Array Traversal Patterns in DSA Python - Execution Trace

Concept Flow - Array Traversal Patterns

Start at index 0

↓

Check: index < length?

No→End traversal

Yes↓

Access element at index

↓

Process element

↓

Update index (e.g., index+1)

↩Back to Check

Start from the first element, check if index is within array bounds, access and process the element, then move to the next index until all elements are visited.

Execution Sample

DSA Python

arr = [10, 20, 30]
index = 0
while index < len(arr):
    print(arr[index])
    index += 1

Traverse the array from start to end, printing each element.

Execution Table

Step	Operation	Index	Condition (index < len(arr))	Element Accessed	Output	Next Index
1	Start traversal	0	0 < 3 = True	arr[0] = 10	10	1
2	Continue traversal	1	1 < 3 = True	arr[1] = 20	20	2
3	Continue traversal	2	2 < 3 = True	arr[2] = 30	30	3
4	Check condition	3	3 < 3 = False	None	None	End

💡 Traversal ends when index reaches array length (3), condition becomes false.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
index	0	1	2	3	3
Output	None	10	20	30	30

Key Moments - 3 Insights

Why does the loop stop when index equals the array length?

Because arrays are zero-indexed, valid indices are 0 to length-1. When index equals length, it is outside the array bounds, so the condition fails (see execution_table row 4).

What happens if we forget to update the index inside the loop?

The index stays the same, causing an infinite loop because the condition never becomes false. The execution_table shows index increasing each step to avoid this.

Why do we access arr[index] only after checking the condition?

To avoid accessing invalid memory or causing errors. The condition ensures index is valid before accessing the element (see execution_table columns 'Condition' and 'Element Accessed').

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'index' after Step 2?

A 2

B 1

C 3

D 0

💡 Hint

Check the 'Next Index' column in execution_table row 2.

At which step does the condition 'index < len(arr)' become false?

A Step 3

B Step 2

C Step 4

D Step 1

💡 Hint

Look at the 'Condition' column in execution_table row 4.

If the array had 5 elements instead of 3, how many steps would the traversal take?

A 5 steps

B 6 steps

C 4 steps

D 3 steps

💡 Hint

Traversal steps = number of elements + 1 for the final condition check (see execution_table rows).

Concept Snapshot

Array Traversal Patterns:
- Start at index 0
- Loop while index < array length
- Access and process element at current index
- Increment index to move forward
- Stop when index reaches array length

Full Transcript

Array traversal means visiting each element in order from start to end. We begin at index zero and check if the index is less than the array's length. If yes, we access the element at that index and do something with it, like printing. Then we increase the index by one and repeat. When the index equals the length, we stop because there are no more elements. This pattern ensures every element is visited exactly once safely.

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