Texas Placement Paper -: DATA STRUCTURES Questions And Answers

Texas Previous Years Solved Sample Placement Papers

1. In the circuit given below, the switch is opened at time t=0. Voltage across the capacitor at t=∞ is:
   

   A. 2V

   B. 3V

   C. 5V

   D. 7V Answer: Option D

2. Solution: At t=∞, the capacitor reaches steady-state voltage determined by the circuit's configuration.
3. What is the functionality represented by the following circuit?
   

   A. y = !(b + ac)

   B. y = !(a + bc)

   C. y = !(a(b + c))

   D. y = !(a + b + c) Answer: Option B

4. Solution: The circuit represents a NOT gate with input logic `a + bc`.
5. The value (0xdeadbeef) needs to be stored at address 0x400. Which of the below ways will the memory look like in a big-endian machine?
   

   A. be ef de ad

   B. ef be ad de

   C. fe eb da ed

   D. ed da eb fe Answer: Option A

6. Solution: In big-endian format, the most significant byte is stored at the lowest memory address.
7. In a given CPU-memory sub-system, all accesses to the memory take two cycles. Which of the following mechanisms guarantees correct data transfer?
   

   A. A read operation followed by a write operation in the next cycle.

   B. A write operation followed by a read operation in the next cycle.

   C. A NOP between every successive reads & writes Answer: Option C

   D. None of the above

8. Solution: Adding a NOP ensures no data conflicts occur between successive read and write operations.
9. An architecture saves 4 control registers automatically on function entry (and restores them on function return). Save of each register costs 1 cycle (so does restore). How many cycles are spent while running the following code with n=5?
   void fib(int n) { if ((n == 0) || (n == 1)) return 1; return(fib(n - 1) + fib(n - 2)); }
   

   A. 120

   B. 80

   C. 125

   D. 128 Answer: Option D

10. Solution: The function call involves repeated saves/restores of control registers. Total cycles depend on the recursion depth and number of function calls.
11. Swapping without using a temporary variable:
    

    (i) x = x + y; y = x - y; x = x - y;

    (ii) x = x ^ y; y = x ^ y; x = x ^ y;

12. Count the number of 1's in a word without using bit-by-bit operations:
    Solution: Use bitwise operations like `value &= value - 1` in a loop to count set bits efficiently.
13. Given two codes:
    Code 1: for (i = 0; i < 1000; i++) for (j = 0; j < 100; j++) x = y;
Code 2: for (i = 0; i < 100; i++) for (j = 0; j < 1000; j++) x = y;
    Which code executes faster?
    

    A. Code 1 and Code 2 are of the same speed

    B. Code 1

    C. Code 2 Answer: Option C

    D. None

14. Solution: Code 2 executes faster because it accesses memory in a cache-friendly manner.
15. Output of the program:
    main() { int a[10] = {1, 2, 3, ..., 10}, i, x = 10, temp; for (i = 0; i < x / 2; i++) { temp = a[i]; a[i] = a[x - i - 1]; a[x - i - 1] = temp; }}
    

    A. All contents of array a are reversed

    B. Only some portions are altered

    C. Remains the same

    D. None Answer: Option A

16. Solution: The array is reversed completely by swapping elements symmetrically.
17. In C, which parameter passing technique is used?
    

    A. Call by value Answer: Option A

    B. Call by reference

    C. Both

18. Solution: C only supports call by value natively. References are emulated using pointers.
19. Output of the program:
    main() { int i = 1; fork(); fork(); printf("\ni = %d\n", i + 1); }
    

    A. 2 printfs will occur

    B. 4 printfs will occur Answer: Option B

    C. 6 printfs will occur

    D. Undefined behavior

20. Solution: Each `fork()` creates a new process, doubling the number of processes. Initially, 1 process becomes 4, each printing the result.