Practice Question

Duration: 7 min

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Module outline

  1. About the Course: About the Course
  2. Section 1 | Language (Hindi): भाषा, लिपि, ध्वनि, वर्ण, अक्षर, स्वर, व्यंजन, मात्रा और संयुक्ताक्षर, वर्तनी, वर्ण-विन्यास, अनुस्वार, अनुनासिक, चंद्रबिंदु और ‘र’ के रूप, शब्द, पद और शब्द-भेद, संज्ञा और सर्वनाम, विशेषण और क्रिया, लिंग, वचन, पुरुष और कारक, काल, वाच्य, वाक्य और वाक्य-भेद, संधि, समास, विलोम शब्द, पर्यायवाची शब्द, सामान्य अशुद्धियाँ, अनेक शब्दों के लिए एक शब्द, मुहावरे और लोकोक्तियाँ, अपठित गद्यांश और अपठित पद्यांश, Multi-Topic Content (Cross-Concept Material), वर्णमाला
  3. Section 1 | Language (उर्दू भाषा — 22 प्रश्न (निश्चित)): Spelling & Diacritics, Grammar (اسم, فعل, صفت, ضمیر), Vocabulary, Idioms & Proverbs, Comprehension
  4. Section 1 | Language (বাংলা / बांग्ला भाषा — 22 प्रश्न (निश्चित)): Bangla Script and Sounds, Bangla Grammar and Usage, Bangla Word Formation and Vocabulary, Reading and Text Interpretation
  5. Section 1 | Language (English): Vocabulary, Noun, Subject Verb Agreement (Verb Noun Agreement), Adjectives, Tenses, Pronoun, Verb, Adverb, Preposition, Articles, Determiner, Conjunction, Modals, Active Passive Voice (Voice), Direct and Indirect Speech, Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Sentence Construction, Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Cloze Test (Cloze Passage), Verbal Analogy (Word Based Analogy), Facts, Inference and Judgements
  6. Section 2 | Primary Maths & Reasoning: Ratio and Proportion (Ratios), Divisibility Rules, Mixture and Alligation, Permutation and Combination, Number System, HCF LCM, Sequence and Series (Series), Speed Time and Distance, Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Percentage, Average, Profit and Loss, Simple Interest and Compound Interest, Time and Work, Mensuration and Geometry, Age Problems, Arithmetic, Powers and Exponents (Surds and Indices), Algebra, Probability, Trigonometry, Syllogisms, Seating Arrangements, Blood Relations, Directions (Direction Test), Analytical Reasoning (Counting Figures Reasoning), Puzzle Solving (Puzzles), Calendar, Clock, Cubes & Dices, Venn Diagram, Data Arrangement, Inequality, Analogy, Ranking, Order and Sequence
  7. Section 2 | General Awareness (GK + Current Affairs): Polity, Physics, Biology, State Wise GK, Chemistry, History, Geography, Indian Economy, Environment, GK- Current Affairs, Static General Knowledge
  8. Section 3 | Computer System and Org. (CSO): Basics of COA, Floating Point Rep, Cache Memory Organization, Instr Formats & Modes, Programming the Basic Computer, Input Output Organisation, Register Transfer and Microoperations, Control Unit Design
  9. Section 3 | Database Management System: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, SQL, File Organization & Indexing, Relational Algebra, Relational Calculus, Transaction Management, Concurrency Control
  10. Section 3 | Operating System: Introduction to OS, Process Management, CPU Scheduling, Threads & Process Creation, Memory Management, Virtual Memory, Disc Scheduling, File Management, Process Synchronization
  11. Section 3 | Computer Networks: Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Transport Layer Services, TL: Congestion & UDP, Application Layer, Hardware basics
  12. Section 3 | Data Structures: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing, Algorithm Analysis, Time Complexity Analysis, Sorting Algorithms, Greedy Algorithms, Minimum Spanning Trees, Shortest Path Algos
  13. Section 3 | Theory Of Computation & Compiler Design: Introduction to TOC, Deterministic FA (DFA), Non-Deterministic FA, Regular Expressions, Grammar, Regular Language Properties, Moore & Mealy Machines, Pushdown Automata & CFG, Turing Machines, Intro to Compilers, Lexical Analysis, Grammar & CFG, Syntax Analysis: Top-Down, Syntax Analysis: Bottom-Up, Semantic Analysis & SDT, Intermediate Code Gen, Code Optimization
  14. Section 3 | Emerging Trends (AI, Societal Ethics & MS Office): Artificial Intelligence, Emerging Tech Trends, Society, Law & Ethics, MS Office Applications
  15. Section 3 | Software Engineering: Fundamentals of Software Engineering, Software Requirements and Quality Assurance, Software Design, Estimation and Metrics, Software Maintenance and Configuration Management, Software Testing
  16. Section 3 | C Programming: C Fundamentals, Control Flow, Functions, Arrays & Pointers, Storage Classes, Structures & Enums, DMA, Macros, Scoping & File Handling
  17. Section 3 | OOPS, C++, JAVA, HTML: OOPS with C++, Java, HTML Basics, CSS
  18. Section 3 | Java: Java Fundamentals, Array, String, IO & References
  19. Section 3 | Python: Introduction to Python, Tokens, Variables, Data Types, I/O, Operators & Expressions, Control Flow & Looping, Collections & Strings, Functions & Modules, File Handling & Data Analysis
  20. Section 3 | Digital Electronics: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits, Number System, Number Representation
  21. Section 3 | Multimedia: Multimedia
  22. BPSC Previous Year Papers:
  23. BPSC Full Mock Tests:
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This educational video segment focuses on solving a practice problem regarding the memory location calculation of two-dimensional array elements. The instructor presents an array VAL with dimensions ranging from 1 to 15 for rows and 1 to 10 for columns. Each element occupies 4 bytes of storage, with a base address of 1500. The core task is to determine the memory location of the specific element VAL[12][9] using two distinct storage methods: Row Major Order (RMO) and Column Major Order (CMO). The session begins by defining the problem parameters, including lower bounds L1 and L2, which are both 1 in this context. The instructor systematically derives the general formula for Row Major Order, Loc(A[i][j]) = BA + W[(i - L1) * #cols + (j - L2)], and applies it to the specific indices. Subsequently, the video transitions to Column Major Order, utilizing a similar but structurally different formula that prioritizes column indices over row indices in the offset calculation. The final output provides precise memory addresses for both storage schemes, demonstrating the arithmetic progression required to solve such data structure problems.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with the instructor displaying a practice question on screen regarding an array VAL[1...15][1...10] stored in memory. The problem statement specifies that each element requires 4 bytes of storage and the base address is 1500. The instructor identifies the task as finding the location of VAL[12][9] using both Row Major Order (RMO) and Column Major Order. Visible text on screen includes the array dimensions, base address value, and element size. The instructor begins writing 'Row wise (RMA)' on the board to initiate the solution process for the first storage method. The setup phase involves identifying the lower bounds L1 and L2 as 1, and calculating the total number of columns as U2 - L2 + 1 = 10. The instructor writes the general formula for Row Major Order: Loc(A[i][j]) = BA + W [(i - L1) x #cols + j - L2]. This section establishes the foundational parameters required for subsequent calculations.

  2. 2:00 – 5:00 02:00-05:00

    The instructor proceeds to solve the Row Major Order (RMO) calculation step-by-step. The screen shows the substitution of specific values into the previously written formula: Loc(A[i][j]) = BA + w[(i-L1)*#cols + (j-L2)]. The target indices are identified as i=12 and j=9. The calculation begins with the substitution: 1500 + 4[(12-1)*10 + (9-1)]. The instructor simplifies the terms inside the brackets, calculating 12 minus 1 to get 11, and 9 minus 1 to get 8. The intermediate step shows the multiplication of 11 by the number of columns (10), resulting in 110, which is then added to 8. The expression becomes 1500 + 4[11*10 + 8], which simplifies to 1500 + 4[118]. The final arithmetic yields 1500 + 472, resulting in a memory address of 1972. The instructor writes the final result clearly on the screen, confirming the location of VAL[12][9] under row-wise storage.

  3. 5:00 – 7:13 05:00-07:13

    The video transitions to the second part of the problem, calculating the memory location using Column Major Order (CMO). The instructor writes the specific formula for column-wise storage: Loc(A[i][j]) = BA + W[(i-Lr) * #rows + (j-Lc)]. The visible text indicates the base address remains 1500 and element size W is 4. The instructor substitutes the indices i=12 and j=9 into this new formula, along with the number of rows (15) instead of columns. The calculation proceeds as 1500 + 4[(9-1) * 15 + (12-1)]. The instructor simplifies the bracketed terms, calculating 9 minus 1 to get 8, and multiplying by 15 rows to obtain 120. The row index offset is calculated as 12 minus 1, which equals 11. Adding these gives 120 + 11 = 131. The final step multiplies 4 by 131 to get 524, which is added to the base address. The screen displays the final result as 1500 + 4 * 131 = 2024, completing the solution for column-wise storage.

The lecture segment effectively demonstrates the application of memory addressing formulas for two-dimensional arrays. The key distinction lies in how the offset is calculated: Row Major Order multiplies the row index difference by the number of columns, while Column Major Order multiplies the column index difference by the number of rows. The problem uses 1-based indexing, requiring the subtraction of lower bounds (L1=1, L2=1) from the target indices before calculation. The arithmetic progression is clearly shown, moving from general formulas to specific substitutions and finally to numerical results. The final addresses differ significantly (1972 vs 2024), highlighting the impact of storage order on memory layout. The instructor's methodical approach ensures clarity in deriving both results from the same initial problem statement.

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