Traversal Forward and Backward in DSA Python - Time & Space Complexity

Time Complexity: Traversal Forward and Backward

O(n)

Understanding Time Complexity

When we move through a list or collection from start to end and then back again, we want to know how long it takes. This helps us understand how the work grows as the list gets bigger.

We ask: How does the time to go forward and backward change with the size of the list?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def traverse_forward_backward(arr):
    # Traverse forward
    for i in range(len(arr)):
        print(arr[i])
    # Traverse backward
    for i in range(len(arr) - 1, -1, -1):
        print(arr[i])

This code goes through the list from the first item to the last, then from the last item back to the first.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two separate loops each visiting every item once.
• How many times: Each loop runs exactly n times, where n is the list size.

How Execution Grows With Input

As the list grows, the total steps grow roughly twice as much because we go forward and backward.

Input Size (n)	Approx. Operations
10	20 (10 forward + 10 backward)
100	200 (100 forward + 100 backward)
1000	2000 (1000 forward + 1000 backward)

Pattern observation: The total steps grow linearly with the list size, doubling because of two passes.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the forward and backward traversal grows in a straight line with the size of the list.

Common Mistake

[X] Wrong: "Because there are two loops, the time complexity is O(n²)."

[OK] Correct: The loops run one after another, not inside each other, so their times add up, not multiply.

Interview Connect

Understanding how multiple passes over data add up helps you explain your code clearly and shows you know how to judge efficiency in real tasks.

Self-Check

"What if we combined the forward and backward traversal into a single loop that prints two items per step? How would the time complexity change?"

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