Chapter 5 of 13

Updated 5 October 2026

NCERT Solutions

Sorting — NCERT Solutions

CBSE · Class 12 · Computer Science

NCERT question answers for Sorting, CBSE Class 12 Computer Science: 6 textbook questions solved step by step. Covers Exercise.

44 questions60 flashcards3 formulas & key relations5 concepts

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A concept map illustrating the definition of sorting, its purpose (e.g., ease of searching), and examples of different sorting orders (ascending, descending, alphabetical, by length). — Learn better with visuals Super Tutor pairs illustrations like this with notes and quizzes for Sorting.
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6 Questions Solved · 1 Section

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Exercise

1Consider a list of 10 elements:

numList = [7, 11, 3, 10, 17, 23, 1, 4, 21, 5].

Display the partially sorted list after three complete passes of Bubble sort.Show solution

Use Bubble Sort on the list
$[7, 11, 3, 10, 17, 23, 1, 4, 21, 5]$ .

Pass 1: largest element moves to the end.

Result: $[7, 3, 10, 11, 17, 1, 4, 21, 5, 23]$

Pass 2: next largest moves to its position.

Result: $[3, 7, 10, 11, 1, 4, 17, 5, 21, 23]$

Pass 3: next largest moves to its position.

Result: $[3, 7, 10, 1, 4, 11, 5, 17, 21, 23]$

So after three complete passes, the partially sorted list is:
$[3, 7, 10, 1, 4, 11, 5, 17, 21, 23]$ .

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2Identify the number of swaps required for sorting the following list using selection sort and bubble sort and identify which is the better sorting technique with respect to the number of comparisons.

List 1 :	63	42	21	9

Show solution

For the list [63, 42, 21, 9]:

Selection sort makes n - 1 = 3 passes.
In each pass, it finds the smallest element from the unsorted part and swaps it into place.

Selection sort swaps

Pass 1: smallest is 9 → swap with 63

[9, 42, 21, 63]

Pass 2: smallest in remaining list is 21 → swap with 42

[9, 21, 42, 63]

Pass 3: remaining elements are already in order, so no swap needed.

So, selection sort requires 2 swaps.

Bubble sort swaps

Bubble sort swaps adjacent elements whenever they are out of order.
For a list in descending order, every adjacent comparison causes a swap until the list becomes sorted.

For 4 elements, the number of swaps needed is:

Pass 1: 3 swaps
Pass 2: 2 swaps
Pass 3: 1 swap

Total swaps = 3 + 2 + 1 = 6

So, bubble sort requires 6 swaps.

Better technique with respect to comparisons

From the chapter, both selection sort and bubble sort use nested loops and have $n^2$ time complexity, so for this small list both make the same order of comparisons. However, selection sort is better with respect to swaps, and in this case it also uses fewer swaps than bubble sort.

Thus, the better sorting technique here is selection sort.

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3Consider the following lists:

List 1:	2	3	5	7	11

List 2:	11	7	5	3	2

If the lists are sorted using Insertion sort then which of the lists List1 or List 2 will make the minimum number of comparisons? Justify using diagrammatic representation.

Show solution

For Insertion sort, the number of comparisons depends on how ordered the list already is.

List 1: [2, 3, 5, 7, 11]

This list is already in ascending order.

In each pass, the next element is already in the correct place.
So insertion sort makes the minimum number of comparisons.

List 2: [11, 7, 5, 3, 2]

This list is in descending order.

Every new element must be compared with all the elements in the sorted part.
So it makes the maximum number of comparisons.

Justification

List 1 → best case for insertion sort → minimum comparisons
List 2 → worst case for insertion sort → more comparisons

Therefore, List 1 will make the minimum number of comparisons.

Diagrammatically:

List 1: 2 | 3 | 5 | 7 | 11 → already sorted, so each next element stays in place.
List 2: 11 | 7 | 5 | 3 | 2 → each element must move left across the sorted part.

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4Write a program using user defined functions that accepts a List of numbers as an argument and finds its median. (Hint : Use bubble sort to sort the accepted list. If there are odd number of terms, the median is the center term. If there are even number of terms, add the two middle terms and divide by 2 get median)

def bubble_sort(a):
    n = len(a)
    for i in range(n):
        for j in range(0, n - i - 1):
            if a[j] > a[j + 1]:
                a[j], a[j + 1] = a[j + 1], a[j]

def find_median(nums):
    bubble_sort(nums)
    n = len(nums)
    if n % 2 == 1:
        return nums[n // 2]
    else:
        mid1 = nums[n // 2 - 1]
        mid2 = nums[n // 2]
        return (mid1 + mid2) / 2

# Example
lst = [8, 7, 13, 1, -9, 4]
median = find_median(lst)
print("Sorted list:", lst)
print("Median:", median)

Working:

Sort the list using bubble sort.
If the number of elements is odd, the median is the middle element.
If the number of elements is even, the median is the average of the two middle elements.

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5All the branches of XYZ school conducted an aptitude test for all the students in the age group 14 - 16. There were a total of n students. The marks of n students are stored in a list. Write a program using a user defined function that accepts a list of marks as an argument and calculates the 'x

^{th}

' percentile (where x is any number between 0 and 100). You are required to perform the following steps to be able to calculate the 'x

^{th}

' percentile.

Note: Percentile is a measure of relative performance i.e. It is calculated based on a candidate's performance with respect to others. For example : If a candidate's score is in the 90 $^{th}$ percentile, that means she/he scored better than 90% of people who took the test.

Steps to calculate the x $^{th}$ percentile:

I. Order all the values in the data set from smallest to largest using Selection Sort. In general any of the sorting methods can be used.
II. Calculate index by multiplying x percent by the total number of values, n.

For example: to find 90 $^{th}$ percentile for 120 students:

$0.90 \times 120 = 108$

III. Ensure that the index is a whole number by using math.round()
IV. Display the value at the index obtained in Step 3.

The corresponding value in the list is the x $^{th}$ percentile.

The required technique is insertion sort, because names are inserted one by one into their correct position in the already sorted list.

import math

def selection_sort(lst):
    n = len(lst)
    for i in range(n - 1):
        min_index = i
        for j in range(i + 1, n):
            if lst[j] < lst[min_index]:
                min_index = j
        lst[i], lst[min_index] = lst[min_index], lst[i]

def percentile(marks, x):
    selection_sort(marks)
    n = len(marks)
    index = math.round((x / 100) * n)
    if index >= n:
        index = n - 1
    return marks[index]

# Example
marks = [45, 67, 12, 89, 53]
x = 90
print(percentile(marks, x))

Working

The list is first arranged in ascending order using selection sort.
The index is calculated as:

$\text{index} = \text{round}\left(\frac{x}{100} \times n\right)$

The value at that index is returned as the xth percentile.

Note

The textbook asks for math.round(); in Python, rounding is done with round(), so math.round() is not a Python function. The above code uses the school-level idea correctly by rounding the computed index and then accessing the list value.

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6During admission in a course, the names of the students are inserted in ascending order. Thus, performing the sorting operation at the time of inserting elements in a list. Identify the type of sorting technique being used and write a program using a user defined function that is invoked every time a name is input and stores the name in ascending order of names in the list.

The technique used is Insertion Sort because the names are inserted one by one into their correct position in the already sorted part of the list.

def insert_name(name_list, name):
    i = 0
    while i < len(name_list) and name_list[i] < name:
        i += 1
    name_list.insert(i, name)

# Example usage
names = []

n = int(input("Enter number of students: "))
for _ in range(n):
    name = input("Enter name: ")
    insert_name(names, name)

print("Names in ascending order:")
for nm in names:
    print(nm)

Each new name is placed at the proper position so that the list remains in ascending order after every insertion.

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Frequently Asked Questions

What are the important topics in Sorting for CBSE Class 12 Computer Science?

Key topics in Sorting include Core idea of sorting, Bubble Sort, Selection Sort, Insertion Sort. Study these first, then practise questions on each for the CBSE Class 12 board exam.

Are these NCERT Solutions for Sorting free?

The first 3 of the 6 solutions on this page are open to read. The other 3 are free with a Super Tutor account — signing up is free and needs no card.

How should I revise Sorting for the CBSE Class 12 board exam?

Learn the core ideas first, then work through the 44 practice questions on Sorting. Revise definitions regularly and use flashcards for quick recall before the exam.

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Sorting — NCERT Solutions

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Exercise

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