Practice question

Duration: 2 min

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Inside: a video lesson and guided study material.

Module outline

  1. C: C Programming by Yash Sir
  2. C++: Programming in C++
  3. Java: Java
  4. Python: Python for Placements by Yash Sir
  5. OOPs Concept: Object Oriented Programming (OOPS)
  6. Pseudo Code: Pseudo Codes
  7. Data Structure: Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
  8. Algorithm: Algorithm Analysis, Sorting Algorithms, Greedy Algorithms, Dynamic Programming, Minimum Spanning Trees, Shortest Path Algos, Time Complexity Analysis, Advanced Algorithms
  9. DSA Implementation: DSA using Python
  10. Competitive Coding: Coding Questions by Yash Sir, Coding Questions Bank
  11. Mega Test:
  12. Grand Test:
  13. Live Class Notes: Placement Live Classes
AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This short practice-question video demonstrates how to compute the memory address of a specific element in a one-dimensional array. The on-screen question states that the base address of the first element is 250, each element occupies 3 bytes, and asks for the address of the fifth element in array a[10]. The instructor first identifies the given values, writing B = 250 and w = 3B in red handwriting. A diagram of the array is drawn as a rectangle with indices 0 through 9 written above it, emphasizing that the fifth element corresponds to index 4 because indexing starts at zero. The solution is then computed manually as 250 + (4 × 3) = 262. A formal array addressing formula, a[k] = B + w * (k - Lb), is also shown and applied with k = 4, B = 250, w = 3, and Lb = 0 to verify the same result. The final answer, 262 bytes, is underlined in red.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The slide displays the orange question: "Let the base address of the first element of the array is 250 and each element of the array occupies 3 bytes in the memory, then address of the fifth element of a one-dimensional array a[10]?" Red handwriting appears showing "B = 250" and "w = 3B", with a red underline under "first element". An empty red rectangle is drawn, then index numbers "0 1 2 3 4 5 6 7 8 9" are written above it to show the fifth element is at index 4. The working expands to "=250 + 4*3B", then "250 + 12", and finally "= 262B" underlined in red. A formal formula, a[k] = B + w * (k - Lb), is shown and applied with k=4, B=250, w=3 to yield 262.

The central concept is array element address calculation using the formula a[k] = B + w * (k - Lb), where B is the base address, w is the word size in bytes, k is the target index, and Lb is the lower bound (starting index). The key teaching point is that "fifth element" means index 4 in a zero-based array, not index 5. The instructor demonstrates both a manual offset calculation (base + index × word size) and the formal formula, arriving at 262 bytes. This reinforces that address computation depends on the index offset from the lower bound multiplied by element size, added to the base address.

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