ITPEC FE Subject B April 2025 Question 11
From the answer group below, select the correct combination of answers to be inserted into A in the description and B in the program. Here, the array index starts at 0.
The procedure sort receives an integer array arr and prints all the integers in arr in ascending order, separated by commas. The number of elements in arr is ≥ 1. The values of all array elements are in the range of 0-10.
If arr is {9, 3, 2, 0, 9, 3, 0, 1, 5, 3, 8}, at the end of the procedure, it outputs "0, 0, 1, 2, 3, 3, 3, 5, 8, 9, 9, " and the values of the elements of array s will be {A}
Answer group
Explanation
A: 2, 1, 1, 3, 0, 1, 0, 0, 1, 2, 0 and B: i — s is a counting array: s[v] holds how many times the value v occurs in arr, and the second loop prints each index as many times as it was counted.
The first loop adds 1 to s[arr[i]] for every element, so each index of s counts one of the values 0–10. For {9, 3, 2, 0, 9, 3, 0, 1, 5, 3, 8}:
The second loop walks i from 0 to 10, and for each non-zero count the inner loop runs s[i] times. What must be printed is the value itself, which is the index i: 0 twice, 1, 2, 3 three times, 5, 8, 9 twice — exactly the expected output.
Answer (g)
Why not others:
- •(a), (b) and (c) — blank A lists the sorted output;
sholds counts, and the program never stores sorted values anywhere, it only prints them - •(d), (e) and (f) — blank A lists the distinct values present; those are the indices where
sis non-zero, not whatscontains - •(h) — prints
j, the repeat counter, so the output is 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 1 instead of the values - •(i) — prints
s[j], a count read at the repeat-counter position (2, 1, 2, 2, …), again a frequency rather than a value
Key rule: In counting sort the index of the count array is the value and the element is its frequency, so output each index as many times as its count.
Sorting · AI-written, may contain errors
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