188 Data Scientist Interview Questions & Answers

Group the Transactions table by YYYY-MM and country. Count every row, conditionally count approved rows, sum every amount, and conditionally sum approved amounts. Keep declined rows in the input so groups with zero approvals still appear, and let NULL countries form one group.

Each row in the table is one transaction. The task is to turn those individual rows into a monthly summary for each country. For every month and country, we need the total number of transactions, the number that were approved, the total amount, and the approved amount. Missing countries must still appear as their own group. A group must also remain when it has transactions but none are approved. The three supplied rows therefore create one US result for December 2018 and one US result for January 2019.

1. Should the amount values be summed exactly as stored, with no currency conversion or unit scaling?

The input grain is one row per transaction, with id as the primary key. The required output grain is one row per calendar month and country.

The analytical client sends the standalone MySQL 8.0 query through a SQL driver or connector. MySQL parses and optimizes the statement, the query executor reads the Transactions rows from the storage engine, computes the grouped aggregates, and returns the result set to the client. Those execution layers do not change the logical meaning of the query.

Use DATE_FORMAT(trans_date, '%Y-%m') to create the requested calendar-month label. Group by that expression together with country. MySQL GROUP BY places NULL country values for the same month into one NULL group, so no special replacement value is needed.

COUNT() gives trans_count because every source row represents one transaction. SUM(state = 'approved') gives approved_count in MySQL 8.0 because the Boolean comparison evaluates to 1 when true and 0 when false. The state column is NOT NULL.

SUM(amount) gives trans_total_amount. For approved_total_amount, use SUM(CASE WHEN state = 'approved' THEN amount ELSE 0 END). The ELSE 0 makes a group with no approved rows return an approved amount of 0.

Do not use WHERE state = 'approved'. That would remove declined rows before aggregation, making trans_count and trans_total_amount incorrect and causing groups with zero approved transactions to disappear.

For the supplied example, December 2018 in the US has trans_count = 2, approved_count = 1, trans_total_amount = 3000, and approved_total_amount = 1000. January 2019 in the US has 1, 1, 2000, and 2000 respectively.

Any row order is allowed. ORDER BY month, country is optional and is included only to make the example output deterministic.

For performance, DATE_FORMAT(trans_date, '%Y-%m') is evaluated as part of grouping. A plain index on (trans_date, country) does not directly provide grouping by that formatted expression. For a large, frequently executed report, inspect the MySQL query plan and measured workload before adding a specialized generated or functional indexing strategy, because indexes add storage and write-maintenance cost.

1. Read all Transactions rows needed for the report without filtering by state.
2. Derive the month with DATE_FORMAT(trans_date, '%Y-%m').
3. Group by the derived month and country.
4. Use COUNT() for trans_count.
5. Use SUM(state = 'approved') for approved_count.
6. Use SUM(amount) for trans_total_amount.
7. Use SUM(CASE WHEN state = 'approved' THEN amount ELSE 0 END) for approved_total_amount.
8. Optionally sort by month and country.

The database must read the relevant transaction rows and place them into month-country groups, so the amount of work generally grows with the number of rows processed. Memory is needed to maintain aggregate groups, and MySQL chooses the physical aggregation strategy. DATE_FORMAT is evaluated for grouping. A plain (trans_date, country) index does not directly provide the required grouping by the formatted month expression. Any extra index also has storage and write-maintenance cost, so performance changes should be based on an observed query plan and workload.

The SQL groups Transactions by the requested YYYY-MM month and country. COUNT() counts every transaction. SUM(state = 'approved') conditionally counts approved rows using MySQL Boolean values. SUM(amount) calculates the total amount. The CASE expression adds amount only for approved rows and contributes 0 otherwise. Because no state filter is applied, declined transactions remain in the overall totals and groups with zero approved transactions remain in the result. NULL countries are grouped together naturally by MySQL.

Filtering with WHERE state = 'approved' is the main mistake because it removes declined transactions and makes the overall count and total wrong; it also removes groups with zero approved rows. Other mistakes include grouping only by month or only by country, grouping by month number without the year, mishandling NULL country values, using a conditional SUM without ELSE 0 for approved_total_amount, or returning the wrong result grain.

Start by stating the result grain: one row per YYYY-MM month and country. Then explain that conditional aggregation is necessary because the query needs overall metrics and approved-only metrics from the same rows. Explicitly call out NULL-country grouping and the zero-approved case.

Aggregate Confirmations by user_id, counting confirmed requests and all requests. LEFT JOIN those counts to Signups so every signup remains. Divide confirmed_cnt by total_cnt with NULLIF for safe division, wrap the rounded result with COALESCE, and return 0.00 when a user has no requests.

We need to give one result for every person who signed up. For each person, look at all confirmation requests and count how many succeeded. Divide the successful count by the total number of requests and show the result with two decimal places. A person who never made a confirmation request must still appear and must receive 0.00. In the example, user 3 succeeded once out of two requests, so the result is 0.50. User 7 had one request and it did not succeed, so the result is 0.00.

1. Is there an expected data-size or performance boundary that should influence whether we discuss additional indexing or query-plan verification?

The required result grain is one row per signup. Signups.user_id is a primary key, so Signups already contains exactly one row per signed-up user. Confirmations has zero or more requests per user and has primary key (user_id, time_stamp).

First, aggregate Confirmations by user_id. In MySQL 8.0, SUM(action = 'confirmed') counts confirmed rows because the comparison evaluates to 1 when true and 0 when false. COUNT(*) counts all confirmation requests. This produces at most one aggregate row per user.

Next, LEFT JOIN that aggregate to Signups. The LEFT JOIN is essential because an INNER JOIN would remove signups that have no confirmation requests. If a signup has no matching aggregate row, confirmed_cnt and total_cnt are NULL.

Calculate c.confirmed_cnt / NULLIF(c.total_cnt, 0). NULLIF prevents division by zero. Then use ROUND(..., 2) and wrap the final expression with COALESCE(..., 0.00). For a signup with no matching confirmations, the division result is NULL and the outer COALESCE returns 0.00.

For the supplied example, user 3 has one confirmed request out of two total requests, so the rate is 0.50. User 7 has zero confirmed requests out of one total request, so the rate is 0.00. Row order is unrestricted, so no ORDER BY is required.

The existing Confirmations primary key (user_id, time_stamp) has user_id as its leftmost column, so it can support access and grouping by user_id. I would not add another index automatically. I would consider table size, data distribution, workload, and the MySQL execution plan before adding an index because extra indexes consume storage and increase write-maintenance cost.

1. Use Signups as the base because every signup must appear exactly once.
2. Group Confirmations by user_id.
3. Compute confirmed_cnt with SUM(action = 'confirmed').
4. Compute total_cnt with COUNT(*).
5. LEFT JOIN the aggregated counts to Signups by user_id.
6. Divide confirmed_cnt by NULLIF(total_cnt, 0).
7. Round the ratio to two decimals.
8. Use outer COALESCE to return 0.00 when no confirmation aggregate exists.
9. Do not add ORDER BY because row order is unrestricted.

Logically, the query processes the signup rows and the confirmation rows needed for aggregation. With S signups and C confirmation rows, the amount of work is roughly proportional to S + C, although MySQL may choose a different physical execution plan. Grouping requires working memory or temporary storage for per-user counts. The existing primary key (user_id, time_stamp) starts with user_id and can support access or grouping by user_id. Additional indexes consume storage and add write cost, so they should be justified by the real workload and execution plan.

The subquery reduces Confirmations to one row per user before the join. SUM(action = 'confirmed') counts confirmed requests in MySQL 8.0, while COUNT(*) counts all requests. LEFT JOIN preserves every signup. NULLIF(total_cnt, 0) protects the denominator, ROUND(..., 2) rounds the rate, and the outer COALESCE converts a missing rate to 0.00. Because the aggregate has at most one row per user and Signups has a primary key on user_id, the final result has exactly one row per signup.

Using INNER JOIN is a major mistake because it removes signups with no confirmation requests. Another mistake is counting all rows after a direct LEFT JOIN with COUNT(*), which can incorrectly count the preserved signup row even when no confirmation exists. Candidates may also divide without safe NULL or zero handling, forget the required 0.00 default, forget to round to two decimals, accidentally change the one-row-per-signup grain, add unnecessary ordering, or recommend another index without considering that the existing (user_id, time_stamp) primary key already begins with user_id.

State the required grain first: one row per signup. Then explain why you aggregate Confirmations before the LEFT JOIN. Walk through the no-request case explicitly: the joined counts are NULL, NULLIF keeps division safe, and outer COALESCE returns 0.00. Finish by checking users 3 and 7 against 0.50 and 0.00.

Use DENSE_RANK() partitioned by departmentId and ordered by salary descending. Tied salaries receive the same rank, so filtering to ranks 1 through 3 keeps the three highest distinct salary levels and every employee tied at those levels. Then join Department to return its name.

The task is to examine each department separately and find its three highest different salary amounts. Every employee earning one of those amounts must be returned. Employees with the same salary belong to the same salary level, so a department can return more than three people. For salaries 100, 90, 90, 80, and 70, the selected levels are 100, 90, and 80. Both employees earning 90 remain in the result. The final output contains the department name, employee name, and salary, with no required row order.

1. Should every employee tied at the third-highest distinct salary be included? Yes, the requirement explicitly says to include all ties.
2. What should happen when a department has fewer than three distinct salary values? Return employees at every salary level that exists.
3. Is a particular final row order required? No. The question says row order is unrestricted.

Use MySQL 8.0's DENSE_RANK() window function. A window function calculates a value across related rows while keeping the original rows instead of collapsing them into groups.

Partition by departmentId, so ranking restarts for every department. Inside each partition, order by salary DESC so the highest salary receives rank 1. DENSE_RANK() assigns the same rank to equal salaries and does not leave gaps between distinct salary levels.

For example, salaries 100, 90, 90, 80, and 70 receive ranks 1, 2, 2, 3, and 4. Filtering to salary_rank <= 3 therefore returns the employee at 100, both employees at 90, and the employee at 80. This exactly matches the requirement to keep the top three distinct salary values while including every tie.

The ranked result stays at employee grain: one row per employee. Join Department using Department.id = Employee.departmentId to obtain the department name. Because every employee has a valid department key, this many-to-one join does not duplicate employee rows. Return d.name AS Department, the employee name as Employee, and salary as Salary.

The schema defines salary and departmentId as NOT NULL, so no special null handling is needed. Employee IDs are primary keys and duplicate employees do not occur, so no final DISTINCT is needed. Row order is unrestricted, so an outer ORDER BY is unnecessary.

For performance, the main cost is partitioning and ordering employees for the window function. An index such as Employee(departmentId, salary DESC, id) can support relevant access and ordering patterns, but indexes consume storage and add write-maintenance cost. MySQL's optimizer chooses the physical execution plan, so use EXPLAIN with representative data when production performance matters.

1. Start with Employee at one row per employee. 2. Partition rows by departmentId. 3. Order salaries descending within each department. 4. Apply DENSE_RANK() so equal salaries receive the same rank and distinct salary levels receive consecutive ranks. 5. Keep rows with salary_rank <= 3. 6. Join Department on Department.id = departmentId. 7. Return Department, Employee, and Salary. 8. Do not add DISTINCT or a final ORDER BY because duplicate employees do not occur and output order is unrestricted.

The main work is ranking employee salaries inside each department. In the general case, ordering N employee rows can require about O(N log N) work, although the actual MySQL execution plan depends on data distribution and available indexes. The window operation may also use working memory or temporary storage. The Department join is efficient because Department.id is a primary key. An index such as Employee(departmentId, salary DESC, id) may help the access and ordering pattern, but it consumes storage and adds maintenance work when Employee rows are inserted or updated. Use EXPLAIN to verify the real plan rather than assuming an index removes every sort.

The common table expression keeps one row per employee and calculates a salary rank within that employee's department. PARTITION BY e.departmentId restarts the ranking for each department, while ORDER BY e.salary DESC ranks higher salaries first. DENSE_RANK() gives employees with equal salaries the same rank and gives the next distinct salary the next consecutive rank. The outer query joins Department to obtain its name and filters to ranks 1, 2, and 3. The schema already excludes null salaries and invalid department keys, duplicate employees do not occur, and the requested output has no required ordering.

Using ROW_NUMBER() is incorrect because employees with equal salaries receive different row numbers, so tied employees can be dropped. Using RANK() with <= 3 is also incorrect for this exact requirement because gaps after ties can prevent the third distinct salary from qualifying; for 100, 90, 90, 80, 70 it produces ranks 1, 2, 2, 4, 5. Using LIMIT 3 returns at most three rows rather than three distinct salary levels. Forgetting PARTITION BY departmentId ranks salaries across the whole company instead of within each department. Forgetting the Department join fails to return the required department name. Adding DISTINCT is unnecessary because duplicate employees do not occur.

Start by saying that the requirement is for three distinct salary levels, not three employee rows. Then explain why DENSE_RANK() matches that rule: tied salaries share one rank and the next distinct salary gets the next consecutive rank. Walk through 100, 90, 90, 80, 70 as ranks 1, 2, 2, 3, 4.

Join Users twice to validate both the client and driver. Filter both to banned = 'No' and keep the required dates. Group by request_at, divide cancelled eligible trips by all eligible trips, round to two decimals, and order by Day ascending.

We need to look at trips made during three specific days and ignore any trip where either person involved is banned. For every remaining day, we count how many trips happened and how many ended because either the client or driver cancelled. We then divide the cancelled count by the total count for that day and round the result to two decimal places. A day appears only when at least one valid trip exists. Finally, the days must be returned from earliest to latest.

1. Should both the client and driver be required to have banned = 'No'? Yes. The question explicitly requires both to be unbanned.
2. Should a date with no eligible trips appear with a zero cancellation rate? No. Only dates with at least one eligible trip should appear.
3. Is request_at already stored as a DATE? Yes. No date-extraction function is needed.

The required result grain is one row per request_at date that has at least one eligible trip. Trips contains one row per trip. Users.users_id is a primary key, so each client or driver identifier can match at most one Users row.

I join Users twice because client_id and driver_id represent two different people. The client alias joins on c.users_id = t.client_id, and the driver alias joins on d.users_id = t.driver_id. I then keep only trips where both c.banned = 'No' and d.banned = 'No'.

The date filter is inclusive: t.request_at BETWEEN '2013-10-01' AND '2013-10-03'. Because request_at is already a DATE, there is no need to wrap it in DATE().

A trip counts as cancelled when status is either cancelled_by_driver or cancelled_by_client. In MySQL, the expression t.status IN (...) evaluates to 1 when true and 0 when false, so summing it counts cancelled eligible trips. COUNT(*) counts every eligible trip in the same daily group. Dividing the cancelled count by COUNT(*) gives the requested cancellation rate, and ROUND(..., 2) rounds it to two decimal places.

There is no divide-by-zero case for an emitted group because a GROUP BY row exists only when at least one eligible trip survived the joins and filters. Finally, ORDER BY t.request_at ASC returns the result from earliest to latest.

The supplied schema represents only the current banned value. It contains no ban-history timestamp, so I would not invent historical ban-state logic.

1. Start from Trips, which has one row per trip.
2. Join Users as the client with c.users_id = t.client_id.
3. Join Users again as the driver with d.users_id = t.driver_id.
4. Keep only rows where c.banned = 'No' and d.banned = 'No'.
5. Keep trips from 2013-10-01 through 2013-10-03 inclusive.
6. Group the eligible trips by t.request_at.
7. Count cancelled trips by summing the condition that status is cancelled_by_driver or cancelled_by_client.
8. Divide cancelled eligible trips by all eligible trips and round to two decimals.
9. Return exactly Day and Cancellation Rate, ordered by day ascending.

The database must find trips in the requested date range, look up the client and driver for each relevant trip, and aggregate the eligible rows by day. The two Users joins are primary-key lookups because users_id is the primary key. An index that starts with request_at may help a large Trips table avoid reading unrelated dates, but the actual benefit should be checked with the query plan. Only a few daily groups are produced here, so aggregation memory is small. No application-side transaction logic is needed for this single read query, and maintenance cost is low because the query follows the supplied schema directly.

The query joins Users twice because every trip has both a client and a driver whose ban status must be checked independently. The WHERE clause applies both eligibility filters and the inclusive three-day date range before aggregation. Because request_at is already a DATE, it is grouped directly. The conditional SUM counts the two cancellation statuses, while COUNT(*) counts every eligible trip in the daily group. Their quotient is rounded to two decimals. The final ORDER BY returns the required ascending day order.

Common mistakes are checking only the client and forgetting that the driver must also be unbanned; joining Users only once; using banned = 0 even though the supplied column is ENUM('Yes','No'); inventing a banned_at column; forgetting one of the two cancellation statuses; multiplying the ratio by 100 even though the question asks for a rate rather than a percentage; wrapping request_at in an unnecessary DATE() call; omitting the inclusive date filter; returning extra columns instead of exactly Day and Cancellation Rate; or forgetting the ascending sort.

State the output grain first: one row per eligible day. Then explain the key insight that Users must be joined twice because both the client and driver must be unbanned. Finish by defining the cancellation numerator, the eligible-trip denominator, the two-decimal rounding, and the ascending order.

Big O notation describes how runtime or memory grows as the input size n becomes larger. I compare growth rates such as O(1), O(log n), O(n), O(n log n), and O(n²). For example, array indexing is O(1), binary search on sorted data is O(log n), scanning rows is O(n), sorting is O(n log n), and comparing every pair is O(n²). Lower growth usually scales better, but constants, data shape, vectorization, and available memory still matter in practice.

The question asks how the amount of work or memory changes when a dataset becomes larger. Big O gives us a simple way to describe that growth. It does not predict an exact number of seconds or bytes. Instead, it shows the trend as the input size n increases. The diagram compares constant, logarithmic, linear, linearithmic, and quadratic growth. It also shows why theoretical growth is only part of the decision because real performance depends on constants, data shape, vectorized execution, and available memory.

1. Should I discuss both runtime complexity and space complexity?
2. Should I compare the five common growth rates with Python examples?
3. Should I also explain practical factors such as vectorization and available RAM?

Big O describes how a cost grows when input size n grows. The cost can be execution time or memory usage. It focuses on the growth trend rather than exact runtime.

Time complexity and space complexity are separate. The same Big O notation can describe either one. An algorithm may have acceptable runtime growth but still need more memory than the machine has.

O(1) is constant complexity. Its cost stays roughly the same as n grows. The diagram shows direct array access with arr = [10, 20, 30, 40] and x = arr[2].

O(log n) is logarithmic complexity. It grows very slowly. The diagram uses binary search on data that is already sorted. With bisect.bisect_left, the search repeatedly narrows the remaining range instead of scanning every item.

O(n) is linear complexity. The work grows in direct proportion to the number of input items. The diagram shows a loop that scans all rows and adds each value to total.

O(n log n) is linearithmic complexity. It grows faster than O(n) but much more slowly than O(n²). The diagram shows sorted_data = sorted(data), representing efficient comparison sorting such as Timsort or merge sort behavior.

O(n²) is quadratic complexity. The diagram shows two nested loops. Each of the n outer iterations performs n inner iterations, so the total number of iterations grows as n².

O(1) examples include array indexing and average-case hash-table lookup. O(log n) appears in binary search on sorted data and balanced-tree operations. O(n) appears when scanning rows, computing column statistics, or performing one feature-engineering pass. O(n log n) appears when sorting for joins or building sorted indexes. O(n²) appears in pairwise similarity calculations and naive algorithms that compare every item with every other item.

For a small dataset, several approaches may all finish quickly. The difference becomes important when n grows to millions or billions of rows. Lower asymptotic growth usually scales better. It can reduce execution time, compute cost, and the risk that an approach becomes impractical as the dataset grows.

Constants matter. For a small n, an operation with a worse Big O class can still be faster because its constant overhead is lower.

Data shape matters too. Sorted, nearly sorted, sparse, duplicated, or skewed data can change real execution time.

Vectorization also matters. NumPy and pandas can move Python-level loops into optimized compiled code. This can reduce constant overhead even when the Big O class does not change.

Memory is another practical limit. Time and space complexity are separate. Either can become the bottleneck. An algorithm can be fast enough in theory but still fail because its data or temporary results do not fit in RAM.

Big O helps compare how algorithms grow as data becomes larger. O(1), O(log n), and O(n) usually scale better than O(n²). But I would not choose an implementation from Big O alone. I would also consider constants, the shape of the data, vectorized execution, and available memory.

This is a complexity-comparison question rather than one problem with one selected algorithm. The key idea is to hold the meaning of n constant as the input size and compare how different operations grow. The diagram uses direct array access for O(1), binary search on already sorted data for O(log n), a full scan for O(n), sorting for O(n log n), and nested all-pairs loops for O(n²). The central invariant is that each complexity label describes the growth trend of the shown operation as n increases. Lower growth usually scales better for very large inputs, but practical performance also depends on constant factors, data shape, vectorization, and memory limits.

There is no single complexity for the whole answer because the diagram compares five different operations. Direct array access is O(1) time because one indexed lookup does not grow with n. Binary search on already sorted data is O(log n) time because each step removes a large part of the remaining search range. Scanning all items is O(n) time because every item is visited once. Sorting is O(n log n) time in the diagram. Two nested loops over n items are O(n²) time because they perform n × n iterations. The small examples shown use O(1) extra working memory apart from the sorted result, while Python sorted creates a new list whose size grows with n. Time and space complexity must be analyzed separately.

A common mistake is treating Big O as an exact runtime instead of a growth trend. Another mistake is counting the sorting step as part of the O(log n) binary-search example even though the diagram says the data is already sorted. Candidates may also assume one Big O value describes both time and memory, but time complexity and space complexity are separate. Another mistake is hiding the O(n log n) cost of sorting. Finally, lower Big O does not guarantee faster execution for every input because constants, data shape, vectorization, and hardware still affect real performance.

Explain the classes in increasing growth order and attach one concrete example to each one: array access, binary search, row scan, sorting, and all-pairs comparison. Then say that Big O describes growth, not exact runtime, and finish with the practical factors shown in the diagram.

I use a hash map that stores each earlier value and its index. For each number, I calculate its complement, which is the target minus the current value. I check for that complement before storing the current number, so I cannot reuse the same element. When the complement exists, I return the two indices immediately. I process each item at most once and stop when the answer is found. This gives O(n) expected time and O(n) auxiliary space.

We are given a list of integers called nums and an integer target. We need the zero-based positions of two different elements whose values add up to target. The list has between 2 and 10,000 elements. Each value and target is between -10^9 and 10^9. Exactly one valid pair is guaranteed. Either index order is accepted. We must return a new two-item list and must not change nums. A dictionary is a good fit because it lets us remember earlier values and quickly find the value needed for the target.

1. Can I return the two valid indices in either order?
2. Is exactly one valid pair always guaranteed, as stated?
3. Should I leave the input list unchanged?

The function receives nums, a list of integers, and target, an integer. It must return a new list containing two distinct zero-based indices. The values at those indices must add to target. The same element cannot be used twice. The input list must stay unchanged. Exactly one valid pair is guaranteed, and either order of the two indices is valid.

I use a dictionary named value_to_index. It acts as a hash map. Each key is a value already processed. Its mapped value is that earlier element's index. Before processing index i, the map contains only values from earlier indices. For the current value x, I calculate complement = target - x. Complement means the value needed with x to reach target. If that complement is already in the map, I have found the required pair.

I start with value_to_index = {}. Nothing has been processed yet, so the empty map is correct. Then I move from left to right through nums. For each element, I get both its index i and its value x.

The example is nums = [2, 7, 11, 15] and target = 9.

At index 0, x = 2. The map before the step is {}. I calculate complement = 9 - 2 = 7. The value 7 is not in the map. I therefore store 2 -> 0. The map becomes {2: 0}.

At index 1, x = 7. The map before the step is {2: 0}. I calculate complement = 9 - 7 = 2. The value 2 is already in the map at index 0. I return [0, 1] immediately. Processing stops, so indices 2 and 3 are not processed.

The returned indices point to nums[0] = 2 and nums[1] = 7. Their values satisfy 2 + 7 = 9. Therefore, [0, 1] is a correct returned pair. [1, 0] is also allowed because the problem accepts either order.

The important invariant is that value_to_index contains only values from earlier indices. I check for the complement before storing the current value. Therefore, when a complement is found, its stored index must be different from the current index. Also, complement + x equals target by definition. Because the problem guarantees one valid pair, the algorithm eventually finds that pair and returns it.

The code creates an empty dictionary. enumerate(nums) gives the current index and value. For each value, the code calculates target - x. It looks for that complement before inserting x. If the complement exists, it returns the earlier index and the current index immediately. Otherwise, it stores the current value and index for later elements. The final return [] is only a defensive fallback. It is not expected for an input that follows the stated contract.

Python dictionary lookup and insertion are O(1) on average. We process the input at most once, so the total expected time is O(n). The dictionary can grow with the number of processed elements, so auxiliary space is O(n). Duplicate values work correctly because the current value is checked against earlier values before it is inserted. Negative numbers and zero also work. The algorithm does not mutate nums.

The key idea is to remember values from earlier indices. The dictionary stores value -> earlier index. For each current value x, calculate complement = target - x. If complement is already stored, return the earlier index and the current index. Otherwise, store x with its index and continue. The central invariant is that the map contains only values from earlier indices. Checking before insertion prevents the current element from matching itself. This avoids the slower approach of checking every possible pair.

Let n be the number of elements in nums. We process the input at most once. For each processed element, Python dictionary lookup and insertion are O(1) on average. Therefore, the total expected time is O(n). The dictionary may store up to O(n) values before the answer is found, so the auxiliary space is O(n). Auxiliary space means the extra memory used by the algorithm.

A common mistake is returning the two values instead of their indices. Another mistake is storing the current value before checking its complement, which can allow the same element to be reused. Sorting nums without preserving original indices can also produce the wrong output. Candidates may continue processing after finding the pair instead of returning immediately. Another mistake is claiming guaranteed O(n) time instead of O(n) expected time for a Python dictionary-based solution.

Explain the invariant before coding: the dictionary contains only values from earlier indices. Then say that you check the complement before storing the current value. This makes it easy to explain why the two returned indices are always distinct.

I would first compare the string lengths. If they differ, I return False. Otherwise, I create two arrays of 26 counters, one for each lowercase English letter. I scan each string and increment the matching counter using ord(ch) - ord('a'). At the end, I compare the two arrays. They are anagrams exactly when every character count matches. This takes O(n) time and O(1) auxiliary space because the arrays always contain exactly 26 counters.

We need to decide whether two lowercase English strings contain exactly the same letters the same number of times. Their order does not matter. Repeated letters do matter. For example, "anagram" and "nagaram" are a match because both contain three "a" characters and one each of "n", "g", "r", and "m". I use two fixed arrays with 26 positions. Each position counts one lowercase letter. This fits the problem because the alphabet size is fixed.

1. Can I assume both strings contain only lowercase English letters, as stated?
2. Should repeated letters count separately? Yes, identical frequencies are required.
3. Do I need to handle inputs outside the stated length and character constraints? No.

The function receives two strings, s and t. Each contains only lowercase English letters and has length from 1 to 50,000. We return True only when every letter appears the same number of times in both strings. We treat both strings as immutable. We do not normalize case or Unicode because inputs outside the stated contract do not need to be handled.

I use two arrays of exactly 26 integer counters. Each array position represents one lowercase English letter. Position 0 represents "a", position 1 represents "b", and so on. The expression ord(ch) - ord('a') converts a character into its array position. The key invariant is that after processing any prefix of a string, its count array stores exactly the letter frequencies seen in that processed prefix.

First, compare len(s) and len(t). If the lengths differ, return False immediately because strings with different numbers of characters cannot have identical character counts. Otherwise, create count_s = [0] * 26 and count_t = [0] * 26. Store ord('a') as offset. Then process every character in s and increment its position in count_s. Next, process every character in t and increment its position in count_t.

For s = "anagram", the characters are processed in this order: a, n, a, g, r, a, m. The important final counts are a = 3, n = 1, g = 1, r = 1, and m = 1. For t = "nagaram", the characters are processed in this order: n, a, g, a, r, a, m. Its final counts are also a = 3, n = 1, g = 1, r = 1, and m = 1. For example, ord('a') - ord('a') = 0, so every "a" increments slot 0. After all three "a" characters are processed, slot 0 contains 3.

After both traversals, each position in count_s equals the frequency of its corresponding lowercase letter in s. The same is true for count_t and t. Therefore, count_s == count_t exactly when every character count is identical. That is exactly the definition required by this problem. For "anagram" and "nagaram", the arrays match, so the function returns True.

After the equal-length check, each length-n string is traversed once. Comparing the two 26-element arrays takes constant work because 26 does not grow with n. The total time is O(n). Auxiliary space is O(1) because the two arrays always contain exactly 26 integers. Unequal lengths return False immediately. Repeated letters are counted exactly. Different character order does not matter. One-character strings work normally.

The key idea is to compare character frequencies instead of character positions. Because the contract limits every character to one of 26 lowercase English letters, two fixed arrays of 26 counters are enough. The expression ord(ch) - ord('a') maps each character to its counter position. The central invariant is: after processing a prefix of a string, its array contains exactly the frequencies of the letters in that processed prefix. After both strings are processed, they are anagrams exactly when the two arrays are equal.

Let n be the length of each string after the initial length check confirms the lengths are equal. We read all n characters in s once and all n characters in t once. We then compare 26 counter positions. Since 26 is a fixed constant, this gives O(n) time. Auxiliary space is O(1) because count_s and count_t always contain exactly 26 integers, regardless of how large n becomes.

A common mistake is checking only whether the same distinct letters appear while ignoring repeated counts. Another mistake is forgetting that different lengths can return False immediately. Candidates may also calculate the wrong array position for a character. Another error is changing the strings even though mutation is unnecessary. Finally, do not claim O(n) auxiliary space. The two counting arrays always have exactly 26 positions, so the auxiliary space is O(1).

Explain the invariant with one concrete mapping: 'a' gives ord('a') - ord('a') = 0, so every 'a' increments slot 0. Then explain that the same idea applies to all 26 lowercase letters. This makes the final count_s == count_t comparison easy to justify.

I would use a stack to keep the opening brackets that still need a match. I process each character at most once and stop immediately if the string becomes invalid. I push each opener. For a closer, the stack must be nonempty, and its top must be the matching opener. After all characters are processed, the stack must be empty. This works because nested brackets close in reverse order. The time complexity is O(n), and the auxiliary space is O(n).

The input is one string containing only round, square, and curly brackets. We need to return True only when every opening bracket has the correct closing bracket and the brackets are nested in the correct order. A closing bracket cannot appear without an opening bracket waiting for it. A stack fits this rule because the most recently opened bracket must be closed first. We read the string from left to right and never modify the input string.

1. Can I rely on the stated guarantee that the input contains only (, ), [, ], {, and } and has length from 1 to 10,000?
2. Is O(n) auxiliary space acceptable for a stack when the input can contain many unmatched opening brackets?

The function is is_valid(s: str) -> bool. The input is a string of length 1 to 10,000 containing only the six allowed bracket characters. We return True exactly when every opening bracket is closed by the same type in properly nested order and no closing bracket appears without a matching opener. The solution uses only the Python standard library and does not mutate any caller-owned object. The diagram's main example is "()[]{}", which returns True.

I use a stack. Each stack entry is an opening bracket that has not been matched yet. The top of the stack is the most recent unmatched opener. I also use the closing-to-opening mapping ')' -> '(', ']' -> '[', and '}' -> '{'.

The central invariant is: after every processed character, the stack contains exactly the unmatched opening brackets from the processed prefix, in nesting order. Because nested brackets must close in reverse order, each closer must match the stack top.

Start with stack = []. Traversal begins at index 0. The stack is empty because no opening bracket has been processed yet. The dictionary pairs tells us which opening bracket is required for each closing bracket.

For s = "()[]{}", the characters are at indices 0 through 5.

At index 0, the character is (. The stack is []. This is an opening bracket, so push it. The stack becomes ['(']. Processing continues.

At index 1, the character is ). The stack is ['(']. It is nonempty, so pop the top value (. The mapping says ) requires (. They match, so the stack becomes []. Processing continues.

At index 2, the character is [. Push it. The stack changes from [] to ['['].

At index 3, the character is ]. Pop [. The mapping says ] requires [. They match, so the stack becomes [].

At index 4, the character is {. Push it. The stack changes from [] to ['{'].

At index 5, the character is }. Pop {. The mapping says } requires {. They match, so the stack becomes [].

All 6 characters were processed. Every closing bracket matched the current stack top, and the stack is empty at the end. Therefore the function returns True. If a closer appeared while the stack was empty, or if a popped opener had the wrong type, the function would return False immediately and no later characters would be processed.

The stack always represents the opening brackets that still need a closing bracket. Its top is the most recent unmatched opener. Checking every closer against that top verifies both bracket type and nesting order. The empty-stack check prevents a closing bracket from appearing without an opener. Finally, an empty stack after the loop proves that no opening bracket was left unmatched.

The code first creates pairs, which maps every closing bracket to the opener it requires. It then creates an empty list called stack. Python lists support the needed stack operations with append and pop.

The loop processes the input from left to right. If ch is one of (, [, or {, the code pushes it. Otherwise, under the stated input guarantee, ch is a closing bracket. The code first checks if not stack so it never pops an empty stack. It then pops the most recent opener and compares it with pairs[ch]. Either failure returns False immediately. After the loop, return not stack returns True exactly when every opener was matched.

Let n be the length of the string. The time complexity is O(n). Each character is processed at most once, and each bracket is pushed or popped at most once. The auxiliary space complexity is O(n) because the stack can hold all opening brackets in the worst case.

Relevant edge cases from the diagram are "(]" -> False for a type mismatch, ")(" -> False for a closing bracket with no opener, "((" -> False because unmatched openers remain at the end, and "{}" -> True for a valid pair.

Bracket nesting follows last-in, first-out order. The most recent unmatched opening bracket must be closed before any earlier opener, so a stack is the correct data structure. Opening brackets are pushed onto the stack. For each closing bracket, the stack must first be nonempty. Then the algorithm pops the most recent opener and checks it against the closing-to-opening mapping. The invariant is that the stack contains exactly the unmatched opening brackets from the processed prefix in nesting order. If the full input is processed and the stack is empty, every bracket was matched correctly.

Let n be the length of the string. The time complexity is O(n). We process each character at most once. Each opening bracket can be pushed once, and each matched opening bracket can be popped once. The auxiliary space is O(n). Auxiliary space means extra memory used by the algorithm. In the worst case, the string can contain many opening brackets before any are closed, so the stack can grow to O(n) entries.

A common mistake is checking only whether the counts of opening and closing brackets are equal. Equal counts do not prove correct type or nesting. Another mistake is popping before checking whether the stack is empty. Candidates may also compare a closing bracket with something other than the stack top, which breaks nesting order. Another error is forgetting the final empty-stack check, which would incorrectly accept inputs such as "((". Finally, once an empty-stack failure or type mismatch is found, continuing to process later characters is unnecessary because the final answer is already False.

Explain the invariant before writing the loop: the stack contains exactly the unmatched opening brackets, and its top is the only opener the next closing bracket may match. Then each push, pop, early return, and final empty-stack check follows naturally.

I would merge from the end of nums1 using three pointers. Pointer i starts at the last valid value in nums1, j starts at the last value in nums2, and k starts at the last position in nums1. I compare nums1[i] and nums2[j], write the larger value at nums1[k], and move the corresponding pointer left. This avoids overwriting values that I still need. The time complexity is O(m + n), and the auxiliary space complexity is O(1).

nums1 already has enough room to hold all values from both arrays. Its first m positions contain sorted values, and its final n positions are placeholders that may be overwritten. nums2 contains n sorted values. We need to combine both sets of values inside nums1, keep the final values in nondecreasing order, leave nums2 unchanged, and return None. The safest approach is to fill nums1 from right to left because the unused positions are at the end. This prevents us from overwriting valid nums1 values before they have been compared.

1. Should I modify nums1 directly and leave nums2 unchanged? Yes. That is the required behavior.
2. Can m or n be zero? Yes. The stated constraints allow either one to be zero.
3. Are the valid values in nums1 and all values in nums2 already sorted in nondecreasing order? Yes.

nums1 has length m + n. Its first m entries are valid sorted values. Its final n entries are zero placeholders that must be overwritten. nums2 contains exactly n sorted values. We must mutate nums1 so that all m + n values are in nondecreasing order. nums2 must remain unchanged, and the function returns None.

I use three indices. i = m - 1 points to the last valid value in nums1. j = n - 1 points to the last value in nums2. k = m + n - 1 points to the last position in nums1. At each step, nums1[i] and nums2[j] are the largest remaining candidates from their arrays. I place the larger one at nums1[k]. Then I move the pointer for the value that was used and move k one position left.

The central invariant is that every position after k already contains its correct final value. The values at or before i in nums1 and at or before j in nums2 are the values that still need to be merged.

For nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], and n = 3, the starting pointers are i = 2, j = 2, and k = 5. Therefore nums1[i] = 3, nums2[j] = 6, and nums1[5] is the first destination position.

Step 1: Compare nums1[2] = 3 with nums2[2] = 6. Since 6 is larger, write 6 at nums1[5]. nums1 becomes [1,2,3,0,0,6]. Move j from 2 to 1 and k from 5 to 4.

Step 2: Compare nums1[2] = 3 with nums2[1] = 5. Since 5 is larger, write 5 at nums1[4]. nums1 becomes [1,2,3,0,5,6]. Move j from 1 to 0 and k from 4 to 3.

Step 3: Compare nums1[2] = 3 with nums2[0] = 2. Since 3 is larger, write 3 at nums1[3]. nums1 becomes [1,2,3,3,5,6]. Move i from 2 to 1 and k from 3 to 2.

Step 4: Compare nums1[1] = 2 with nums2[0] = 2. The condition is nums1[i] > nums2[j], so equal values go through the else branch. Write nums2[0] = 2 at nums1[2]. nums1 becomes [1,2,2,3,5,6]. Move j from 0 to -1 and k from 2 to 1.

Now j < 0, so every value from nums2 has been placed. The remaining values from nums1 are already in their correct positions. The final nums1 is [1,2,2,3,5,6].

At every iteration, nums1[i] and nums2[j] are the largest remaining values in their respective unmerged parts. The algorithm places the larger one into the rightmost unfilled position k. Therefore each position filled from right to left receives its correct final value. If nums2 is exhausted first, the remaining nums1 values are already correctly positioned. If nums1 is exhausted first, the remaining nums2 values are copied into the open positions.

The code initializes i, j, and k exactly as shown in the diagram. The loop continues while j >= 0 because every nums2 value must eventually be placed into nums1. If i is still valid and nums1[i] > nums2[j], the code writes nums1[i] at nums1[k] and moves i left. Otherwise, it writes nums2[j] and moves j left. After either action, k moves left. When j becomes negative, the merge is complete. nums1 has been modified in place, nums2 has not been changed, and Python returns None implicitly.

The time complexity is O(m + n). Each source pointer moves only left, so each value is processed at most once. The auxiliary space complexity is O(1) because only three integer indices are stored. If n = 0, the loop does not run. If m = 0, all values from nums2 are copied into nums1. Duplicate values, negative values, and zero values are handled correctly because the algorithm relies only on the sorted order.

The key insight is to use the free positions at the end of nums1. Filling from the front could overwrite a valid nums1 value before it has been compared. Filling from the back avoids that problem. Set i = m - 1, j = n - 1, and k = m + n - 1. Compare nums1[i] and nums2[j], place the larger remaining value at nums1[k], and move the appropriate pointer left. Then move k left. The invariant is that every position after k is already in its correct final position. The loop only needs to continue while j >= 0 because any values left in nums1 are already correctly placed.

Let m be the number of valid starting values in nums1 and n be the number of values in nums2. The time complexity is O(m + n). Each value is handled at most once because i and j only move from right to left. The auxiliary space complexity is O(1). Auxiliary space means extra memory used by the algorithm. The solution stores only the three integer indices i, j, and k. It does not create another array whose size grows with the input.

A common mistake is merging from the front of nums1. That can overwrite a valid value before it has been processed. Another mistake is forgetting to check i >= 0 before reading nums1[i]. A candidate may also stop when i becomes negative, even though values can still remain in nums2. Moving the wrong source pointer after writing a value is another common bug. It is also easy to forget to decrement k after each write. Finally, creating another merged array gives correct values but does not match the O(1) auxiliary-space approach shown in the diagram.

Before writing code, explain why the merge must go from right to left. Say that nums1 has free space at the end, so filling those positions first avoids overwriting valid values that still need to be compared. Then define i, j, and k clearly.

I would scan the prices from left to right while keeping the lowest price seen so far and the best profit found so far. If the current price is a new minimum, I update min_price. Otherwise, I calculate the profit from selling today after buying at that earlier minimum. I keep the largest profit and return it. This works because every sell day is compared with its best earlier buy price. The time complexity is O(n), and the auxiliary space is O(1).

We have a list of daily stock prices. We may make at most one trade. That means we can buy once and sell once on a later day, or make no trade. We want the largest possible profit. If every possible trade gives no positive profit, we return 0. The simple idea is to move from left to right, remember the lowest price seen so far, and compare each later selling price with that earlier low price. This gives the best answer in one pass without changing the input.

1. Can I assume the list always contains at least one price, as stated in the contract?
2. Should I return only the maximum profit, not the buy and sell days?
3. If no profitable trade exists, should I return 0?

The input is prices: list[int]. prices[i] is the nonnegative stock price on day i. The list length is from 1 to 100,000, and each price is at most 10,000. We may buy at most once and sell at most once. If we trade, the buy day must be strictly before the sell day. We return only the largest profit as an integer. If no profitable trade exists, we return 0. We do not modify the input list.

We keep two main values. min_price is the lowest price seen so far. It represents the best buying price available before a later sale. max_profit is the largest valid profit found so far. We scan from left to right. If today's price is lower than min_price, it becomes the new minimum. Otherwise, we calculate price - min_price and compare that profit with max_profit.

The diagram starts with min_price = float('inf') and max_profit = 0. Positive infinity is larger than every valid stock price, so the first price becomes the first minimum. Because max_profit starts at 0, the function naturally returns 0 when no profitable trade exists.

Use prices = [7, 1, 5, 3, 6, 4].

Day 0 has price 7. Since 7 is lower than positive infinity, set min_price = 7. max_profit stays 0.

Day 1 has price 1. Since 1 is lower than 7, set min_price = 1. max_profit stays 0.

Day 2 has price 5. It is not a new minimum. The profit from selling today is 5 - 1 = 4. Set max_profit = 4.

Day 3 has price 3. The profit is 3 - 1 = 2. This is smaller than 4, so max_profit stays 4.

Day 4 has price 6. The profit is 6 - 1 = 5. This is better, so set max_profit = 5.

Day 5 has price 4. The profit is 4 - 1 = 3. The best profit remains 5.

The final result is 5. The illustrated best trade buys at price 1 on day 1 and sells at price 6 on day 4.

The key invariant is that min_price is always the lowest buying price seen so far. When we consider a later selling price, subtracting min_price gives the best profit possible for selling on that day. max_profit keeps the largest profit from all selling days processed so far. Because we consider the prices from left to right, the buy price used for a positive trade comes before its sell price.

The function initializes the running minimum and best profit. It then processes each price once. A lower price updates min_price. Otherwise, the code calculates the profit from selling at the current price after buying at the stored minimum. If that profit is larger than the current max_profit, the function saves it. After the loop finishes, it returns max_profit.

The function scans the list once, so the time complexity is O(n), where n is the number of prices. It uses only a few variables, so the auxiliary space complexity is O(1). A one-element list returns 0 because no later sell day exists. A strictly decreasing list returns 0 because no profitable trade exists. Equal prices are also handled correctly. Although the diagram notes an empty-list result of 0, empty input is outside the stated contract and does not need special handling.

The key insight is that for any possible selling day, the best buying price is the lowest price that appeared before it. We therefore scan from left to right and maintain min_price, the lowest price seen so far, and max_profit, the best valid profit found so far. If the current price is lower than min_price, we update the minimum. Otherwise, price - min_price is the best profit available if we sell today. The central invariant is that min_price represents the cheapest buying opportunity seen so far and max_profit represents the best completed trade seen so far.

Let n be the number of prices. The loop processes each price once, so the time complexity is O(n). The algorithm does not create another list, map, table, or other structure that grows with n. It stores only a few variables such as min_price, max_profit, and profit. Therefore, the auxiliary space complexity is O(1).

1. Comparing only neighboring days instead of allowing the buy and sell days to be farther apart. 2. Sorting the prices and losing the original day order, which can violate the buy-before-sell rule. 3. Tracking only the smallest and largest values without checking whether the smaller value occurs first. 4. Returning a negative number when prices only decrease instead of returning 0. 5. Claiming O(n²) time for the one-pass implementation or claiming extra space grows with n when it does not.

State the invariant early: min_price is the lowest buying price seen so far, and max_profit is the best valid profit seen so far. Then trace [7, 1, 5, 3, 6, 4] and show why buying at 1 and selling at 6 gives the final profit of 5.