Traversal Forward and Backward in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print all elements of the list forward.

DSA Python

for i in range(len(lst)):
    print(lst[[1]])

Drag options to blanks, or click blank then click option'

Alen(lst)

Bi

C0

Dlst[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using the list itself as an index like lst[lst[i]]

Using a fixed index like 0 instead of the loop variable

We use the loop variable 'i' to access each element in the list by index.

2fill in blank

medium

Complete the code to print all elements of the list backward.

DSA Python

for i in range(len(lst)-1, [1], -1):
    print(lst[i])

Drag options to blanks, or click blank then click option'

A1

Blen(lst)

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as stop index which excludes index 0

Using len(lst) which is out of range

To go backward through the list, we start from the last index and go down to -1 (exclusive) to include index 0.

3fill in blank

hard

Fix the error in the code to traverse the list forward and print elements.

DSA Python

i = 0
while i < [1]:
    print(lst[i])
    i += 1

Drag options to blanks, or click blank then click option'

Alen(lst)

Blst

Ci

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing 'i' with the list itself instead of its length

Using a fixed number instead of the list length

The loop should run while 'i' is less than the length of the list to cover all elements.

4fill in blank

hard

Fill both blanks to create a dictionary of word lengths for words longer than 3 characters.

DSA Python

{word: [1] for word in words if len(word) [2] 3}

Drag options to blanks, or click blank then click option'

Alen(word)

B<

C>

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as the value instead of its length

Using '<' instead of '>' in the condition

We want the length of each word as the value and only include words longer than 3 characters.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase keys and values for positive values only.

DSA Python

result = {{ [1]: [2] for k, v in data.items() if v [3] 0 }}

Drag options to blanks, or click blank then click option'

Ak.upper()

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using original keys instead of uppercase

Using '<' instead of '>' in the condition

Using keys as values

Keys are uppercase versions of k, values are v, and we include only positive values (v > 0).

Practice

1. What does traversal forward mean when moving through a list?

easy

A. Visiting elements randomly

B. Visiting elements from the last to the first

C. Visiting elements from the first to the last

D. Skipping every other element

Solution

1. Step 1: Understand traversal direction

   Traversal forward means starting at the beginning of the list and moving step-by-step to the end.

2. Step 2: Compare options

   Only Visiting elements from the first to the last describes visiting elements from first to last, which matches forward traversal.

3. Final Answer:

   Visiting elements from the first to the last -> Option C

4. Quick Check:

   Forward traversal = first to last [OK]

Hint: Forward means start to end in order [OK]

Common Mistakes:

• Confusing forward with backward traversal
• Thinking traversal means skipping elements
• Assuming traversal is random

2. Which of the following Python loops correctly traverses a list items backward?

easy

A. for i in range(len(items)): print(items[i])

B. for i in range(len(items)-1, -1, -1): print(items[i])

C. for i in items: print(i)

D. for i in range(0, len(items), 2): print(items[i])

Solution

1. Step 1: Identify backward traversal syntax

   Backward traversal uses a loop that starts from the last index and moves to the first, decreasing the index.

2. Step 2: Check each option

   for i in range(len(items)-1, -1, -1): print(items[i]) uses range(len(items)-1, -1, -1), which counts backward correctly.

3. Final Answer:

   for i in range(len(items)-1, -1, -1): print(items[i]) -> Option B

4. Quick Check:

   Backward loop uses decreasing range [OK]

Hint: Use range with step -1 for backward loops [OK]

Common Mistakes:

• Using range(len(items)) for backward traversal
• Forgetting to set step as -1
• Using direct iteration which is forward only

3. What is the output of this code?

items = [10, 20, 30]
for i in range(len(items)-1, -1, -1):
    print(items[i], end=' ')

medium

A. 30 20 10

B. 10 20 30

C. 0 1 2

D. SyntaxError

Solution

1. Step 1: Understand the loop range

   The loop runs from index 2 down to 0 (inclusive), so it visits indices 2, 1, 0 in order.

2. Step 2: Print elements in backward order

   items[2] = 30, items[1] = 20, items[0] = 10, printed with spaces.

3. Final Answer:

   30 20 10 -> Option A

4. Quick Check:

   Backward print = 30 20 10 [OK]

Hint: Backward range prints list in reverse order [OK]

Common Mistakes:

• Confusing index order and printing forward
• Misreading range parameters
• Expecting index numbers instead of list values

4. Find the error in this backward traversal code:

items = [1, 2, 3, 4]
for i in range(len(items)):
    print(items[len(items)-i])

medium

A. IndexError due to out-of-range index

B. SyntaxError because of missing colon

C. Prints elements forward instead of backward

D. No error, prints backward correctly

Solution

1. Step 1: Analyze index calculation

   Index used is len(items)-i. For i=0, index is 4, which is out of range (max index 3).

2. Step 2: Identify error type

   Accessing index 4 causes IndexError because list indices go from 0 to 3.

3. Final Answer:

   IndexError due to out-of-range index -> Option A

4. Quick Check:

   Index out of range = IndexError [OK]

Hint: Check if index goes beyond list length [OK]

Common Mistakes:

• Using len(items)-i instead of len(items)-1-i
• Ignoring zero-based indexing
• Assuming no error if code runs partially

5. Given a doubly linked list with nodes head and tail, which approach correctly prints all elements forward then backward?

hard

A. Traverse from tail to head, then from head to tail

B. Traverse only from head to tail twice

C. Traverse randomly from any node

D. Traverse from head to tail printing, then from tail to head printing

Solution

1. Step 1: Understand doubly linked list traversal

   Doubly linked lists allow moving forward from head to tail and backward from tail to head using next and prev pointers.

2. Step 2: Identify correct traversal order

   To print forward then backward, first traverse from head to tail, then from tail to head.

3. Final Answer:

   Traverse from head to tail printing, then from tail to head printing -> Option D

4. Quick Check:

   Forward then backward traversal = head to tail, tail to head [OK]

Hint: Use head-to-tail then tail-to-head for full traversal [OK]

Common Mistakes:

• Trying to traverse backward without tail pointer
• Traversing forward twice instead of backward
• Assuming singly linked list can traverse backward