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DSA Typescriptprogramming~5 mins

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

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Time Complexity: Traversal Forward and Backward
O(n)
Understanding Time Complexity

When we move through a list or array from start to end and then back again, we want to know how long this takes as the list grows.

We ask: How does the time to go forward and backward change when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


function traverseForwardBackward(arr: number[]): void {
  for (let i = 0; i < arr.length; i++) {
    console.log(arr[i]);
  }
  for (let j = arr.length - 1; j >= 0; j--) {
    console.log(arr[j]);
  }
}
    

This code goes through the array from start to end, then from end back to start, printing each element.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Two separate loops each visiting every element once.
  • How many times: Each loop runs n times, where n is the array length.
How Execution Grows With Input

As the array gets bigger, the total steps grow roughly twice as fast because we go forward and backward.

Input Size (n)Approx. Operations
1020 (10 forward + 10 backward)
100200 (100 forward + 100 backward)
10002000 (1000 forward + 1000 backward)

Pattern observation: Operations grow linearly with input size, doubling because of two passes.

Final Time Complexity

Time Complexity: O(n)

This means the time grows directly in proportion to the size of the array, even though we do two passes.

Common Mistake

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

[OK] Correct: The loops run one after another, not nested. So the total steps add up, not multiply.

Interview Connect

Understanding how multiple passes over data add up helps you explain and optimize code clearly in interviews.

Self-Check

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

Practice

(1/5)
1. What does forward traversal in a linked list mean?
easy
A. Visiting nodes randomly without order
B. Visiting nodes from the tail to the head in order
C. Skipping every other node while traversing
D. Visiting nodes from the head to the tail in order

Solution

  1. Step 1: Understand linked list structure

    A linked list has nodes connected from head to tail in one direction.
  2. Step 2: Define forward traversal

    Forward traversal means visiting nodes starting at the head and moving to the tail in order.
  3. Final Answer:

    Visiting nodes from the head to the tail in order -> Option D
  4. Quick Check:

    Forward traversal = head to tail [OK]
Hint: Forward means start at head and go next nodes [OK]
Common Mistakes:
  • Confusing forward with backward traversal
  • Thinking traversal is random
  • Assuming skipping nodes is traversal
2. Which TypeScript syntax correctly traverses an array forward using a for loop?
easy
A. for(let i = arr.length; i >= 0; i--) { console.log(arr[i]); }
B. for(let i = 0; i < arr.length; i++) { console.log(arr[i]); }
C. for(let i = 1; i <= arr.length; i++) { console.log(arr[i]); }
D. for(let i = arr.length - 1; i > 0; i--) { console.log(arr[i]); }

Solution

  1. Step 1: Identify forward traversal loop

    Forward traversal starts at index 0 and goes up to length-1.
  2. Step 2: Check loop conditions

    for(let i = 0; i < arr.length; i++) { console.log(arr[i]); } loops from 0 to less than arr.length, printing each element correctly.
  3. Final Answer:

    for(let i = 0; i < arr.length; i++) { console.log(arr[i]); } -> Option B
  4. Quick Check:

    Forward loop index from 0 to length-1 [OK]
Hint: Start index 0, end before length for forward loop [OK]
Common Mistakes:
  • Starting index at 1 instead of 0
  • Using i <= arr.length causing out-of-bounds
  • Looping backward for forward traversal
3. What is the output of this TypeScript code?
const arr = [10, 20, 30];
for(let i = arr.length - 1; i >= 0; i--) {
  console.log(arr[i]);
}
medium
A. 30 20 10
B. 10 20 30
C. 30 20
D. Error: Index out of bounds

Solution

  1. Step 1: Understand loop start and end

    The loop starts at index 2 (arr.length - 1) and goes down to 0.
  2. Step 2: Trace output values

    It prints arr[2] = 30, arr[1] = 20, arr[0] = 10 in that order.
  3. Final Answer:

    30 20 10 -> Option A
  4. Quick Check:

    Backward traversal prints 30 20 10 [OK]
Hint: Backward loop starts at length-1 down to 0 [OK]
Common Mistakes:
  • Thinking output is forward order
  • Missing last element at index 0
  • Assuming error due to loop
4. Identify the error in this backward traversal code for a linked list:
let current = head;
while(current !== null) {
  console.log(current.value);
  current = current.prev;
}
medium
A. It should use current.next instead of current.prev
B. No error, code is correct
C. It should start from tail, not head
D. It should check current.next !== null in while condition

Solution

  1. Step 1: Understand backward traversal in linked list

    Backward traversal uses prev pointers starting from the tail node.
  2. Step 2: Check starting node

    The code starts at head, but to go backward, it must start at tail.
  3. Final Answer:

    It should start from tail, not head -> Option C
  4. Quick Check:

    Backward traversal starts at tail [OK]
Hint: Backward traversal must start at tail node [OK]
Common Mistakes:
  • Starting backward traversal at head
  • Using next instead of prev for backward
  • Wrong loop condition check
5. Given a doubly linked list, how would you print all values backward starting from the head node?
hard
A. Traverse forward to tail, then traverse backward from tail to head
B. Traverse backward directly from head using prev pointers
C. Use only next pointers starting from head to print backward
D. Reverse the list first, then traverse forward

Solution

  1. Step 1: Understand doubly linked list structure

    Each node has next and prev pointers; head has no prev, tail has no next.
  2. Step 2: Plan backward print from head

    To print backward from head, first move forward to tail using next pointers.
  3. Step 3: Traverse backward from tail

    From tail, use prev pointers to print nodes backward to head.
  4. Final Answer:

    Traverse forward to tail, then traverse backward from tail to head -> Option A
  5. Quick Check:

    Backward print from head requires forward then backward traversal [OK]
Hint: Go forward to tail, then backward to print [OK]
Common Mistakes:
  • Trying to go backward directly from head
  • Using next pointers to print backward
  • Reversing list unnecessarily
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard