Practice Question

Duration: 7 min

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Module outline

  1. About Course: About the Course
  2. Paper 2 | Digital Logic: Digital Systems & Boolean Basics, Logic Gates & Hardware, Boolean Expression, Boolean Minimization, Combinational Circuit, Sequential Circuits, Number System, Number Representation, Floating Point Rep
  3. Paper 2 | Computer Organization And Architecture: Basics of COA, Cache Memory Organization, Input Output Organisation, Pipelining, Control Unit Design, Instr Formats & Modes, Programming the Basic Computer
  4. Paper 2 | Analog And Digital Communication: Introduction to Communication Systems & Random Signals, Analog Communication Systems, Information Theory & Digital Communication Fundamentals, Digital Modulation & Detection Techniques, Error Control, Synchronization & Multiple Access Techniques
  5. Paper 2 | Programming in C: C Fundamentals, Control Flow, Functions, Arrays & Pointers, Storage Classes, Structures & Enums, DMA, Macros, Scoping & File Handling
  6. Paper 2 | Data Structures And Programming: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing, Java, OOPS with C++, Java Fundamentals, Array, String, IO & References
  7. Paper 2 | Algorithms: Algorithm Analysis, Sorting Algorithms, Time Complexity Analysis, Greedy Algorithms, Dynamic Programming, Minimum Spanning Trees, Shortest Path Algos
  8. Paper 2 | Compiler Design: Intro to Compilers, Lexical Analysis, Grammar & CFG, Syntax Analysis: Top-Down, Syntax Analysis: Bottom-Up, Semantic Analysis & SDT, Intermediate Code Gen, Code Optimization, Run Time Environment
  9. Paper 2 | Operating System: Introduction to OS, Process Management, CPU Scheduling, Process Synchronization, Threads & Process Creation, Deadlock, Memory Management, Virtual Memory, Disc Scheduling, File Management, Unix OS, Linux OS
  10. Paper 2 | Databases: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, Relational Algebra, SQL, Relational Calculus, File Organization & Indexing, Transaction Management, Concurrency Control
  11. Paper 2 | Information Systems and Software Engineering: Fundamentals of Software Engineering, Software Requirements and Quality Assurance, Software Design, Estimation and Metrics, Software Testing, Software Maintenance and Configuration Management
  12. Paper 2 | Computer Networks: Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, DLL: Framing, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Net Layer:Routing Protocol, Transport Layer Services, TL: Congestion & UDP, Application Layer, Hardware Basics
  13. Paper 2 | Internet & Web Technologies: Website, Browser & Services, Internet Fundamentals, Web Design & Development, Protocols & Services, Web Publishing & Hosting, Server Side & Security, Emerging Technologies, HTML Basics, JavaScript-Basics, CSS, XML, HTML5
  14. Paper 2 | Cyber Security And Emerging Technologies: Emerging Tech Trends
  15. Paper 2 | Cloud Technology: Cloud Computing
  16. Paper 2 | Probability and statistics: Probability, Statistics
  17. Paper-1 English: Vocabulary, Preposition, Articles, Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Cloze Test (Cloze Passage), Sentence Construction
  18. Paper 1 General Mental Ability & Logical reasoning: Ratio and Proportion (Ratios), Divisibility Rules, Data Interpretation, Mixture and Alligation, Permutation and Combination, Logarithm, Number System, HCF LCM, Sequence and Series (Series), Speed Time and Distance, Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Data Sufficiency, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Percentage, Average, Powers and Exponents (Surds and Indices), Algebra, Profit and Loss, Simple Interest and Compound Interest, Time and Work, Pipe and Cistern, Mensuration and Geometry, Mental Ability, Age Problems, Partnership, Arithmetic, Speed Mathematics and Basic Mathematics, Syllogisms, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Seating Arrangements, Blood Relations, Directions (Direction Test), Analytical Reasoning (Counting Figures Reasoning), Puzzle Solving (Puzzles), Calendar, Clock, Cubes & Dices, Selection Decision Table (Decision Making), Logical Sequence Of Words, Venn Diagram, Analogy
  19. Paper-1 General Knowledge & Current Affairs: State Wise GK, History, Geography, GK- Current Affairs, Static General Knowledge
  20. Full Length Mock Test:
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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