Array Traversal Patterns in DSA Python

Mental Model

Go through each item in the list one by one to look at or use it.

Analogy: Like reading a book page by page from start to finish to understand the story.

Index: 0   1   2   3   4
Array: [5] -> [3] -> [8] -> [1] -> [7]

Dry Run Walkthrough

Input: array: [5, 3, 8, 1, 7], print each element

Goal: Visit every element in the array and print it in order

Step 1: Start at index 0, read value 5

Index: 0 ↑ 1   2   3   4
Array: [5] -> [3] -> [8] -> [1] -> [7]

Why: We begin at the first element to start traversal

Step 2: Move to index 1, read value 3

Index: 0   1 ↑ 2   3   4
Array: [5] -> [3] -> [8] -> [1] -> [7]

Why: Next element must be visited to continue traversal

Step 3: Move to index 2, read value 8

Index: 0   1   2 ↑ 3   4
Array: [5] -> [3] -> [8] -> [1] -> [7]

Why: Continue visiting elements in order

Step 4: Move to index 3, read value 1

Index: 0   1   2   3 ↑ 4
Array: [5] -> [3] -> [8] -> [1] -> [7]

Why: Keep moving forward to cover all elements

Step 5: Move to index 4, read value 7

Index: 0   1   2   3   4 ↑
Array: [5] -> [3] -> [8] -> [1] -> [7]

Why: Last element to visit before finishing traversal

Result:

5 -> 3 -> 8 -> 1 -> 7

Annotated Code

DSA Python

def traverse_array(arr):
    for i in range(len(arr)):
        print(arr[i], end=' -> ' if i < len(arr) - 1 else '\n')

# Driver code
array = [5, 3, 8, 1, 7]
traverse_array(array)

for i in range(len(arr)):

loop through each index from start to end to visit every element

print(arr[i], end=' -> ' if i < len(arr) - 1 else '\n')

print current element and add arrow except after last element

Output

5 -> 3 -> 8 -> 1 -> 7

Complexity Analysis

Time: O(n) because we visit each of the n elements once

Space: O(1) because we only use a few variables regardless of input size

vs Alternative: Compared to random access, traversal is simple and direct with no extra cost

Edge Cases

empty array []

No output is printed because there are no elements to visit

DSA Python

for i in range(len(arr)):

array with one element [10]

Prints the single element without arrow after it

DSA Python

print(arr[i], end=' -> ' if i < len(arr) - 1 else '\n')

When to Use This Pattern

When you need to process or check every item in a list in order, use array traversal to visit each element one by one.

Common Mistakes

Mistake: Stopping traversal early or skipping elements by wrong loop conditions
Fix: Use correct loop bounds from 0 to length-1 to cover all elements

Mistake: Printing arrow after the last element causing extra symbol
Fix: Add condition to print arrow only if not at last element

Summary

It visits each element in an array one by one in order.

Use it when you want to look at or use every item in a list.

The key is to move step-by-step from the first to the last element without skipping.

Practice

1. What does the following code do?
for i in range(len(arr)):
  print(arr[i])
Assuming arr is a list of numbers.

easy

A. Prints only the first element of the array.

B. Prints all elements of the array from first to last.

C. Prints the indices of the array elements.

D. Prints the array in reverse order.

Solution

1. Step 1: Understand the loop range

   The loop runs from 0 to len(arr)-1, covering all valid indices of the array.

2. Step 2: Access and print each element

   Inside the loop, arr[i] accesses each element in order, printing them one by one.

3. Final Answer:

   Prints all elements of the array from first to last. -> Option B

4. Quick Check:

   Loop over indices prints elements in order [OK]

Hint: Loop over range(len(arr)) accesses all elements by index [OK]

Common Mistakes:

• Thinking it prints indices instead of elements
• Assuming it prints in reverse order
• Confusing range(len(arr)) with just range(arr)

2. Which of the following is the correct syntax to traverse all elements in a list arr using a for-each style loop in Python?

easy

A. for i to len(arr): print(arr[i])

B. for i in range(arr): print(i)

C. for element in arr: print(element)

D. for element in range(len(arr)): print(element)

Solution

1. Step 1: Identify correct for-each syntax

   Python uses for variable in iterable to loop over elements directly.

2. Step 2: Check each option

   for element in arr: print(element) uses for element in arr, which correctly loops over elements. Others have syntax errors or misuse range.

3. Final Answer:

   for element in arr: print(element) -> Option C

4. Quick Check:

   For-each loop uses 'for element in arr' [OK]

Hint: Use 'for element in arr' to loop elements directly [OK]

Common Mistakes:

• Using range(arr) instead of range(len(arr))
• Writing 'for i to len(arr)' which is invalid syntax
• Confusing indices with elements in for-each loops

3. What is the output of the following code?

arr = [10, 20, 30, 40]
for i in range(1, len(arr)):
   print(arr[i])

medium

A. 20 30 40

B. 40

C. 10 20 30

D. 10 20 30 40

Solution

1. Step 1: Analyze the loop range

   The loop runs from 1 to len(arr)-1, which is 1 to 3 for this array.

2. Step 2: Print elements at indices 1, 2, 3

   Elements are arr[1]=20, arr[2]=30, arr[3]=40, printed in order.

3. Final Answer:

   20 30 40 -> Option A

4. Quick Check:

   Range(1,4) prints elements from index 1 to 3 [OK]

Hint: range(1, len(arr)) skips first element [OK]

Common Mistakes:

• Including the first element (index 0) mistakenly
• Confusing range end as inclusive
• Printing indices instead of elements

4. Find the error in this code snippet that tries to print all elements of arr:

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

medium

A. No error, prints all elements correctly.

B. SyntaxError due to incorrect for loop syntax.

C. Prints elements in reverse order.

D. IndexError because i+1 goes out of range on last iteration.

Solution

1. Step 1: Check loop indices and access

   Loop runs i from 0 to 3 (len(arr)=4). Accessing arr[i+1] means indices 1 to 4.

2. Step 2: Identify out of range access

   When i=3, arr[4] is accessed, which is outside the array bounds causing IndexError.

3. Final Answer:

   IndexError because i+1 goes out of range on last iteration. -> Option D

4. Quick Check:

   Accessing arr[i+1] with i at last index causes error [OK]

Hint: Check index +1 can exceed array length [OK]

Common Mistakes:

• Ignoring index out of range error
• Thinking syntax is wrong instead
• Assuming it prints reversed elements

5. Given a list arr = [3, 5, 7, 9, 11], write code to print elements at even indices only using array traversal patterns. Which code snippet does this correctly?

hard

A. for i in range(0, len(arr), 2): print(arr[i])

B. for i in range(1, len(arr), 2): print(arr[i])

C. for element in arr: if element % 2 == 0: print(element)

D. for i in range(len(arr)): if i % 2 == 1: print(arr[i])

Solution

1. Step 1: Understand even index traversal

   Even indices are 0, 2, 4 for this list. Step size 2 starting at 0 covers these.

2. Step 2: Check each option

   for i in range(0, len(arr), 2): print(arr[i]) uses range(0, len(arr), 2) which correctly iterates over even indices. Others either use odd indices or check element values incorrectly.

3. Final Answer:

   for i in range(0, len(arr), 2): print(arr[i]) -> Option A

4. Quick Check:

   Step 2 in range picks even indices [OK]

Hint: Use range(start=0, end, step=2) for even indices [OK]

Common Mistakes:

• Using odd indices instead of even
• Checking element value parity instead of index
• Forgetting step argument in range