Lists — NCERT Solutions

Given: stRecord = ['Raman', 'A-36', [56, 98, 99, 72, 69], 78.8]

Structure of the list:

• Index 0 → Name: 'Raman'
• Index 1 → Roll No.: 'A-36'
• Index 2 → Marks list: [56, 98, 99, 72, 69]
• Index 3 → Percentage: 78.8

a) Percentage of the student:

print(stRecord[3])

Output: 78.8

b) Marks in the fifth subject:

• The marks list is at index 2. The fifth subject is at index 4 of the inner list.

print(stRecord[2][4])

Output: 69

c) Maximum marks of the student:

• Use max() on the marks list at index 2.

print(max(stRecord[2]))

Output: 99

d) Roll no. of the student:

print(stRecord[1])

Output: A-36

e) Change the name from 'Raman' to 'Raghav':

• Lists are mutable, so we can directly assign a new value at index 0.

stRecord[0] = 'Raghav'
print(stRecord)

Output: ['Raghav', 'A-36', [56, 98, 99, 72, 69], 78.8]

Concept: Use the built-in count() method of lists which returns the number of times a specified element appears in the list.

Program:

# Program to find the number of times an element occurs in a list

# Input: Read the list
n = int(input("Enter the number of elements: "))
myList = []
for i in range(n):
    element = int(input("Enter element: "))
    myList.append(element)

print("List:", myList)

# Input: Element to search
item = int(input("Enter the element to count: "))

# Count occurrences using count()
count = myList.count(item)

print("The element", item, "occurs", count, "time(s) in the list.")

Sample Run:

Enter the number of elements: 6
Enter element: 1
Enter element: 2
Enter element: 3
Enter element: 2
Enter element: 1
Enter element: 2
List: [1, 2, 3, 2, 1, 2]
Enter the element to count: 2
The element 2 occurs 3 time(s) in the list.

Concept: Traverse the original list and check each element. If it is greater than 0, add to the positive list; if less than 0, add to the negative list.

Program:

# Program to separate positive and negative numbers into two lists

n = int(input("Enter the number of elements: "))
originalList = []

for i in range(n):
    num = int(input("Enter element: "))
    originalList.append(num)

positiveList = []
negativeList = []

for num in originalList:
    if num > 0:
        positiveList.append(num)
    elif num < 0:
        negativeList.append(num)

print("Original List:  ", originalList)
print("Positive Numbers:", positiveList)
print("Negative Numbers:", negativeList)

Sample Run:

Enter the number of elements: 7
Enter element: 3
Enter element: -5
Enter element: 8
Enter element: -2
Enter element: 0
Enter element: 7
Enter element: -9
Original List:   [3, -5, 8, -2, 0, 7, -9]
Positive Numbers: [3, 8, 7]
Negative Numbers: [-5, -2, -9]

Concept: Traverse the list and keep track of the maximum element found so far. Return it at the end. (We avoid using the built-in max() to demonstrate the logic.)

Program:

# Function to return the largest element of a list

def findLargest(lst):
    """Returns the largest element in the list."""
    largest = lst[0]          # Assume first element is largest
    for i in range(1, len(lst)):
        if lst[i] > largest:
            largest = lst[i]  # Update if a larger element is found
    return largest

# Main program
n = int(input("Enter the number of elements: "))
myList = []
for i in range(n):
    num = int(input("Enter element: "))
    myList.append(num)

result = findLargest(myList)
print("The largest element is:", result)

Sample Run:

Enter the number of elements: 5
Enter element: 34
Enter element: 12
Enter element: 78
Enter element: 45
Enter element: 23
The largest element is: 78

Concept: Sort the list in descending order and return the element at index 1. Alternatively, find the largest, remove it, and find the largest again. We must handle duplicates (e.g., if the largest appears multiple times).

Program:

# Function to return the second largest number from a list

def secondLargest(lst):
    """Returns the second largest unique element in the list."""
    # Create a sorted list of unique elements in descending order
    uniqueSorted = sorted(set(lst), reverse=True)
    if len(uniqueSorted) < 2:
        return None  # No second largest exists
    return uniqueSorted[1]

# Main program
n = int(input("Enter the number of elements: "))
myList = []
for i in range(n):
    num = int(input("Enter element: "))
    myList.append(num)

result = secondLargest(myList)
if result is not None:
    print("The second largest element is:", result)
else:
    print("Second largest does not exist.")

Sample Run:

Enter the number of elements: 5
Enter element: 10
Enter element: 45
Enter element: 23
Enter element: 45
Enter element: 8
The second largest element is: 23

Concept:

• Sort the list.
• If the number of elements $n$ is odd, the median is the middle element at index $n // 2$.
• If $n$ is even, the median is the average of the two middle elements at indices $n//2 - 1$ and $n//2$.

Program:

# Program to find the median of a list of n integers

def findMedian(lst):
    """Returns the median of the list."""
    lst.sort()          # Sort the list in ascending order
    n = len(lst)
    if n % 2 != 0:      # Odd number of elements
        median = lst[n // 2]
    else:               # Even number of elements
        mid1 = lst[n // 2 - 1]
        mid2 = lst[n // 2]
        median = (mid1 + mid2) / 2
    return median

# Main program
n = int(input("Enter the number of elements: "))
myList = []
for i in range(n):
    num = int(input("Enter element: "))
    myList.append(num)

result = findMedian(myList)
print("Sorted List:", myList)
print("Median:", result)

Sample Run 1 (Odd n):

Enter the number of elements: 5
Enter element: 7
Enter element: 2
Enter element: 9
Enter element: 4
Enter element: 1
Sorted List: [1, 2, 4, 7, 9]
Median: 4

Sample Run 2 (Even n):

Enter the number of elements: 4
Enter element: 3
Enter element: 8
Enter element: 1
Enter element: 6
Sorted List: [1, 3, 6, 8]
Median: 4.5

Concept: Traverse the list and build a new list that contains only the first occurrence of each element. We check if the element is already present before adding it.

Program:

# Program to remove duplicate elements from a list

n = int(input("Enter the number of elements: "))
myList = []
for i in range(n):
    num = int(input("Enter element: "))
    myList.append(num)

print("Original List:", myList)

# Remove duplicates
newList = []
for item in myList:
    if item not in newList:
        newList.append(item)

# Update the original list
myList = newList

print("List after removing duplicates:", myList)

Sample Run:

Enter the number of elements: 8
Enter element: 1
Enter element: 2
Enter element: 3
Enter element: 2
Enter element: 4
Enter element: 1
Enter element: 5
Enter element: 3
Original List: [1, 2, 3, 2, 4, 1, 5, 3]
List after removing duplicates: [1, 2, 3, 4, 5]

Concept: Use the insert(index, element) method of lists inside a user-defined function. The position entered by the user is 1-based (natural), so we convert it to 0-based index.

Program:

# Program to insert an element at a desired position in a list

def insertElement(lst, pos, elem):
    """Inserts elem at position pos (1-based) in lst."""
    if pos < 1 or pos > len(lst) + 1:
        print("Invalid position!")
    else:
        lst.insert(pos - 1, elem)  # Convert to 0-based index
    return lst

# Main program
n = int(input("Enter the number of elements: "))
myList = []
for i in range(n):
    num = int(input("Enter element: "))
    myList.append(num)

print("Original List:", myList)

elem = int(input("Enter the element to insert: "))
pos  = int(input("Enter the position (1-based) to insert at: "))

myList = insertElement(myList, pos, elem)
print("List after insertion:", myList)

Sample Run:

Enter the number of elements: 4
Enter element: 10
Enter element: 20
Enter element: 40
Enter element: 50
Original List: [10, 20, 40, 50]
Enter the element to insert: 30
Enter the position (1-based) to insert at: 3
List after insertion: [10, 20, 30, 40, 50]

Concept: Use pop(index) to delete the element at a given index. The user provides a 1-based position which is converted to a 0-based index.

Program:

# Program to delete an element at a given position from a list

def deleteByPosition(lst, pos):
    """Deletes element at position pos (1-based) from lst."""
    if pos < 1 or pos > len(lst):
        print("Invalid position!")
    else:
        removed = lst.pop(pos - 1)  # Convert to 0-based index
        print("Deleted element:", removed)
    return lst

# Main program
n = int(input("Enter the number of elements: "))
myList = []
for i in range(n):
    num = int(input("Enter element: "))
    myList.append(num)

print("Original List:", myList)

pos = int(input("Enter the position (1-based) of element to delete: "))
myList = deleteByPosition(myList, pos)
print("List after deletion:", myList)

Sample Run:

Enter the number of elements: 5
Enter element: 10
Enter element: 20
Enter element: 30
Enter element: 40
Enter element: 50
Original List: [10, 20, 30, 40, 50]
Enter the position (1-based) of element to delete: 3
Deleted element: 30
List after deletion: [10, 20, 40, 50]

Concept: Use remove(value) to delete the first occurrence of the specified value. If the value is not present, handle the error gracefully.

Program:

# Program to delete an element by value from a list

def deleteByValue(lst, val):
    """Deletes the first occurrence of val from lst."""
    if val in lst:
        lst.remove(val)
        print("Element", val, "deleted successfully.")
    else:
        print("Element", val, "not found in the list.")
    return lst

# Main program
n = int(input("Enter the number of elements: "))
myList = []
for i in range(n):
    num = int(input("Enter element: "))
    myList.append(num)

print("Original List:", myList)

val = int(input("Enter the value of element to delete: "))
myList = deleteByValue(myList, val)
print("List after deletion:", myList)

Sample Run:

Enter the number of elements: 5
Enter element: 10
Enter element: 20
Enter element: 30
Enter element: 20
Enter element: 50
Original List: [10, 20, 30, 20, 50]
Enter the value of element to delete: 20
Element 20 deleted successfully.
List after deletion: [10, 30, 20, 50]

Concept: To reverse a list in-place without creating a new list, we swap elements from both ends moving towards the centre. We use two pointers: left starting at index $0$ and right starting at index $n-1$, swapping and moving them towards each other.

Program:

# Program to reverse a list in-place without creating a new list

def reverseList(lst):
    """Reverses the list in-place using two-pointer swapping."""
    left = 0
    right = len(lst) - 1
    while left < right:
        # Swap elements at left and right indices
        lst[left], lst[right] = lst[right], lst[left]
        left  += 1
        right -= 1
    # No return needed as list is modified in-place

# Main program
n = int(input("Enter the number of elements: "))
myList = []
for i in range(n):
    num = int(input("Enter element: "))
    myList.append(num)

print("Original List:", myList)

reverseList(myList)   # List is modified in-place

print("Reversed List:", myList)

Sample Run:

Enter the number of elements: 6
Enter element: 10
Enter element: 20
Enter element: 30
Enter element: 40
Enter element: 50
Enter element: 60
Original List: [10, 20, 30, 40, 50, 60]
Reversed List: [60, 50, 40, 30, 20, 10]

Note: The function modifies the original list directly (in-place) because lists are passed by reference in Python. No new list is created.