DBMS & SQL: 5-Chapter Comprehensive Practice & 75 Verified MCQs

Question 1

Which normal form requires a table to be in 1NF and additionally ensures that no non-prime attribute is partially dependent on any candidate key?

Accepted Answer

B. Second Normal Form (2NF). Second Normal Form (2NF) mandates that the relation is in 1NF and contains zero partial functional dependencies. A partial dependency occurs when a non-prime attribute depends on only a subset of a composite candidate key.

Question 2

What is the defining condition for a relation R to satisfy Third Normal Form (3NF) for every non-trivial functional dependency X -> Y?

Accepted Answer

B. Either X is a superkey of R, OR Y is a prime attribute (part of a candidate key).. 3NF eliminates transitive dependencies. For every non-trivial FD X -> Y, either X is a superkey (meaning X uniquely identifies rows), or every attribute in Y is prime (belongs to some candidate key).

Question 3

Which ACID property guarantees that all database updates within a transaction are either completely committed to disk or entirely rolled back if an error occurs?

Accepted Answer

A. Atomicity. Atomicity enforces the 'all-or-nothing' principle: either every statement within the transaction executes successfully, or the entire transaction is aborted and the database reverts to its prior state via undo logs.

Question 4

Which transaction isolation level allows a transaction to read uncommitted changes made by concurrent transactions, exposing it to 'Dirty Reads'?

Accepted Answer

D. Read Uncommitted. 'Read Uncommitted' is the lowest ANSI SQL isolation level. It permits dirty reads because transactions read data directly from the buffer pool without checking whether the modifying transaction has committed or will eventually roll back.

Question 5

In Boyce-Codd Normal Form (BCNF), what strict rule must hold for every non-trivial functional dependency X -> Y?

Accepted Answer

B. X must be a superkey of the relation.. BCNF is a stricter variant of 3NF that drops the alternative clause allowing Y to be a prime attribute. In BCNF, for every functional dependency X -> Y, the determinant X must strictly be a superkey of the relation.

Question 6

What is the primary role of Write-Ahead Logging (WAL) in database crash recovery mechanisms?

Accepted Answer

B. It guarantees that log records describing data changes are flushed to stable non-volatile storage before corresponding dirty database pages are written to disk.. The Write-Ahead Logging protocol dictates that log records recording any change must reach durable disk storage before the corresponding modified table data pages are written. This ensures the database can perform REDO or UNDO operations following an unexpected server crash.

Question 7

In relational algebra, what operation is represented by the Greek letter Sigma (σ)?

Accepted Answer

B. Selection (filtering rows satisfying a predicate condition). Sigma (σ) denotes the Selection operation in relational algebra, which filters tuples (rows) from a relation based on a specified boolean predicate condition (equivalent to SQL's WHERE clause). Projection is denoted by Pi (π).

Question 8

Under the Two-Phase Locking (2PL) protocol, what characterizes the 'shrinking phase' of a transaction?

Accepted Answer

B. The transaction may only release existing locks and cannot acquire any new locks.. Two-Phase Locking divides concurrency into two distinct phases: 1) Growing phase (locks are acquired, none released), and 2) Shrinking phase (locks are released, no new locks can be obtained). Once a transaction releases a single lock, it enters the shrinking phase and can never acquire another lock.

Question 9

What is a 'Phantom Read' anomaly in database transactions?

Accepted Answer

B. A transaction re-executes a range query with identical search criteria and discovers new rows inserted and committed by another concurrent transaction.. A Phantom Read occurs when transaction T1 executes a range query (e.g., 'SELECT WHERE age > 30'), and a concurrent transaction T2 inserts new rows satisfying that condition and commits. When T1 re-runs the range query, new 'phantom' rows appear.

Question 10

How does a Clustered Index differ physically from a Non-Clustered (Secondary) Index in relational databases like SQL Server and MySQL (InnoDB)?

Accepted Answer

B. A clustered index dictates the physical sorting and storage order of actual data rows in leaf pages; non-clustered indexes maintain a separate B+Tree containing index keys pointing back to data rows.. In a clustered index, the leaf nodes of the B+Tree ARE the actual data rows; hence a table can have only ONE clustered index because rows can be physically ordered in only one sequence. Non-clustered indexes are secondary trees whose leaves contain the indexed column plus a row locator (RID or clustered key).

Question 11

What type of dependency is eliminated when moving from Second Normal Form (2NF) to Third Normal Form (3NF)?

Accepted Answer

B. Transitive dependency. 3NF specifically resolves transitive dependencies: situations where a non-prime attribute depends on another non-prime attribute (X -> Z where X -> Y and Y -> Z, with X being candidate key and Y non-prime).

Question 12

In SQL, what is the consequence of executing 'DELETE FROM table_name' compared to 'TRUNCATE TABLE table_name'?

Accepted Answer

B. DELETE removes rows row-by-row logging each deletion in the transaction log; TRUNCATE deallocates entire data pages with minimal logging and resets identity seeds.. DELETE is a DML command that evaluates where-clauses, fires delete triggers, and logs individual row deletions, making it slower for massive datasets. TRUNCATE is a DDL operation that deallocates entire data storage pages, logs page deallocations minimally, resets identity seeds, and bypasses row-level triggers.

Question 13

What is the primary characteristic of an outer join (e.g., LEFT OUTER JOIN) compared to an INNER JOIN in SQL?

Accepted Answer

B. It preserves all rows from the specified table (e.g. left), filling columns from the non-matching side with NULL values when join criteria fail.. A LEFT OUTER JOIN returns every record from the left table. If a matching record exists in the right table, right columns are populated; if no match exists, all columns from the right table are returned as NULL.

Question 14

In database normalization, when is a decomposition of a relation schema R into R1 and R2 guaranteed to be a 'Lossless-Join' decomposition?

Accepted Answer

B. When the intersection of attributes (R1 ∩ R2) forms a superkey of at least one of the decomposed relations (R1 or R2).. By definition, a decomposition of R into (R1, R2) is lossless-join with respect to functional dependencies F if and only if the common attributes (R1 ∩ R2) functionally determine all attributes in R1 ((R1 ∩ R2) -> R1) OR all attributes in R2 ((R1 ∩ R2) -> R2).

Question 15

What is the purpose of the SQL 'HAVING' clause, and how does it fundamentally differ from the 'WHERE' clause?

Accepted Answer

B. WHERE filters individual rows prior to GROUP BY aggregation; HAVING filters aggregated groups based on aggregate function results (e.g., COUNT(), AVG()).. In SQL logical query processing, WHERE filters raw rows before aggregation occurs. The GROUP BY clause then groups surviving rows into aggregated buckets. Finally, the HAVING clause filters those aggregated buckets based on aggregate conditions (e.g., 'HAVING COUNT(*) > 5').

Question 16

What is the primary objective of normalizing a relational database schema?

Accepted Answer

B. To eliminate data anomalies (insertion, deletion, update) and minimize unnecessary data redundancy. Normalization is the systematic process of decomposing relations to minimize redundancy and eradicate update, insertion, and deletion anomalies without losing informational dependency.

Question 17

A relational table is in First Normal Form (1NF) if and only if:

Accepted Answer

B. Every attribute domain contains only atomic (indivisible) values and there are no repeating groups. 1NF requires that each column contain only single scalar, atomic values (e.g., no comma-separated arrays or nested composite structures) and each record is unique.

Question 18

Under what condition is a relation in 1NF guaranteed to already be in Second Normal Form (2NF)?

Accepted Answer

B. When the primary key consists of a single (simple) attribute rather than a composite key. 2NF requires that no non-prime attribute be partially dependent on any candidate key. If the primary key is a single attribute, partial dependency is mathematically impossible, so it is automatically in 2NF.

Question 19

What defines Third Normal Form (3NF) in relational schema theory?

Accepted Answer

A. The relation is in 2NF and has no transitive dependencies for non-prime attributes. A relation is in 3NF if it is in 2NF and for every functional dependency X -> A, either X is a superkey or A is a prime attribute (member of a candidate key), preventing X -> Y -> Z transitive chains among non-prime attributes.

Question 20

How does Boyce-Codd Normal Form (BCNF) differ strictly from Third Normal Form (3NF)?

Accepted Answer

B. In BCNF, for every non-trivial functional dependency X -> A, X must strictly be a Superkey (removing 3NF's exception where A can be a prime attribute). 3NF allows X -> A if A is a prime attribute even when X is not a superkey. BCNF eliminates this loophole: for every functional dependency X -> A, X MUST be a superkey without exception.

Question 21

Which normal form specifically addresses and eliminates Multivalued Dependencies (MVDs)?

Accepted Answer

C. Fourth Normal Form (4NF). Fourth Normal Form (4NF) handles multivalued dependencies (X ->-> Y). A relation is in 4NF if it is in BCNF and for every non-trivial multivalued dependency X ->-> Y, X is a superkey.

Question 22

Fifth Normal Form (5NF), also known as Project-Join Normal Form (PJNF), eliminates anomalies caused by:

Accepted Answer

A. Join Dependencies that cannot be inferred from candidate keys. 5NF deals with Join Dependencies. A relation is in 5NF if every non-trivial join dependency in it is implied by the candidate keys, guaranteeing lossless join reconstruction when decomposed.

Question 23

What is the concept of 'Lossless-Join Decomposition' in relational database design?

Accepted Answer

B. Decomposing relation R into R1 and R2 guarantees that natural joining R1 and R2 reconstructs the exact original relation R without generating spurious tuples. Lossless join ensures that R1 ⨝ R2 = R. According to Heath's theorem, decomposition into R1 and R2 is lossless if and only if (R1 ∩ R2) -> R1 or (R1 ∩ R2) -> R2.

Question 24

When is a decomposition of a relation considered 'Dependency Preserving'?

Accepted Answer

B. When every functional dependency in the original relation can be checked and verified by examining individual decomposed relations without computing multi-table joins. Dependency preservation guarantees that the union of dependencies of all decomposed sub-relations logically entails the complete set of original dependencies F (i.e. (F1 ∪ F2)+ = F+).

Question 25

Given Relation R(A, B, C, D) with FDs: {A -> B, B -> C, C -> D}. What is the Candidate Key of R?

Accepted Answer

A. {A}. Computing attribute closure: A+ = {A, B, C, D}. Because A derives all attributes and no proper subset of A can do so, {A} is the minimal candidate key.

Question 26

In relation R(A, B, C) where A is Primary Key and FD is {B -> C}. In which normal form is R?

Accepted Answer

A. 1NF and 2NF, but NOT 3NF (transitive dependency exists). The primary key is single-column A, so 2NF holds (no partial dependency). However, B is a non-prime attribute determining another non-prime attribute C (transitive dependency A -> B -> C), violating 3NF.

Question 27

What is Armstrong's Axiom of 'Augmentation' for functional dependencies?

Accepted Answer

A. If X -> Y, then XZ -> YZ for any attribute set Z. Augmentation states that adding common attributes to both sides preserves validity: If X -> Y holds, then XZ -> YZ holds for any attribute set Z.

Question 28

What is a 'Canonical Cover' (Minimal Cover) of a set of functional dependencies?

Accepted Answer

B. A minimal, simplified equivalent set of functional dependencies having no redundant dependencies or extraneous attributes. A canonical cover Fc for F has the same closure as F (Fc+ = F+), contains only singleton right-hand sides, has no extraneous attributes on left or right, and contains no redundant FDs.

Question 29

What is the primary operational trade-off of applying intentional 'Denormalization' in production OLAP data warehouses?

Accepted Answer

B. Introduces controlled data redundancy to avoid expensive multi-table JOINs, accelerating analytical read query throughput. Denormalization deliberately combines related tables (e.g., Star Schema / Snowflake Schema) to eliminate JOIN overhead during complex analytical queries, trading off write performance and disk storage for read speed.

Question 30

If relation R(A, B, C) has Candidate Keys {A, B} and {B, C}, and FD {C -> A}, why is R in 3NF but NOT in BCNF?

Accepted Answer

B. In C -> A, C is not a superkey, but A is a prime attribute (part of candidate key {A, B}), satisfying 3NF but violating BCNF. In 3NF, C -> A is permitted because attribute A is prime. However, BCNF strictly demands that the left-hand determinant C must be a superkey. Because C alone cannot determine all attributes, it fails BCNF.

Question 31

What is the difference between RANK() and DENSE_RANK() window functions in SQL?

Accepted Answer

B. RANK() leaves gaps in sequence ranks after tied values; DENSE_RANK() assigns consecutive rank numbers without skipping gaps. If two rows tie for 1st place, RANK() gives both 1 and assigns 3 to the next row (gap). DENSE_RANK() gives both 1 and assigns 2 to the next row (no gaps).

Question 32

What is the result of applying a LEFT JOIN between Table A (10 rows) and Table B (5 rows) if none of the join conditions match?

Accepted Answer

B. 10 rows returned, with NULL values populated for all Table B columns. A LEFT JOIN preserves every row from the left table (Table A). When no matching record is found in Table B, NULL is substituted for all columns belonging to Table B.

Question 33

How does the 'HAVING' clause differ from the 'WHERE' clause in SQL execution order?

Accepted Answer

A. WHERE filters individual rows before grouping; HAVING filters aggregated groups after GROUP BY evaluation. WHERE filters base records prior to GROUP BY aggregation. Aggregate functions like SUM() or COUNT() cannot appear in WHERE; HAVING filters the groups generated by GROUP BY.

Question 34

What is a Common Table Expression (CTE) defined with the 'WITH' keyword in SQL?

Accepted Answer

A. A temporary named result set whose scope is confined to the execution of a single primary query statement. A CTE provides a readable, modular temporary result set defined using 'WITH cte_name AS (...)'. Recursive CTEs can also traverse hierarchical tree and graph structures.

Question 35

What does the SQL window function LAG(column_name, 1) OVER (ORDER BY date) return?

Accepted Answer

B. The value of column_name from the row preceding the current row in the partitioned sort order. LAG() accesses data from a previous row at a specified physical offset without requiring a self-join. LEAD() similarly accesses data from subsequent rows.

Question 36

What is the difference between UNION and UNION ALL operators in SQL?

Accepted Answer

A. UNION combines results and performs a deduplication sort pass to remove duplicate rows; UNION ALL concatenates results directly without eliminating duplicates. UNION executes an implicit DISTINCT sort/hash step to remove duplicates, incurring CPU overhead. UNION ALL simply appends results together, running substantially faster.

Question 37

In SQL, what is a Correlated Subquery?

Accepted Answer

A. A subquery that references columns from the outer enclosing query and is evaluated once for each row processed by the outer query. A correlated subquery cannot be executed independently because it depends on values passed in from the outer query row-by-row, which can result in O(N*M) performance if unoptimized.

Question 38

What is the output of the expression 'SELECT COUNT(*), COUNT(col) FROM tbl' if 'col' contains 5 non-null numbers and 3 NULL values?

Accepted Answer

B. 8, 5. COUNT(*) counts all rows regardless of NULL content (8 total rows). COUNT(column_name) counts only rows where the specified column is NOT NULL (5 rows).

Question 39

What does a CROSS JOIN between Table X (m rows) and Table Y (n rows) produce?

Accepted Answer

A. A Cartesian product containing m * n total rows. CROSS JOIN produces the Cartesian product, pairing every row of Table X with every row of Table Y, generating exactly m * n rows.

Question 40

How does NULL behave in boolean ternary logic comparison (e.g. 'WHERE salary = NULL')?

Accepted Answer

C. Evaluates to UNKNOWN, which causes the row to be excluded by the WHERE filter. In SQL three-valued logic (3VL), comparing anything with NULL using equality (= or !=) yields UNKNOWN. Only 'IS NULL' or 'IS NOT NULL' evaluates to TRUE/FALSE.

Question 41

What does the SQL COALESCE(val1, val2, val3, ...) function do?

Accepted Answer

B. Returns the first non-NULL expression from its argument list, or NULL if all arguments are NULL. COALESCE evaluates arguments from left to right and returns the first non-null value found, functioning as an ANSI-standard null fallback mechanism.

Question 42

What is the purpose of the 'PARTITION BY' clause inside a SQL OVER() window definition?

Accepted Answer

B. It divides query result rows into separate subsets/partitions across which the window function is independently calculated. PARTITION BY divides rows into groups (e.g. PARTITION BY department_id) so the window function calculates ranks or aggregates separately for each department without collapsing rows.

Question 43

What happens when an INNER JOIN condition matches multiple rows in both tables (a 1-to-many or many-to-many match)?

Accepted Answer

B. Output rows are multiplied, producing all combination pairs matching the join predicate. INNER JOIN outputs every pair of rows that satisfies the join predicate. If row A matches 3 rows in B, 3 result rows are emitted.

Question 44

What does the 'EXISTS' operator evaluate to in an SQL query filter?

Accepted Answer

A. TRUE if the subquery returns at least one row; short-circuits as soon as a single match is found. EXISTS tests for the existence of rows in a subquery. The database engine short-circuits evaluation as soon as the first matching row is discovered, making it highly efficient.

Question 45

What is an SQL View, and does a standard (non-materialized) View store data on disk?

Accepted Answer

A. A view is a virtual table defined by a stored SELECT query; standard views do NOT store data on disk, re-executing the underlying query upon access. A standard View is metadata storing a query definition. Data is dynamically read from base tables at runtime. In contrast, Materialized Views persist query output to disk.

Question 46

Why do relational database storage engines (InnoDB, PostgreSQL) predominantly utilize B+ Trees over standard Binary Search Trees for disk-based indexing?

Accepted Answer

B. B+ Trees have very high fan-out (hundreds of keys per node), keeping tree height low (3-4 levels) to minimize expensive disk I/O operations. Disk seeks are thousands of times slower than RAM. High fan-out B+ trees store millions of keys with height 3-4, meaning any lookup requires at most 3-4 disk block reads. BSTs would require log2(N) levels (~20 disk reads).

Question 47

What is the primary architectural difference between a Clustered Index and a Non-Clustered (Secondary) Index?

Accepted Answer

B. The leaf pages of a Clustered Index contain the actual physical table data rows; secondary indexes store index keys and pointer references to the clustered index key. A Clustered Index dictates the physical ordering of data rows on disk (leaf level = actual row data). Therefore, a table can have exactly ONE clustered index (usually the Primary Key).

Question 48

What is a 'Covering Index' in relational database performance tuning?

Accepted Answer

B. An index that contains all columns requested by a SELECT query, allowing the engine to satisfy the query entirely from the index without reading the base table pages. When an index contains every column referenced in SELECT, WHERE, and ORDER BY clauses, the database performs an 'Index Only Scan', completely avoiding random I/O row lookups.

Question 49

How does the 'Leftmost Prefix Rule' apply to a composite index created on (col_a, col_b, col_c)?

Accepted Answer

A. The index can only be used if col_a is included in the query search condition. A composite B-Tree index sorts data first by col_a, then by col_b, then by col_c. If a query does not specify col_a, the index cannot be traversed hierarchically, resulting in a full table scan.

Question 50

What is the primary drawback of adding excessive indexes to high-write (OLTP) tables?

Accepted Answer

B. Every INSERT, UPDATE, and DELETE operation must update all corresponding index B+ Trees, substantially degrading write throughput and causing page splits. Indexes accelerate reads but penalize writes. Every write operation must maintain B+ Tree invariants, balancing nodes and performing costly disk page splits across every index on the table.

Question 51

Why are standard Hash Indexes unsuitable for SQL queries containing 'WHERE age BETWEEN 20 AND 30'?

Accepted Answer

B. Hash indexes distribute keys uniformly across buckets using a hash function, destroying ordering and preventing range scans. Hash indexes offer O(1) point lookups for exact equality (=, IN) but cannot satisfy range operators (<, >, BETWEEN) because hashed keys do not preserve lexical or numeric order.

Question 52

What does a 'Full Table Scan' (Seq Scan in PostgreSQL) signify in a database EXPLAIN query plan?

Accepted Answer

B. The storage engine read every page of the table sequentially from disk/cache to evaluate query predicates because no suitable index was matched or cost was deemed lower. A full table scan inspects every row in the table. For small tables, sequential scans are often faster than random index lookups; for large tables, missing indexes cause severe latency spikes.

Question 53

What causes an 'Index Page Split' in a B+ Tree data structure?

Accepted Answer

B. Inserting a new key into an index leaf page that is already 100% full, forcing the engine to allocate a new page and redistribute half the keys. When an inserted key lands on a leaf page with no remaining free space, the engine splits the page into two 50% full pages, updating parent pointers. Random UUID keys cause severe page split thrashing.

Question 54

How does an LSM Tree (Log-Structured Merge-tree, used in RocksDB/Cassandra) differ fundamentally from a B+ Tree?

Accepted Answer

A. LSM Trees write sequentially to an append-only in-memory MemTable and commit log, flushing sorted SSTables to disk to maximize write throughput. B+ Trees perform random in-place page writes. LSM Trees convert all mutations into sequential append writes in RAM and log, flushing immutable SSTables and merging them via background compaction.

Question 55

What is Index Cardinality in database statistics?

Accepted Answer

B. The uniqueness of values in a column; high cardinality means high percentage of unique values (e.g. email, user_id). High cardinality columns (e.g. SSN, email) make excellent index candidates because an index can quickly filter down to a tiny fraction of records. Low cardinality columns (e.g. gender: M/F) are rarely indexed alone.

Question 56

What happens when an expression or function wraps an indexed column in the WHERE clause (e.g. 'WHERE UPPER(last_name) = "SMITH"')?

Accepted Answer

B. The standard index is rendered unusable, triggering a full table scan unless a Function-Based Index is defined. The B-Tree index stores raw last_name values, not UPPER(last_name). Applying functions prevents binary search traversal unless a functional/expression index on UPPER(last_name) exists.

Question 57

What is the purpose of the Database Buffer Pool (Buffer Cache)?

Accepted Answer

B. To cache recently accessed data and index pages in main RAM, avoiding disk I/O for frequently read tables. The buffer pool caches database disk pages in RAM using LRU eviction policies. Queries finding target pages in the buffer pool achieve sub-millisecond memory-speed execution (cache hits).

Question 58

What is a 'Bitmap Index' and in which environment is it most effectively utilized?

Accepted Answer

B. An index that uses bit arrays (0s and 1s) for low-cardinality columns, highly effective in OLAP data warehouses for fast bitwise AND/OR query filtering. Bitmap indexes represent column values as strings of bits. Bitwise operations (AND, OR, NOT) evaluate complex multi-attribute queries in microseconds. However, row locking makes them unviable for OLTP writes.

Question 59

What is the difference between Clustered Index implementation in MySQL InnoDB vs PostgreSQL?

Accepted Answer

A. InnoDB always organizes table storage as a clustered index B+ Tree (IOT); PostgreSQL uses Heap Tables where rows are unordered and all indexes are secondary pointer structures. InnoDB tables ARE clustered indexes (Index-Organized Tables). In PostgreSQL, all tables are unordered Heap structures, and the CLUSTER command is a one-time sort that does not maintain continuous clustering.

Question 60

What is the role of the Cost-Based Optimizer (CBO) in a relational database engine?

Accepted Answer

B. Evaluates multiple candidate execution plans (JOIN algorithms, index paths) and selects the plan with the lowest estimated disk I/O and CPU cost. The CBO uses table statistics, histogram distributions, and hardware parameters to estimate costs for candidate query execution plans, selecting the mathematically cheapest execution strategy.

Question 61

What does the 'A' (Atomicity) in ACID transaction properties guarantee?

Accepted Answer

B. All operations within a transaction succeed completely, or the entire transaction is rolled back leaving the database state unmodified ('All or Nothing'). Atomicity ensures that a transaction containing multiple SQL statements is treated as an indivisible atomic unit. If a crash or error occurs mid-way, all changes are completely aborted and undone.

Question 62

What phenomenon is referred to as a 'Dirty Read' in database concurrency?

Accepted Answer

B. Transaction T1 reads data modified by uncommitted concurrent transaction T2, which subsequently rolls back. A Dirty Read occurs when a transaction views uncommitted changes of another transaction. If the modifying transaction aborts, the first transaction has acted upon phantom, non-existent data.

Question 63

In ANSI SQL isolation levels, what concurrency anomaly is permitted under 'Read Committed' but prevented under 'Repeatable Read'?

Accepted Answer

B. Non-Repeatable Read (Fuzzy Read). Non-Repeatable Read occurs when T1 reads a row, T2 updates/deletes that row and commits, and T1 re-reads the row obtaining different values. Repeatable Read ensures T1 re-reads identical row values.

Question 64

What is a 'Phantom Read' anomaly?

Accepted Answer

B. Transaction T1 queries a range of rows; concurrent transaction T2 inserts new matching rows and commits; T1 re-runs the range query and discovers newly appeared 'phantom' records. Phantom Reads involve set-based range queries where concurrent committed INSERTs cause new rows to appear between consecutive range scans within the same transaction.

DBMS & SQL: 5-Chapter Comprehensive Practice & 75 Verified MCQs

DBMS & SQL: 5-Chapter Comprehensive Practice & 75 Verified MCQs

Explore Other Comprehensive 75-Question Subject Tracks

Queryiest

Queryiest – Technology Writer | Software Developer | Digital Learning Enthusiast

What is a database?

How to rotate an array in-place with O(1) space and ...

What steps can businesses take to identify the most valuable ...

Queryiest

Anonymous

paperubofficial