31 Amazon Python Developer Interview Questions & Answers

I would use the existing dictionary as the hash based lookup structure and retrieve each requested field by its key. I assume that specified values means the values associated with specified keys. Each lookup is usually constant time on average. I would also use a unique marker for missing keys so a stored None value is not mistaken for a missing field.

Use the existing dictionary and look up each requested field by its key. I assume that specified values means the values associated with specified keys. A Python dictionary uses a hash table. Python calculates the key hash, finds a likely storage position, and then checks key equality before returning the value. Hash collisions are handled internally.

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 function first confirms that the log is a dictionary. It then loops through the requested keys and calls get with a private marker object. This marker distinguishes a missing key from a key whose stored value is None. Found entries go into a result dictionary. Missing keys go into a separate list.

In the example, level and request_id are found, while user_id is missing. The original dictionary is not copied or changed. The function allocates only the returned dictionary, the missing list, and one marker object.

This approach is suitable when the log has already been parsed into a dictionary. Raw JSON text must be parsed first. Requested keys must be hashable, so strings are a safe choice. Each lookup is usually O(1) on average, but worst case lookup can be O(n).

The code receives one log dictionary and a list of requested keys. It validates that the log is a dictionary and that every requested key is hashable. It uses dictionary get with one private marker object. This correctly separates a missing key from a key whose value is None. For the example, the function returns level and request_id in the result dictionary and places user_id in the missing list. It does not copy or modify the input dictionary. If k keys are requested, average time is O(k) because each dictionary lookup is usually O(1). Worst case time is O(k multiplied by n), where n is the dictionary size, although this is uncommon. Extra memory is O(k) for the returned collections.

Common mistakes include scanning every dictionary item for each requested key, using direct indexing without handling KeyError, and using None as the missing marker even though None may be a valid stored value. Other mistakes include treating raw JSON text as an already parsed dictionary, passing an unhashable key such as a list, claiming lookup is always constant time, and saying the input dictionary is copied when the code only creates result collections.

State the assumption first. Then explain that the dictionary already provides hash based lookup, show safe missing key handling, mention that keys must be hashable, and give both average and worst case costs.

I would stream the log file one record at a time instead of loading the whole file into memory. Python file objects are iterators, so a for loop reads lines as needed. I would parse each JSON record, update small counters, track malformed records, and use gzip.open for compressed files. This keeps memory tied to the largest record and the number of distinct summary keys rather than the total file size.

I would process the file as a stream and keep only the current record plus compact summary data in memory. I assume the input uses JSON Lines, which means one JSON object per line. Python file objects are iterators, so a for loop requests each line as needed instead of creating a list of all lines. A context manager closes the file even if parsing fails. gzip.open provides the same text iteration model for compressed files.

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 each line, I would call json.loads, confirm that the result is a dictionary, extract eventName and responseCode, and update Counter objects. Blank lines are ignored. Invalid JSON and unexpected record types are counted instead of stopping the job.

Processing takes O(B) time, where B is the total input size, because JSON parsing examines the text. Memory is O(U + L), where U is the number of distinct counter keys and L is the largest line. A huge line or many unique values can still use substantial memory. For one large JSON array, I would use a streaming JSON parser. In production, I would add metrics, structured error logs, format specific tests, retry handling, and record size controls.

The program accepts a local plain text or gzip file containing one JSON object per line. open_log_file selects Path.open or gzip.open and returns a text stream. analyze_log iterates directly over that stream, so it does not create a list containing the complete file. It ignores blank lines, counts invalid JSON and non dictionary records, and normalizes missing or null eventName and responseCode values to UNKNOWN. It updates two Counter objects and returns the valid record count, malformed record count, and ten most common values from each counter. Processing time is O(B + U log 10), where B is the total text size and U is the total number of distinct counter keys. Memory is O(U + L), where L is the largest line. The counters are exact, so many unique values can still require substantial memory.

Common mistakes include calling read, readlines, list on the file iterator, or json.load on a very large document. These choices can place most or all input data in memory. Other mistakes include assuming every AWS service uses JSON Lines, claiming that streaming gives constant memory, ignoring the largest line size, forgetting that exact counters grow with the number of unique values, stopping on the first malformed record, and reporting O(N) time without accounting for the amount of text parsed.

Start with the practical decision to stream records instead of loading the file. State the JSON Lines assumption, explain direct file iteration and gzip support, and give the precise O(B) time and O(U + L) memory costs. Then mention malformed records, high variety fields, and the separate parser needed for one large JSON array.

I would use a sliding window with two pointers and a hash map. The left and right pointers define the current substring. The map stores each character and its most recent index. I move right through the string. If the current character already appears inside the window, I move left just past its previous index. After each step, I update the maximum length. This gives O(n) expected time because Python dictionary operations are O(1) on average, with O(min(n, k)) auxiliary space.

The problem asks for the length of the longest contiguous substring that has no repeated character. A substring uses neighboring characters from the original string. A sliding window fits this problem because it keeps one valid substring while moving from left to right. A hash map lets us find the most recent position of a repeated character quickly.

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 string.

The output is one integer. It is the maximum length of any contiguous substring containing only unique characters.

For the example input "abcabcbb", the expected output is 3. Valid longest substrings include "abc", "bca", and "cab". Each has length 3.

I use two pointers named left and right.

They define the current window s[left:right+1].

The dictionary last_seen stores character to most recent index. For example, last_seen["a"] = 3 means that the most recent "a" was found at index 3.

The invariant is that after duplicate handling at each iteration, the current window contains no repeated characters.

Set left = 0.

Set max_len = 0.

Start with an empty dictionary: last_seen = {}.

The right pointer then moves from index 0 through index 7.

The input is "abcabcbb".

At right = 0, the character is "a". It is not repeated inside the window. Store a:0. The window is "a". Its length is 1, so max_len becomes 1.

At right = 1, the character is "b". It is not repeated inside the window. Store b:1. The window is "ab". Its length is 2, so max_len becomes 2.

At right = 2, the character is "c". It is not repeated inside the window. Store c:2. The window is "abc". Its length is 3, so max_len becomes 3.

At right = 3, the character is "a". Its previous index is 0, which is inside the current window because 0 >= left. Move left from 0 to 1. Update a to index 3. The new window is "bca". Its length is 3, so max_len remains 3. The map is now {a:3, b:1, c:2}.

At right = 4, the character is "b". Its previous index is 1, which is inside the current window. Move left from 1 to 2. Update b to index 4. The new window is "cab". Its length is 3, so max_len remains 3. The map is now {a:3, b:4, c:2}.

At right = 5, the character is "c". Its previous index is 2, which is inside the current window. Move left from 2 to 3. Update c to index 5. The new window is "abc". Its length is 3, so max_len remains 3. The map is now {a:3, b:4, c:5}.

At right = 6, the character is "b". Its previous index is 4, which is inside the current window. Move left from 3 to 5. Update b to index 6. The new window is "cb". Its length is 2, so max_len remains 3. The map is now {a:3, b:6, c:5}.

At right = 7, the character is "b". Its previous index is 6, which is inside the current window. Move left from 5 to 7. Update b to index 7. The new window is "b". Its length is 1, so max_len remains 3. The map is now {a:3, b:7, c:5}.

All eight characters are processed. The function returns 3.

After duplicate handling, s[left:right+1] contains no repeated characters.

When the current character already appears inside the window, left moves to one position after its previous occurrence. This removes that duplicate. The left pointer never moves backward.

For every right position, the code measures the valid window ending at that position. max_len stores the largest valid window length seen so far. Therefore, the final value is the length of the longest substring without repeating characters.

The function receives a string and returns an integer.

The dictionary starts empty. left and max_len both start at 0.

The enumerate loop gives the current index as right and the current character as c.

The condition checks two things. The character must already be in last_seen, and its saved index must be greater than or equal to left. This means the earlier occurrence is still inside the current window.

When the condition is true, left becomes last_seen[c] + 1.

The code then stores the current index in last_seen[c]. It calculates the current window length as right - left + 1 and updates max_len.

After the loop finishes, it returns max_len.

The expected time complexity is O(n). The right pointer processes each character once. Python dictionary lookup and insertion take O(1) time on average.

The auxiliary space complexity is O(min(n, k)), where n is the string length and k represents the number of distinct characters that can be stored for the input.

For an empty string, the result is 0. For "a", the result is 1. For "aaaa", the result is 1. For "abcdef", the result is 6.

The key insight is to keep one valid sliding window instead of generating every possible substring. The right pointer expands the window. When a repeated character appears inside the current window, the dictionary gives its most recent index, so left can jump directly past it. The central invariant is that after duplicate handling, s[left:right+1] contains no repeated characters. Updating max_len after the window is valid records the best length without checking all O(n²) substrings.

The expected time complexity is O(n). The right pointer visits each character once, and the left pointer only moves forward. Python dictionary lookup and insertion take O(1) time on average, not guaranteed O(1) time in every possible case. The auxiliary space complexity is O(min(n, k)), where n is the string length and k is the number of distinct characters represented in the input. The dictionary stores at most one latest index for each distinct character.

A common mistake is treating a substring as a subsequence. The characters must be contiguous. Another mistake is moving left whenever a character exists in the dictionary, even when its saved index is outside the current window. The check last_seen[c] >= left prevents left from moving backward. Candidates may also move left by only one position instead of jumping to last_seen[c] + 1. Another mistake is updating the answer before duplicate handling. It must be updated after the window is valid. It is also incorrect to describe Python dictionary operations as guaranteed O(1).

State the invariant before writing the loop: after duplicate handling, s[left:right+1] contains no repeated characters. Then explain why left moves to last_seen[c] + 1 and never moves backward.

I would use merge sort and count inversions during each merge. Every recursive call returns a sorted subarray and the number of inversions inside it. When the current right-side value is smaller than the current left-side value, it is also smaller than every remaining left-side value, so I add len(left) - i. This counts each inversion exactly once. The time complexity is O(n log n). The auxiliary space is O(n), with O(log n) recursion depth.

The input is an array, and the output is the number of index pairs (i, j) where i < j and nums[i] > nums[j]. Checking every pair would take O(n²) time. The diagram uses merge sort because its merge step compares two sorted halves. This lets us count several cross-half inversions with one comparison while still producing a sorted result.

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?

An inversion is a pair of indices (i, j) where i comes before j, but nums[i] is greater than nums[j].

For the input [2, 4, 1, 3, 5], the inversion index pairs are (0, 2), (1, 2), and (1, 3). Their value pairs are (2, 1), (4, 1), and (4, 3). Therefore, the returned count is 3.

Each recursive call returns two values:

1. The sorted version of its subarray.
2. The inversion count inside that subarray.

The recursive calls count inversions fully inside the left and right halves. The merge step counts inversions that cross from the left half to the right half.

The central invariant is that both halves are sorted before every merge comparison. If right[j] < left[i], then right[j] is smaller than every value in left[i:]. Therefore, we add len(left) - i inversions at once.

A subarray with zero or one element is already sorted and has 0 inversions.

The algorithm splits [2, 4, 1, 3, 5] into smaller halves until it reaches single-element arrays. It then merges those arrays back together while counting inversions.

First, merge [2] and [4]. Since 2 <= 4, take 2 and then append 4. The merged result is [2, 4], and 0 inversions are added.

Next, merge [3] and [5]. Since 3 <= 5, take 3 and then append 5. The merged result is [3, 5], and 0 inversions are added.

Next, merge [1] and [3, 5]. Since 1 <= 3, take 1 and then append 3 and 5. The merged result is [1, 3, 5], and 0 inversions are added.

Now merge [2, 4] and [1, 3, 5]. Compare 2 and 1. Since 2 > 1, take 1 from the right half. Both remaining left-side values, 2 and 4, are greater than 1. Add len([2, 4]) - 0 = 2 inversions: (2, 1) and (4, 1). The partial merged result is [1], and the running total is 2.

Compare 2 and 3. Since 2 <= 3, take 2 from the left half. The partial result becomes [1, 2], and the total stays 2.

Compare 4 and 3. Since 4 > 3, take 3 from the right half. One left-side value remains, so add len([4]) - 0 = 1 inversion: (4, 3). The partial result becomes [1, 2, 3], and the total becomes 3.

Compare 4 and 5. Since 4 <= 5, take 4 and then append 5. The final sorted array is [1, 2, 3, 4, 5], and the returned inversion count is 3.

The recursive calls count every inversion that is completely inside one half. The merge counts every inversion whose first value is in the left half and whose second value is in the right half.

When right[j] < left[i], the sorted order of the left half proves that right[j] is smaller than every remaining left value. Adding len(left) - i therefore counts exactly those new cross-half inversions.

Every inversion belongs to one recursive half or one merge step. No inversion is skipped or counted twice.

The helper function sort_count returns a tuple containing a sorted list and its inversion count. It first handles the base case. It then splits the list, recursively processes both halves, and merges the returned sorted lists.

The indices i and j point to the current values in the left and right lists. The variable split_count stores inversions found during the current merge. The helper returns merged and left_count + right_count + split_count. The outer function ignores the final sorted array and returns only the total count.

The time complexity is O(n log n). Merge sort has O(log n) recursive levels, and all merges on one level process O(n) values.

The peak auxiliary space is O(n) for merged lists and copied slices. The recursion depth is O(log n).

An empty array or one element has 0 inversions. An already sorted array has 0 inversions. A reverse-sorted array has n(n - 1) / 2 inversions. Equal values are not inversions because the condition is strictly greater than. Negative values require no algorithm change.

The key insight is to count inversions while merging two sorted halves. A direct solution compares every pair and takes O(n²) time. Merge sort is more suitable because sorted halves let the algorithm count several inversions at once. The invariant is that both halves are sorted before each merge comparison. When right[j] < left[i], the right-side value is smaller than every remaining value in left[i:], so len(left) - i new inversions are added. Recursive calls count inversions within each half, and the merge counts inversions crossing between the halves.

The time complexity is O(n log n). The recursion creates O(log n) levels. Across each level, the merge work processes O(n) values. The peak auxiliary space is O(n) because the code creates sorted slices and merged lists whose total live size grows linearly with the input. The recursion also uses O(log n) stack depth. The output is only one integer, so output space is O(1).

Candidates may count only adjacent out-of-order values, but an inversion can use any two indices. Another mistake is adding only 1 when a right-side value is smaller. The correct addition is len(left_sorted) - i. Using < instead of <= in the left branch incorrectly treats equal values as inversions. Other common mistakes are forgetting the recursion base case, omitting left_count or right_count from the final total, confusing value pairs with index pairs, or claiming O(1) auxiliary space even though the implementation creates slices and merged lists.

Before coding, state the merge invariant clearly: if right[j] < left[i], then right[j] is smaller than every remaining value in the sorted left half, so add len(left) - i.

I would use a dictionary that maps each original node object to its clone. In the first pass, I create one clone for every node. In the second pass, I connect each clone’s next and prev pointers. In the third pass, I connect each random pointer through the same dictionary. Because node objects are keys, duplicate values stay separate. This takes O(n) expected time and O(n) auxiliary space.

The input is the head of a doubly linked list. Every node has val, next, prev, and random fields. We must return the head of a deep copy. The copied nodes must keep the same links, but they must be different objects. A dictionary fits this problem because it lets us translate every original node reference into its matching clone.

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 output must have the same node values and the same next, prev, and random relationships as the input.

It must be a deep copy. This means no cloned pointer may refer to an original node.

Duplicate values are allowed. In the example, nodes A and C both store 7, but they are different node objects.

I use a dictionary named original_to_clone.

Each key is an original node object. Each value is the new clone of that exact object.

The main invariant is: every processed original node reference maps to exactly one distinct clone. Equal values do not share a dictionary entry.

The dictionary starts empty. Traversal starts at the original head and follows next pointers.

The example is:

A(7) ⇄ B(13) ⇄ C(7) ⇄ D(10)

The first pass creates the clones in this order:

Step 1: create A′ with value 7. The map becomes {A → A′}. Step 2: create B′ with value 13. The map becomes {A → A′, B → B′}. Step 3: create C′ with value 7. The map becomes {A → A′, B → B′, C → C′}. Step 4: create D′ with value 10. The map becomes {A → A′, B → B′, C → C′, D → D′}.

A and C have the same value, but they remain separate keys and receive separate clones.

I traverse the original list again. For each original node, I find its clone and translate its next and prev targets through the dictionary.

The executed assignments are:

A′.next = B′ and A′.prev = None. B′.next = C′ and B′.prev = A′. C′.next = D′ and C′.prev = B′. D′.next = None and D′.prev = C′.

After this pass, the cloned doubly linked chain is complete.

I traverse the original list a third time. For each node, I translate its random target through the same dictionary.

The example uses these relationships:

A.random = B, so A′.random = B′. B.random = A, so B′.random = A′. C.random = C, so C′.random = C′. D.random = B, so D′.random = B′.

A self-reference, such as C.random = C, works because C already maps to C′.

The original head is A, so the returned result is original_to_clone[A], which is A′.

Every original node has exactly one clone. Every next, prev, and random target is replaced by the clone of that target. Therefore, the cloned structure has the same relationships as the original while sharing no nodes with it.

The code makes three passes over n nodes. Python dictionary lookup and insertion are O(1) on average, so the total expected time is O(n).

The dictionary stores one entry per original node, so the auxiliary space is O(n).

An empty list returns None. A single node works when random is None or points to itself. Random may point to any node in the list. Duplicate values are safe because dictionary keys are node references, not values.

The key insight is to create all cloned nodes before connecting any cloned pointers. A dictionary stores original node reference → cloned node reference. Once this map exists, each original next, prev, or random target can be translated into the matching cloned target. The invariant is that each original node object maps to exactly one distinct clone. This is why two nodes with the same value, such as A(7) and C(7), are copied correctly and remain separate.

Let n be the number of nodes. The algorithm walks through the list three times. Each walk handles n nodes. Python dictionary lookup and insertion are O(1) on average, so the total expected time is O(n). The dictionary stores one mapping for each original node, so the auxiliary space is O(n). The returned cloned list also contains n new nodes, but those nodes are the required output rather than temporary working memory.

1. Using node values as dictionary keys. This merges different nodes when duplicate values exist.
2. Copying next but forgetting prev or random.
3. Assigning a cloned pointer directly to an original node instead of translating the target through the dictionary.
4. Trying to connect pointers before all clones exist.
5. Claiming guaranteed O(n) time instead of O(n) expected time for Python dictionary operations.

Say the mapping direction before coding: original node reference → cloned node reference. Then name the three passes: create clones, connect next and prev, and connect random. This makes the duplicate-value case easy to justify.

I would scan the array from right to left and use a monotonic stack. The stack stores useful values from the right side in strictly decreasing order from bottom to top. For each current value, I pop every value that is less than or equal to it. If the stack is not empty, its top is the next greater value. Then I push the current value. Each value is pushed once and popped at most once, so the time is O(n) and the auxiliary space is O(n).

The problem asks us to find the first strictly greater value to the right of every array element. If no greater value exists, the answer for that position is -1. For nums = [2, 1, 2, 4, 3], the result is [4, 2, 4, -1, -1]. A monotonic stack works well because it removes values that can no longer be useful and avoids repeatedly scanning the same elements.

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 values.

For each index i, we need the first value to the right of nums[i] that is strictly greater than nums[i].

The output contains values, not indices.

If no greater value exists to the right, the output contains -1 for that position.

For the example nums = [2, 1, 2, 4, 3], the correct returned array is [4, 2, 4, -1, -1].

We use a monotonic stack and process the array from right to left.

The stack stores candidate next-greater values from positions that have already been processed. Those positions are strictly to the right of the current index.

The stack is strictly decreasing from bottom to top. The rightmost entry is the top.

Before processing index i, the stack contains only useful candidates from the right side. We pop values that are less than or equal to the current value because they cannot be a strictly greater answer.

Let n = 5 for the example.

Create result = [-1, -1, -1, -1, -1]. The value -1 is the default when no greater value exists.

Create an empty stack: stack = [].

Begin at index 4 and move left until index 0.

At index 4, current = 3. The stack before processing is []. There is nothing to pop. The stack is empty, so result[4] remains -1. Push 3. The stack becomes [3].

At index 3, current = 4. The stack before processing is [3]. Pop 3 because 3 <= 4. The stack becomes empty, so result[3] remains -1. Push 4. The stack becomes [4].

At index 2, current = 2. The stack before processing is [4]. The top value 4 is greater than 2, so result[2] = 4. Push 2. The stack becomes [4, 2].

At index 1, current = 1. The stack before processing is [4, 2]. The top value 2 is greater than 1, so result[1] = 2. Push 1. The stack becomes [4, 2, 1].

At index 0, current = 2. The stack before processing is [4, 2, 1]. Pop 1 because 1 <= 2. Pop the equal value 2 because 2 is not strictly greater than 2. The remaining top is 4, so result[0] = 4. Push 2. The stack becomes [4, 2].

The final returned array is [4, 2, 4, -1, -1]. The relationships are 2 → 4, 1 → 2, 2 → 4, 4 → -1, and 3 → -1.

Before index i is processed, every value in the stack comes from a position strictly to its right.

Popping values less than or equal to the current value is safe. Those values cannot be the answer for the current position because they are not strictly greater. They are also blocked by the current value for positions farther left.

After the pops, the remaining top, when present, is the nearest surviving value that is greater than the current value. Therefore, it is the correct next greater value.

The function creates a result list filled with -1 and an empty list used as a stack.

The for loop visits indices from n - 1 down to 0.

The while loop pops values while stack[-1] <= current.

If the stack is not empty after the pops, the code writes stack[-1] into result[i].

The code then pushes the current value so it can help elements farther left.

After all positions are processed, the function returns the result list.

The time complexity is O(n). Each value is pushed once and popped at most once.

The auxiliary space is O(n) for the monotonic stack. The returned result list uses O(n) output space.

An empty list returns []. One element returns [-1]. A strictly decreasing list such as [5, 4, 3] returns [-1, -1, -1]. A strictly increasing list such as [1, 2, 3] returns [2, 3, -1]. Equal values are popped because equal values do not count as greater.

The key insight is to process the input from right to left. This makes every value already stored in the stack come from the current element's right side. The stack contains unresolved candidate values and stays strictly decreasing from bottom to top. Before processing index i, the stack contains only useful candidates from positions strictly to the right of i. We remove every value less than or equal to the current value because it cannot be the current element's strictly greater answer and cannot help an earlier element before the current value would help it. If a value remains, the top is the nearest surviving greater value. A direct nested-loop solution may inspect many right-side values for each position and take O(n^2) time. The monotonic stack avoids that repeated work.

Let n be the number of input values. The time complexity is O(n). Each value is pushed onto the stack exactly once. A value can also be popped at most once. This means the total number of stack operations is proportional to n. The auxiliary space is O(n) because the stack may contain up to n values. The returned result array also uses O(n) output space.

A common mistake is scanning from left to right while using this value-stack lookup rule. The candidates must come from elements already processed on the right. Another mistake is popping only values smaller than the current value. The condition must also pop equal values because the answer must be strictly greater. Candidates may also read stack[-1] without first checking whether the stack is empty. Another mistake is returning indices even though this version of the problem asks for values. Finally, the auxiliary space is not O(1) because the stack can grow with the input.

State the invariant before writing code: before processing index i, the stack contains useful next-greater candidates from positions strictly to the right of i.

I treat each recipe’s ingredient list as a set. For each recipe, I remove repeated ingredients and sort the remaining names. I convert that sorted list to a tuple and use it as a dictionary key. The dictionary value is a list of recipe IDs. Recipes with the same ingredient set produce the same key, so they enter the same group. For n recipes with at most k ingredients each, this takes O(n × k log k) expected time and O(n × k) auxiliary space.

The problem asks us to group recipe IDs when their ingredient lists represent the same set. The order of ingredients does not matter. Repeated ingredients also do not matter. We solve this by creating one normalized key for each recipe. We remove duplicates, sort the remaining ingredient names, and use the sorted tuple as a dictionary 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 input contains recipe IDs and their ingredient lists.

The example input is:

R1: [egg, milk, flour, sugar] R2: [milk, flour, egg, sugar] R3: [flour, sugar, egg, milk] R4: [egg, butter, flour] R5: [butter, egg, flour, butter] R6: [chicken, salt, pepper]

The output groups recipe IDs that have the same normalized ingredient set.

The result is:

[egg, flour, milk, sugar] → [R1, R2, R3] [butter, egg, flour] → [R4, R5] [chicken, pepper, salt] → [R6]

For each recipe, we create a normalized key. Normalized means that equivalent ingredient lists get one standard form.

First, we remove duplicate ingredients. Next, we sort the remaining names. Finally, we convert the sorted list to a tuple because a tuple can be used as a dictionary key.

The dictionary maps each normalized key to a list of recipe IDs.

The invariant is: after each recipe is processed, every processed recipe ID is stored under the key for its exact ingredient set.

We start with an empty dictionary.

R1 has [egg, milk, flour, sugar]. Removing duplicates changes nothing. Sorting gives [egg, flour, milk, sugar]. The key is new, so the dictionary stores [R1].

R2 has [milk, flour, egg, sugar]. Its normalized key is also [egg, flour, milk, sugar]. The key already exists, so R2 is appended. The group becomes [R1, R2].

R3 has [flour, sugar, egg, milk]. Its normalized key is again [egg, flour, milk, sugar]. R3 is appended. The group becomes [R1, R2, R3].

R4 has [egg, butter, flour]. Its normalized key is [butter, egg, flour]. The key is new, so the dictionary stores [R4].

R5 has [butter, egg, flour, butter]. The repeated butter is removed. Sorting gives [butter, egg, flour]. The key already exists, so R5 is appended. The group becomes [R4, R5].

R6 has [chicken, salt, pepper]. Its normalized key is [chicken, pepper, salt]. The key is new, so the dictionary stores [R6].

After R6, all recipes have been processed, so the algorithm returns the dictionary.

Two recipes belong in the same group exactly when they have the same ingredient set.

Removing duplicates makes repeated ingredients irrelevant. Sorting places equal sets in the same order. Therefore, recipes with equal ingredient sets create equal tuple keys.

Recipes with different ingredient sets create different tuple keys. The dictionary therefore places exactly the correct recipe IDs in each group.

The function creates an empty dictionary named groups.

It loops through the recipes in input order. For each ingredient list, set removes duplicates. sorted creates a consistent order. tuple creates a hashable key.

setdefault creates an empty list when the key is first seen. The current recipe ID is then appended to that list.

When the loop finishes, the function returns the grouped dictionary.

Let n be the number of recipes. Let k be the maximum number of ingredients in one recipe.

Removing duplicates takes O(k) expected time. Sorting takes O(k log k) time. This is done for every recipe, so the total time is O(n × k log k) expected time.

The dictionary, normalized keys, and recipe ID lists use O(n × k) auxiliary space in the worst case.

An empty ingredient list creates an empty tuple key. Recipes with empty ingredient lists join the same group. A repeated ingredient appears only once in the normalized key. A single recipe forms a group containing only its own ID.

The key insight is to convert every ingredient list into one canonical representation. Canonical means one standard form. We remove duplicate names, sort the remaining names, and convert them to a tuple. That tuple becomes the dictionary key, and the dictionary value is the list of recipe IDs with that key. The invariant is that every processed recipe is stored under the normalized key for its exact ingredient set. Equal sets create equal keys, while different sets create different keys.

Let n be the number of recipes and k be the maximum number of ingredients in one recipe. Creating a set takes O(k) expected time because Python set insertion is O(1) on average. Sorting the unique ingredients takes O(k log k) time. We repeat this for all n recipes, so the total time is O(n × k log k) expected time. Dictionary lookup and insertion are also O(1) on average. The normalized keys, dictionary entries, and recipe ID lists use O(n × k) auxiliary space in the worst case.

One mistake is sorting the original list without removing duplicates. Then R5 would not match R4 because R5 contains butter twice. Another mistake is using a mutable set directly as a dictionary key. A normal set cannot be a key, so the result must be converted to a tuple or another hashable value. Candidates may also compare the original ingredient order, which incorrectly separates R1, R2, and R3. Another mistake is overwriting an existing group instead of appending the recipe ID. Finally, do not forget the O(k log k) sorting cost when explaining complexity.

Explain the normalization rule before writing code: remove duplicates, sort the ingredient names, convert them to a tuple, and use that tuple as the dictionary key. This makes the code and correctness argument easy to follow.

I would use Kahn’s algorithm for topological sorting. I build a directed adjacency list from each prerequisite to the courses that depend on it. I also keep an indegree array that counts each course’s remaining prerequisites. I put every indegree-0 course into a deque, process ready courses from the front, and unlock dependent courses. If all courses are processed, I return the order. Otherwise, I return an empty list because a cycle exists. Time is O(V + E), and auxiliary space is O(V + E).

The problem asks us to return an order in which all courses can be completed. Each prerequisite pair creates a directed dependency between two courses. A course can be processed only after all its prerequisites are complete. Kahn’s algorithm fits this problem because it repeatedly processes courses with no remaining prerequisites. If some courses never become ready, the dependency graph contains a cycle.

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?

numCourses is the total number of courses. The courses are numbered from 0 to numCourses - 1.

Each pair [course, prereq] means that prereq must be completed before course.

We must return one valid order containing all courses. More than one order may be valid. If no valid order exists, we return an empty list.

The diagram uses this example:

numCourses = 4

prerequisites = [[1, 0], [2, 0], [3, 1], [3, 2]]

One valid returned order is [0, 1, 2, 3].

I build a directed adjacency list from each prerequisite to its dependent courses.

graph[prereq] stores every course that directly depends on prereq.

For the example, the adjacency list is:

0 -> [1, 2]

1 -> [3]

2 -> [3]

3 -> []

I also build an indegree array. indegree[course] is the number of prerequisites that the course still needs.

The initial indegree array is [0, 1, 1, 2].

Course 0 has no prerequisites. Courses 1 and 2 each need course 0. Course 3 needs both courses 1 and 2.

I put every course whose indegree is 0 into a deque.

For this example, the initial deque is [0].

The initial order is [].

The central invariant is that every course in the deque is ready. Its remaining prerequisite count is 0, so processing it is safe.

Step 1 starts with deque [0] and order [].

I pop course 0 and append it to the order. The order becomes [0].

Course 0 has dependent courses 1 and 2. I reduce indegree[1] from 1 to 0 and indegree[2] from 1 to 0.

Both courses are now ready, so I append them to the deque in adjacency-list order.

The deque becomes [1, 2]. The indegree array becomes [0, 0, 0, 2].

Step 2 starts with deque [1, 2] and order [0].

I pop course 1 and append it to the order. The order becomes [0, 1].

Course 1 points to course 3. I reduce indegree[3] from 2 to 1.

Course 3 is still blocked, so it is not added to the deque.

The deque becomes [2]. The indegree array becomes [0, 0, 0, 1].

Step 3 starts with deque [2] and order [0, 1].

I pop course 2 and append it to the order. The order becomes [0, 1, 2].

Course 2 also points to course 3. I reduce indegree[3] from 1 to 0.

Course 3 is now ready, so I append it to the deque.

The deque becomes [3]. The indegree array becomes [0, 0, 0, 0].

Step 4 starts with deque [3] and order [0, 1, 2].

I pop course 3 and append it to the order. The order becomes [0, 1, 2, 3].

Course 3 has no dependent courses, so no indegree values change.

The deque becomes empty. All four courses were processed.

A course enters the deque only when its indegree becomes 0. This means all its prerequisites have already been processed.

Therefore, every course is added to the returned order only after all courses it depends on.

In [0, 1, 2, 3], course 0 appears before courses 1 and 2. Courses 1 and 2 both appear before course 3.

This proves that [0, 1, 2, 3] is a valid order. It is one valid order, not necessarily the only valid order.

If fewer than numCourses courses are processed, some courses never reach indegree 0. A directed cycle is blocking them, so the correct result is [].

The code creates an empty adjacency list for each course and an indegree array filled with zeros.

For each [course, prereq] pair, it adds course to graph[prereq] and increases indegree[course] by one.

It then creates a deque containing every course with indegree 0.

While the deque is not empty, it removes the leftmost ready course, appends it to the order, and reduces the indegree of every dependent course.

When a dependent course reaches indegree 0, the code appends it to the deque.

Finally, the function returns the order only when its length equals numCourses. Otherwise, it returns [].

Let V be the number of courses and E be the number of prerequisite pairs.

Building the graph and indegree array takes O(V + E) time. Each course is enqueued and dequeued once, and each directed edge is processed once. The total time is O(V + E).

The adjacency list uses O(V + E) space. The indegree array, deque, and result list use O(V) space. Therefore, the total auxiliary space is O(V + E).

Relevant edge cases are no prerequisites, multiple valid orders, a directed cycle, and disconnected groups of courses.

The key insight is to repeatedly process courses that have no remaining prerequisites. The adjacency list stores prerequisite-to-dependent edges. The indegree array stores how many prerequisites each course still needs. The invariant is that every course in the deque has indegree 0, so all its prerequisites have already been completed. Processing a ready course decreases the indegrees of its dependents. Any dependent that reaches indegree 0 becomes ready. If all V courses are processed, the result is a valid topological order. If fewer than V courses are processed, a cycle prevents a valid order.

Let V be the number of courses and E be the number of prerequisite pairs. Creating the adjacency list and indegree array takes O(V + E) time. Each course enters and leaves the deque at most once, so course processing takes O(V) time. Each prerequisite edge is examined once, so edge processing takes O(E) time. The total time is O(V + E). The adjacency list, indegree array, deque, and order list use O(V + E) auxiliary space in total.

A common mistake is reversing the graph edges. The correct edge is prereq -> course. Another mistake is increasing the indegree of the prerequisite instead of the dependent course. Some candidates enqueue a course before its indegree reaches exactly 0, or enqueue it more than once. Another mistake is returning a partial order without checking len(order) == numCourses. Using list.pop(0) instead of deque.popleft() makes front removal slower. It is also incorrect to claim that [0, 1, 2, 3] is the only valid order.

Say the invariant before writing code: every course in the deque has indegree 0, so all its prerequisites are complete. Then make the graph direction, indegree updates, queue operations, and final length check match that invariant.

I would subtract the target average k from every number. Then a non-empty subset has average k exactly when its transformed values sum to

1. I use dynamic programming with a Counter that maps each reachable sum to the number of subsets producing it. For every value, I copy the current Counter to count the skip choice, then add shifted counts for the take choice. Finally, I return dp[0] -
2. The time is O(n × M), and the auxiliary space is O(M).

The problem asks us to count non-empty subsets whose average equals k. The main idea is to replace the average condition with a zero-sum condition. We subtract k from every number. A subset has average k exactly when its transformed values add up to 0. We then use dynamic programming to count how many subsets produce each reachable transformed sum.

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?

For any non-empty subset S:

average(S) = k

This is equivalent to:

sum(num - k for num in S) = 0

For the example:

nums = [1, 2, 3] k = 2

The transformed array is:

[-1, 0, 1]

The task is now to count non-empty subsets of this transformed array whose sum is 0.

I use a Counter named dp.

dp[s] stores the number of subsets from the processed values whose transformed sum is s.

The invariant is: after processing the first i transformed values, dp[s] equals the number of subsets of those i values whose sum is s.

I initialize:

dp = {0: 1}

This represents the empty subset. It produces sum 0 in one way.

For every transformed value, each existing subset has two choices.

It can skip the value. I count this by copying dp into next_dp.

It can take the value. For every current_sum and count in dp, I add count to next_dp[current_sum + value].

After all transitions for the current value are complete, I assign dp = next_dp.

Start with:

transformed = [-1, 0, 1] dp = {0: 1}

Process -1:

The skip choice keeps sum 0 with count 1. The take choice creates sum -1 because 0 + (-1) = -1. The new state is:

{-1: 1, 0: 1}

Process 0:

Copying the old state counts the skip choices. Taking 0 from sum -1 adds one more way to sum -1. Taking 0 from sum 0 adds one more way to sum 0. The new state is:

{-1: 2, 0: 2}

Process 1:

Copying the old state counts the skip choices. Taking 1 from sum -1 adds two ways to sum

1. Taking 1 from sum 0 adds two ways to sum
2. The final state is:

{-1: 2, 0: 4, 1: 2}

The value dp[0] is 4. It counts the empty subset and three non-empty subsets. The valid original subsets are [2], [1, 3], and [1, 2, 3]. Their averages are all 2.

Therefore, the returned result is:

4 - 1 = 3

Before each value is processed, dp[s] contains the correct number of subsets of the processed prefix that produce sum s.

Copying dp into next_dp counts every subset that skips the current value.

Adding each count to next_dp[s + value] counts every subset that takes the current value.

Every subset makes exactly one skip-or-take choice for each value. Therefore, every subset is counted exactly once. dp[0] counts all zero-sum subsets, including the empty subset, so the answer is dp[0] - 1.

The code first creates the transformed array. It initializes Counter({0: 1}) for the empty subset. For each transformed value, it copies the current Counter to next_dp. It then adds every take transition to next_dp. After processing that value, it assigns dp = next_dp. At the end, it returns dp[0] - 1.

Let n be the number of input values. Let M be the maximum number of reachable transformed sums stored at one step.

The time complexity is O(n × M). The auxiliary space complexity is O(M). Python Counter lookup and insertion are O(1) on average.

This is a pseudo-polynomial counting solution because M can depend on the numeric value range.

For one element, the result is 1 when that element equals k. Otherwise, it is 0. If every element equals k, every non-empty subset is valid, so the result is 2^n - 1. If no non-empty subset has average k, dp[0] remains 1 from the empty subset, so the returned result is 0. Negative values work normally after the transformation.

The key insight is to convert the average condition into a zero-sum condition. For a non-empty subset S, average(S) = k exactly when sum(num - k for num in S) = 0. After transforming the input, a Counter-based dynamic program counts how many subsets create each reachable sum. The invariant is that dp[s] equals the number of subsets of the processed prefix whose transformed sum is s. Copying dp counts subsets that skip the current value. Adding counts to shifted sums counts subsets that take it. Every subset is therefore counted exactly once.

Let n be the number of input values. Let M be the maximum number of different transformed sums stored at one step. For each of the n values, the algorithm visits up to M stored sums, so the time complexity is O(n × M). The current Counter and the copied next Counter each contain at most O(M) states, so the auxiliary space complexity is O(M). Counter lookup and insertion are O(1) on average in Python. The solution is pseudo-polynomial because M can grow with the numeric value range.

A common mistake is forgetting to subtract one for the empty subset. Another mistake is updating dp directly while iterating over it, which can use the current value more than once. Candidates may also forget that copying dp represents the skip choice. Some people count contiguous subarrays instead of arbitrary subsets. Another mistake is using the wrong transformed condition. The correct condition is sum(num - k) = 0. Finally, do not claim O(n) time because the number of reachable sums can grow.

Start by showing the equation average(S) = k if and only if sum(num - k for num in S) = 0. Then define exactly what dp[s] means before writing the recurrence.