stack = [40], out_stack = [30, 20].

out_stack is not empty, so no transfer is needed. The code pops 20 directly.

After: in_stack = [40], out_stack = [30].

Returned value: 20.

The returned values are [10, 20]. The remaining queue is [30, 40] from front to back.

The central invariant is that when out_stack is not empty, its top is the oldest value currently in the queue.

Newer values stay in in_stack. When out_stack becomes empty, moving all values from in_stack reverses their order. This puts the oldest waiting value on top of out_stack.

Therefore, values leave in the same order in which they entered. That is exactly FIFO behavior.

The constructor creates two empty lists.

enqueue uses append on in_stack.

dequeue transfers values only when out_stack is empty. This avoids repeating work.

After the optional transfer, dequeue checks out_stack again. If it is still empty, the queue has no values, so the method raises IndexError.

Otherwise, it pops and returns the top value from out_stack.

Enqueue takes O(1) time because it performs one append.

A dequeue can take O(n) time when it transfers n values. However, each value is transferred from in_stack to out_stack at most once. Across many operations, dequeue therefore takes amortized O(1) time.

The two stacks can store up to n values in total, so auxiliary space is O(n).

Important edge cases are an empty queue, one value, many enqueues before the first dequeue, and interleaved enqueue and dequeue operations.

The key idea is to split the queue state between two stacks. in_stack accepts new values in arrival order. out_stack exposes values in removal order. The central invariant is that when out_stack is not empty, its top is the current front of the queue. When out_stack is empty, moving every value from in_stack to out_stack reverses the order. The oldest waiting value then becomes the top of out_stack. We transfer only when needed, so each value is moved from in_stack to out_stack at most once. This gives the standard optimal amortized solution.

enqueue takes O(1) time because it only appends one value to in_stack. A single dequeue can take O(n) time when out_stack is empty and all n waiting values must be transferred. However, each value moves from in_stack to out_stack only once. Because of this, dequeue is amortized O(1). Amortized means the average cost across a sequence of operations is constant, even though one operation may be expensive. The two stacks hold at most n values in total, so auxiliary space is O(n).

A common mistake is transferring values on every dequeue. The transfer should happen only when out_stack is empty. Another mistake is moving values in the wrong direction. Values must move from in_stack to out_stack. Some candidates pop from in_stack directly, which returns the newest value and breaks FIFO order. Others forget to raise an error when both stacks are empty. It is also incorrect to claim that every dequeue is always O(1). One dequeue may take O(n), while the amortized cost is O(1).

State the invariant before writing code: when out_stack is not empty, its top is the queue front. Then explain that a transfer happens only when out_stack is empty. This makes both correctness and the amortized O(1) dequeue cost easy to justify.

I would use BFS with a deque and a color map. The map stores each node with color 1 or -1. I loop through every node because the graph may be disconnected. For each uncolored node, I start BFS and give every uncolored neighbor the opposite color. If an edge connects two nodes with the same color, I return False immediately. If every component finishes without a conflict, I return True. The time complexity is O(V + E), and the auxiliary space is O(V).

The input is an undirected graph stored as an adjacency list. We must decide whether its nodes can be divided into two color groups so that every edge connects nodes from different groups. This is called checking whether the graph is bipartite, or two-colorable. BFS works well because we can color each newly discovered neighbor with the opposite color and detect a conflict immediately.

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 undirected, unweighted graph represented by an adjacency list. Each dictionary key is a node. Its value is the list of nodes connected to it.

The output is a Boolean value. We return True if every connected component can be colored with two colors so that adjacent nodes have different colors. We return False when we find an edge whose two endpoints have the same color.

The graph may be disconnected. One BFS can visit only the component containing its start node, so an outer loop must check every node.

The color map stores node to color. Each color is either 1 or -1.

The deque is the BFS queue. It stores discovered nodes that still need to be processed.

When the current node has one color, each uncolored neighbor receives the opposite color. If a neighbor already has the same color as the current node, the graph is not two-colorable.

The central invariant is that every processed edge connects nodes with opposite colors.

We begin with an empty color map.

The outer loop visits every node in the graph. If a node is already in the color map, BFS has already reached it, so we skip it.

If a node is uncolored, we start a new component. We assign it color 1 and place it in a deque.

For the diagram's example, iteration begins at node 0 because the dictionary keys are listed in numeric order.

The graph has these edges:

(0,1), (1,2), (2,3), (3,0), (4,5), (5,6), and (6,4).

It contains two disconnected components.

Step 1: Start component A at node 0. The color map becomes {0: 1}. The queue becomes [0].

Step 2: Remove node 0 from the queue. Nodes 1 and 3 are uncolored, so both receive color -1. The queue becomes [1, 3]. The color map becomes {0: 1, 1: -1, 3: -1}.

Step 3: Remove node 1. Neighbor 0 already has the opposite color. Node 2 is uncolored, so it receives color 1. The queue becomes [3, 2].

Step 4: Remove node 3. Neighbors 0 and 2 already have the opposite color, so no state changes.

Step 5: Remove node 2. Neighbors 1 and 3 already have the opposite color. Component A finishes successfully.

Step 6: The outer loop reaches uncolored node 4. We assign node 4 color 1 and place it in the queue.

Step 7: Remove node 4. Nodes 5 and 6 are uncolored, so both receive color -1. The queue becomes [5, 6].

Step 8: Remove node 5. Neighbor 4 has the opposite color, which is valid. Neighbor 6 already has color -1, which is the same color as node 5. Edge (5, 6) is a conflict, so the function returns False immediately.

Node 6 was discovered and colored, but it is not removed from the queue after the conflict. The nodes removed before stopping are 0, 1, 3, 2, 4, and 5.

Whenever BFS colors a new neighbor, it assigns the negative of the current node's color. Therefore, the edge between them connects opposite colors.

If an examined edge connects two nodes with the same color, the current component cannot satisfy the two-color rule. In this example, nodes 4, 5, and 6 form a triangle. A triangle is an odd cycle, and an odd cycle cannot be colored with only two colors.

The outer loop starts BFS from every remaining uncolored node. Therefore, every disconnected component is checked.

The function creates an empty dictionary named color.

The outer for loop processes every node key. It starts BFS only when the node is not already colored.

Inside BFS, popleft removes the next node from the deque. The inner loop reads each neighbor from the adjacency list.

If a neighbor is uncolored, the code assigns -color[node] and adds the neighbor to the deque. If the neighbor already has the same color, the code returns False immediately.

If all components finish without a conflict, the function returns True.

The time complexity is O(V + E). V is the number of nodes, and E is the number of edges. Each node is colored at most once. Each undirected edge is examined from both endpoints, which is still O(E) total work.

The auxiliary space is O(V). The color map can store all nodes, and the deque can also contain up to V nodes.

An empty graph returns True. An isolated node is valid because it has no conflicting edge. Disconnected graphs are handled by the outer loop. A self-loop returns False because a node cannot have a different color from itself. Any odd cycle also returns False.

Use BFS separately on every disconnected component. A dictionary stores each node's color as either 1 or -1. Start each uncolored component with color 1. During BFS, assign every uncolored neighbor the opposite color and add it to the deque. If an already colored neighbor has the same color as the current node, return False immediately. The invariant is that every processed edge connects nodes with opposite colors. If all components preserve this invariant, the graph is two-colorable.

The time complexity is O(V + E). V is the number of nodes, and E is the number of edges. Each node receives a color at most once. Each undirected edge is checked from both endpoints, so all edge checks together take O(E) time. The algorithm may stop early when it finds a conflict. The auxiliary space is O(V) because the color dictionary and the deque can each grow with the number of nodes.

A common mistake is starting BFS only from node 0, which can miss a conflicting disconnected component. Another mistake is adding a node to the queue before assigning its color, which can cause repeated discovery. Some candidates check only whether a neighbor was visited and forget to compare the two colors. Others fail to return immediately after finding a same-color edge. It is also incorrect to claim O(1) auxiliary space because the color map and deque can grow to O(V).

State the invariant before writing code: every processed edge must connect nodes with opposite colors. Then mention that the outer loop is required for disconnected components. These two points explain almost the entire solution.

I scan the Roman numeral from right to left and use a fixed lookup map for the seven Roman symbols. I keep a running total and the value of the symbol immediately to the right. If the current value is smaller than that previous value, I subtract it. Otherwise, I add it. This handles pairs such as IV, XC, and CM correctly. The solution takes O(n) time and O(1) auxiliary space because the lookup map has a fixed size.

The input is a valid Roman numeral string, and the output is its integer value. I scan from right to left because this makes subtractive pairs easy to detect. When the current symbol is smaller than the symbol immediately to its right, I subtract it. Otherwise, I add it.

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 function receives a valid Roman numeral string such as "MCMXCIV". It must return the integer 1994.

The question guarantees that the Roman numeral is valid. Therefore, this solution does not need to reject invalid formats.

The symbol values are I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, and M = 1000.

I use a fixed lookup map from each Roman symbol to its integer value.

I scan the string from right to left. I keep two variables:

total stores the value calculated so far.

prev_value stores the value of the symbol immediately to the right of the next symbol being processed.

If current_value < prev_value, the current symbol is part of a subtractive pair, so I subtract it. Otherwise, I add it.

The invariant is that before each new character is processed, total equals the value of the suffix already processed, and prev_value equals the value of that suffix's leftmost symbol, which is immediately to the right of the next character.

Set total = 0 and prev_value = 0.

The scan starts at the rightmost character. Every Roman symbol has a positive value, so the first value is added because it is greater than or equal to zero.

The example is MCMXCIV, with indices 0 through 6.

At index 6, the character is V and its value is 5. The previous value is 0. Since 5 is greater than or equal to 0, add 5. The state becomes total = 5 and prev_value = 5.

At index 5, the character is I and its value is 1. Since 1 is smaller than 5, subtract 1. The state becomes total = 4 and prev_value = 1. This is how IV contributes 4.

At index 4, the character is C and its value is 100. Since 100 is greater than or equal to 1, add 100. The state becomes total = 104 and prev_value = 100.

At index 3, the character is X and its value is 10. Since 10 is smaller than 100, subtract 10. The state becomes total = 94 and prev_value = 10. This handles XC.

At index 2, the character is M and its value is 1000. Since 1000 is greater than or equal to 10, add 1000. The state becomes total = 1094 and prev_value = 1000.

At index 1, the character is C and its value is 100. Since 100 is smaller than 1000, subtract 100. The state becomes total = 994 and prev_value = 100. This handles CM.

At index 0, the character is M and its value is 1000. Since 1000 is greater than or equal to 100, add 1000. The state becomes total = 1994 and prev_value = 1000.

All seven characters have now been processed. The function returns 1994.

Roman numeral values are normally added. A smaller symbol that appears immediately before a larger symbol must be subtracted.

Scanning from right to left lets the algorithm compare each symbol with the symbol immediately to its right. The algorithm subtracts exactly when the current value is smaller. In every other case, it adds. Therefore, after every step, total remains equal to the value of the suffix already processed.

The dictionary maps each Roman character to its integer value. The loop uses reversed(s) to visit the string from right to left.

For each character, the code reads current_value. It compares that value with prev_value, adds or subtracts it, and then updates prev_value.

After the loop finishes, total contains the full integer value and is returned.

The algorithm processes each of the n characters once, so the time complexity is O(n).

The lookup map always contains seven entries. The other state consists of two integer variables. Therefore, the auxiliary space complexity is O(1).

Relevant cases include a single symbol such as V, repeated symbols such as III, and valid subtractive pairs such as IV, IX, XL, XC, CD, and CM. Invalid-format validation is outside this solution because the input is guaranteed to be valid.

The key insight is to scan from right to left. This makes the symbol immediately to the right available through prev_value. A fixed lookup map converts each Roman character to its integer value. The invariant is that total equals the value of the suffix already processed. If the current value is smaller than prev_value, Roman numeral rules require subtraction. Otherwise, the value is added. Updating prev_value after every character keeps the comparison correct for the next step.

Let n be the number of characters in the Roman numeral. The loop processes every character once, so the time complexity is O(n). The lookup map always has the same seven entries, and the algorithm uses only total, prev_value, and current_value. Its memory use does not grow with n. Therefore, the auxiliary space complexity is O(1).

A common mistake is comparing the current symbol with the wrong neighbor. Another mistake is forgetting to update prev_value after each character. Some candidates subtract a symbol because it is smaller than any later symbol, but this method compares it with the symbol immediately to its right. Another mistake is changing the traversal direction without changing the comparison logic. Candidates may also claim O(n) extra space because a dictionary is used, but this dictionary always has seven entries, so the auxiliary space is O(1).

State the invariant before writing the loop: total is the value of the suffix already processed, and prev_value is the value immediately to the right of the next character. This makes the subtraction rule easy to explain.

I would use the Sieve of Eratosthenes. I create a boolean array where each index represents the same number. I mark 0 and 1 as not prime. Then I process possible base primes starting from 2. For each confirmed prime p, I mark its multiples from p squared as composite. I stop when p squared is greater than n. The remaining True indices are the primes. The time complexity is O(n log log n), and the auxiliary space complexity is O(n).

The problem asks us to return every prime number from 2 through n. If n is less than 2, we return an empty list. The Sieve of Eratosthenes is a good fit because it finds all primes in the range together. It removes composite numbers by marking multiples of confirmed primes.

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 one integer named n. The output is a list containing every prime number from 2 through n in ascending order.

A prime number has exactly two positive factors: 1 and itself. If n is less than 2, there are no prime numbers in the range, so the function returns an empty list.

For the example n = 20, the output is [2, 3, 5, 7, 11, 13, 17, 19].

I use the Sieve of Eratosthenes and a boolean array named is_prime. Each array index represents the number with the same value. For example, is_prime[7] tells us whether 7 is still considered prime.

The central invariant is: any value already marked False has a smaller prime factor. Therefore, that value is composite.

This method is more suitable than checking every number separately when we need all primes up to the same limit.

I create is_prime with n + 1 entries and set every entry to True. The valid indices are 0 through n.

I then set is_prime[0] and is_prime[1] to False because 0 and 1 are not prime.

The variable p starts at 2. The loop continues while p * p <= n.

The array covers indices 0 through 20. Indices 0 and 1 are False. Indices 2 through 20 begin as True.

For p = 2, the condition 2 * 2 <= 20 is true, and is_prime[2] is True. I start at 2 squared, which is 4. I mark 4, 6, 8, 10, 12, 14, 16, 18, and 20 as False. The values still marked True are 2, 3, 5, 7, 9, 11, 13, 15, 17, and 19.

For p = 3, the condition 3 * 3 <= 20 is true, and is_prime[3] is True. I start at 3 squared, which is 9. I mark 9, 12, 15, and 18 as False. The values still marked True are 2, 3, 5, 7, 11, 13, 17, and 19.

For p = 4, the condition 4 * 4 <= 20 is true, but is_prime[4] is False. I skip the marking step because 4 is already known to be composite. The state does not change.

The next value is p = 5. Now 5 * 5 is 25, which is greater than 20. The loop stops. I collect every index from 2 through 20 whose value is still True. The returned result is [2, 3, 5, 7, 11, 13, 17, 19].

Every composite number up to n has at least one prime factor no greater than the square root of n.

For each confirmed prime p, the algorithm marks its composite multiples. It starts at p squared because every smaller multiple of p already has a smaller prime factor and was handled earlier.

After all required base values have been processed, every composite number is False. Therefore, every index that remains True is prime.

The function first handles the n < 2 case. It then creates the is_prime array and marks indices 0 and 1 as False.

A while loop processes p while p * p <= n. If is_prime[p] is True, a for loop marks multiples from p * p through n. The loop uses a step of p. After each outer-loop iteration, p increases by 1.

Finally, a list comprehension returns every index from 2 through n whose entry is still True.

The time complexity is O(n log log n). The sieve marks multiples of primes in an efficient combined process.

The auxiliary space complexity is O(n) because the boolean array contains n + 1 entries.

Important edge cases are n < 2, which returns []; n = 2, which returns [2]; and small limits such as 3 or 4. The result stays in ascending order because the final list scans indices from 2 to n.

The key idea is to mark composite numbers instead of testing every number independently. The algorithm creates a boolean array where index i represents the number i. It initially treats every number as a possible prime, except 0 and 1. For each confirmed prime p, it marks multiples from p squared as False. Starting at p squared avoids repeated work because smaller multiples already have smaller prime factors. The invariant is that every value marked False has a discovered smaller prime factor. After processing base values up to the square root of n, the indices still marked True are exactly the prime numbers.

The time complexity is O(n log log n). The sieve reaches this time by marking multiples of each prime instead of checking every number with repeated division. The auxiliary space complexity is O(n). Auxiliary space means the extra memory used by the algorithm. The is_prime array contains n + 1 boolean entries, so its size grows in direct proportion to n.

A common mistake is treating 0 or 1 as prime. Another mistake is using range(p * p, n, p), which fails to mark n when n is a multiple of p. Some candidates start at 2 * p instead of p * p. That can still produce the correct result, but it repeats work already done by smaller primes. Another mistake is using the wrong stopping condition instead of p * p <= n. It is also incorrect to claim O(n) time or O(1) auxiliary space for this implementation.

Explain why marking starts at p squared. Say that every smaller multiple of p already has a smaller prime factor and was handled earlier. This shows that you understand both the correctness and the efficiency of the sieve.