ITPEC FE Subject B April 2025 Question 11

Counting sort uses a frequency array where each index represents a value and each element stores its occurrence count.

The procedure implements counting sort:

• •s[0..10] is a frequency array initialized to all zeros
• •First loop: for each element arr[i], increment s[arr[i]] — this counts how many times each value (0–10) appears
• •Second loop: iterate i from 0 to 10; if s[i] > 0, output i exactly s[i] times

Blank A (description): asks for the contents of array s after the counting phase:
- s = {2, 1, 1, 3, 0, 1, 0, 0, 1, 2, 0} — each position holds the frequency of that value in arr

Blank B (program): the value printed is i (the index of the frequency array = the actual number being sorted)

Why not others:
- a/b/c) show the sorted output {0,0,1,2,3,3,3,5,8,9,9} — that is the printed result, not the contents of array s
- d/e/f) show unique sorted values {0,1,2,3,5,8,9} — neither s contents nor full sorted output
- h) correct s but B = j — j is just the loop counter for repetition, not the value to print
- i) correct s but B = s[j] — would print the frequency count, not the value itself

Key rule: In counting sort, the frequency array index is the value; you output the index i repeatedly s[i] times. The array s itself holds counts, not sorted values.

AI-generated — may contain errors