Class 12 Computer Science Chapter 5: Sorting – NCERT Solutions and Key Answers

1. Consider 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.

   
   

   Answer:
   After each complete pass of Bubble sort, the largest unsorted element is placed at its correct position towards the end of the list.
   
   Original List: [7, 11, 3, 10, 17, 23, 1, 4, 21, 5]
   
   After Pass 1: [7, 3, 10, 11, 17, 1, 4, 21, 5, 23]
   After Pass 2: [3, 7, 10, 11, 1, 4, 17, 5, 21, 23]
   After Pass 3: [3, 7, 10, 1, 4, 11, 5, 17, 21, 23]
   
   Thus, after three passes: [3, 7, 10, 1, 4, 11, 5, 17, 21, 23]

   
   

2. Identify 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

   
   

   Answer:
   

   • Selection Sort:
     • Pass 1: 9 is placed at index 0 (swap 63 and 9) → [9,42,21,63]
     • Pass 2: 21 is placed at index 1 (swap 42 and 21) → [9,21,42,63]
     • Pass 3: 42 at index 2 (no swap needed)
     Total Swaps: 2
   • Bubble Sort:
     • Pass 1: [42,21,9,63] (swap 63-42, 42-21, 21-9): 3 swaps
     • Pass 2: [21,9,42,63] (swap 42-21, 21-9): 2 swaps
     • Pass 3: [9,21,42,63] (swap 21-9): 1 swap
     Total Swaps: 6
   Selection sort performs fewer swaps (2) than bubble sort (6). In terms of number of comparisons, both perform similar comparisons (for four elements, 6 comparisons), but selection sort is better regarding swaps.
   
   

3. Consider 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, which of the lists List1 or List2 will make the minimum number of comparisons? Justify using diagrammatic representation.

   
   

   Answer:
   Insertion Sort on List 1 (Already Sorted):
   

   • Pass 1: Compare 3 with 2 (1 comparison, no shift)
   • Pass 2: Compare 5 with 3 (1), with 2 (no shift)
   • Pass 3: Compare 7 with 5 (1), (no shift)
   • Pass 4: Compare 11 with 7 (1), (no shift)
     Total comparisons: 4
   List 1 requires the minimum comparisons as it is already sorted (best case for insertion sort).
   
   Insertion Sort on List 2 (Reversely Sorted):
   
   • Pass 1: Compare 7 with 11 (1, need to shift)
   • Pass 2: 5 with 7, 11 (2, need to shift both)
   • Pass 3: 3 with 5, 7, 11 (3, shift all three)
   • Pass 4: 2 with 3, 5, 7, 11 (4, shift all four)
   • Total comparisons: 10
   List 2 requires maximum comparisons and shifts (worst case).
   
   Conclusion: List 1 will make the minimum number of comparisons (best case for insertion sort), while List 2 (reverse order) will take the maximum. Diagrammatically, the forward-sorted list requires only one comparison for each element, while the reverse requires comparing against all sorted elements for each insert.
   
   

4. Write 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)

   
   

   Answer:
   

   def bubble_sort(lst):
       n = len(lst)
       for i in range(n):
           for j in range(0, n-i-1):
               if lst[j] > lst[j+1]:
                   lst[j], lst[j+1] = lst[j+1], lst[j]
   
   def find_median(lst):
       bubble_sort(lst)
       n = len(lst)
       if n % 2 == 1:
           return lst[n // 2]
       else:
           return (lst[(n // 2) - 1] + lst[n // 2]) / 2
   
   # Example usage:
   numbers = [7, 11, 3, 10, 17]
   result = find_median(numbers)
   print("Median is:", result)
   # For odd length, returns middle value. For even, average of two middle values.
       

   
   

5. All 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 ‘xth’ 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 90th percentile, that means she/he scored better than 90% of people who took the test.

   
   Steps:

   1. Order all the values from smallest to largest using Selection Sort (or any sorting method).
   2. 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 * 120 = 108
   3. Ensure that the index is a whole number using math.round().
   4. Display the value at that index — it is the x^th percentile.
   Note: Percentile is a measure of relative performance. For example, a score in the 90th percentile means better than 90% of people who took the test.
   
   

   Answer:
   

   import math
   
   def selection_sort(lst):
       n = len(lst)
       for i in range(n):
           min_idx = i
           for j in range(i+1, n):
               if lst[j] < lst[min_idx]:
                   min_idx = j
           lst[i], lst[min_idx] = lst[min_idx], lst[i]
   
   def percentile(lst, x):
       n = len(lst)
       selection_sort(lst)
       index = int(round((x / 100) * n))
       if index == 0:
           index = 1
       if index > n:
           index = n
       return lst[index-1]
   
   # Example usage:
   marks = [45, 67, 89, 34, 56, 78, 90]
   x = 90
   result = percentile(marks, x)
   print(f"{x}th percentile is:", result)
       

   
   

6. During 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.

   
   

   Answer:
   The sorting technique being used is Insertion Sort, as each new element (name) is inserted at its correct position to maintain the order.
   
   

   def insert_name(sorted_list, name):
       sorted_list.append(name)
       i = len(sorted_list) - 2
       while i >= 0 and sorted_list[i] > name:
           sorted_list[i + 1] = sorted_list[i]
           i -= 1
       sorted_list[i + 1] = name
   
   # Example usage:
   names = []
   insert_name(names, "Amit")
   insert_name(names, "Neha")
   insert_name(names, "Bharat")
   print(names)  # Output: ['Amit', 'Bharat', 'Neha']
       

   
   

Understand Sorting Techniques Easily with NCERT Solutions Class 12 Computer Science Chapter 5

Understanding bubble sort, selection sort, and insertion sort is key for Class 12 Computer Science. With NCERT Solutions for Chapter 5 Sorting (2025-26), you can learn these algorithms step-by-step and build a strong foundation for coding and logical reasoning in your final exams.

Practicing NCERT exercise-based questions for Sorting ensures you're familiar with the logic, code, and applications of sort algorithms. Focus on each pass and understand how comparing and swapping elements helps arrange data both in ascending and descending order.

Revise time complexity concepts and attempt all the programming activities and exercises for better scores. Remember, consistent practice of code-based and conceptual questions boosts your confidence and enhances your problem-solving skills in computer science!