Practice Question

Duration: 3 min

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Module outline

  1. About Course: About the Course
  2. Paper-2 | Programming & Data Structures: C Fundamentals, Control Flow, Functions, Arrays & Pointers, Storage Classes, Structures & Enums, DMA, Macros, Scoping & File Handling, Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
  3. Paper-2 | Databases / DBMS & SQL: Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Decomposition Properties & 4NF, File Organization & Indexing, Relational Algebra, SQL, Relational Calculus, Transaction Management, Concurrency Control
  4. Paper-2 | Algorithms: Algorithm Analysis, Time Complexity Analysis, Sorting Algorithms, Greedy Algorithms, Dynamic Programming, Minimum Spanning Trees, Shortest Path Algos
  5. Paper-2 | Computer Networks & Security: 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, Network Access Technologies, Network Security
  6. 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
  7. 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
  8. Paper-2 | Theory of Computation: Introduction to TOC, Deterministic FA (DFA), Non-Deterministic FA, Regular Expressions, Grammar, Regular Language Properties, Moore & Mealy Machines, Pushdown Automata & CFG, Turing Machines
  9. Paper-2 | Computer Organization & Architecture: Basics of COA, Cache Memory Organization, Input Output Organisation, Pipelining, Instr Formats & Modes, Control Unit Design, Programming the Basic Computer
  10. 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
  11. Paper-1 | General Awareness: Polity, Physics, Biology, Chemistry, Static General Knowledge, GK- Current Affairs, History, Geography, Indian Economy, Environment
  12. Paper-1 | Logical Reasoning: Series (Number and Letter Series) (Numerical Relations and Reasoning), Coding Decoding, Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Data Interpretation, Blood Relations, Syllogisms, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Selection Decision Table (Decision Making), Clock, Calendar, Cubes & Dices, Directions (Direction Test), Analogy, Seating Arrangements, Analytical Reasoning (Counting Figures Reasoning)
  13. Paper-1 | Quantitative Aptitude: Ratio and Proportion (Ratios), Mixture and Alligation, Permutation and Combination, Number System, HCF LCM, Speed Time and Distance, Average, Age Problems, Percentage, Profit and Loss, Time and Work, Pipe and Cistern, Simple Interest and Compound Interest, Powers and Exponents (Surds and Indices), Mensuration and Geometry, Probability, Statistics, Arithmetic, Algebra
  14. Paper-1 | English: Vocabulary, 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), Sentence Construction, Cloze Test (Cloze Passage)
  15. Coal India Previous Year Paper:
  16. Engineering Mathematics: Set Theory, Relations, Functions, Graph Theory, Group Theory, Propositional and Predicate Logic, Linear Algebra, Calculus
  17. Coal India Mock Test:
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This educational video demonstrates how to calculate the memory address of a specific element within a two-dimensional array using the Row Major Order storage scheme. The instructor presents a specific problem where the array A has non-standard indexing ranges: rows from 5 to 15 and columns from -8 to 8. The total dimensions are given as A[11][17]. The problem asks for the address of element A[8][5]. Essential parameters are provided: a Base Address of 800 and an element size of 4 memory cells. The lecture walks through the derivation of the address calculation formula and applies it to solve the numerical problem, highlighting the handling of negative lower bounds.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The session begins with the problem statement displayed on screen: Q A[5.....15, -8.....8] is A[11][17] and we are supposed to find A[8][5]. The instructor identifies the given values: Row Major Order, Base Address = 800, and element size w = 4. He writes the general formula for Row Major Order on the board: A[i, j] = B + w * [(i - li) (ul - ll + 1) + (j - lj)]. He explicitly defines the lower and upper limits for rows (li=5, ui=15) and columns (lj=-8, uj=8). He notes that the term (ul - ll + 1) represents the total number of columns, which is 17. He sets up the substitution for the target element A[8][5], identifying i=8 and j=5.

  2. 2:00 – 3:14 02:00-03:14

    The instructor performs the arithmetic substitution. He writes 800 + 4 * [(8 - 5) * 17 + (5 - (-8))]. He calculates the row difference (8 - 5) as 3. He calculates the column difference (5 - (-8)) as 13. He multiplies the row difference by the column count: 3 * 17 = 51. He adds the column difference to this product: 51 + 13 = 64. This sum represents the total number of elements skipped. He multiplies this by the element size: 4 * 64 = 256. Finally, he adds this offset to the base address: 800 + 256, resulting in the final address 1056.

The video provides a clear, step-by-step guide to memory address calculation for 2D arrays. It emphasizes the formula's structure, particularly how the row index is weighted by the total column count. The example with negative lower bounds serves as a practical test of understanding the (index - lower_limit) component of the formula. The final calculation confirms the address is 1056, reinforcing the method for handling non-zero-based indexing in programming and data structures.

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