Array Traversal Patterns in DSA Java - Execution Trace

Concept Flow - Array Traversal Patterns

Start at index 0

↓

Check: index < array.length?

No→End traversal

Yes↓

Access element at index

↓

Process element

↓

Increment index

↩Back to Check

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

Execution Sample

DSA Java

int[] arr = {10, 20, 30};
for (int i = 0; i < arr.length; i++) {
    System.out.print(arr[i] + " ");
}

Traverse the array from start to end and print each element.

Execution Table

Step	Operation	Index	Condition (i < length)	Element Accessed	Output	Next Index
1	Initialize i	0	0 < 3 = true	arr[0] = 10	10	1
2	Increment i	1	1 < 3 = true	arr[1] = 20	20	2
3	Increment i	2	2 < 3 = true	arr[2] = 30	30	3
4	Increment i	3	3 < 3 = false	N/A	End	N/A

💡 Index i reaches 3, condition i < 3 is false, traversal ends.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4
i	undefined	0	1	2	3
Output	""	"10 "	"10 20 "	"10 20 30 "	"10 20 30 "

Key Moments - 3 Insights

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

Because the condition i < arr.length becomes false at step 4 in the execution_table, preventing out-of-bounds access.

Why do we start indexing at 0 instead of 1?

Arrays in Java are zero-indexed, so the first element is at index 0 as shown in step 1 where i is 0 and arr[0] is accessed.

What happens if we try to access arr[i] when i equals arr.length?

It causes an error because arr.length is out of bounds; the loop condition prevents this by stopping before i reaches arr.length (see step 4).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of i at step 3?

A 2

B 3

C 1

D 0

💡 Hint

Check the 'Index' column in the execution_table at step 3.

At which step does the loop condition become false?

A Step 2

B Step 4

C Step 3

D Step 1

💡 Hint

Look at the 'Condition (i < length)' column where it shows false.

If the array length was 4, how many times would the loop run?

A 3 times

B 5 times

C 4 times

D 2 times

💡 Hint

Loop runs while i < array length, so it runs once per index from 0 to length-1.

Concept Snapshot

Array Traversal Pattern:
- Start at index 0
- Loop while index < array length
- Access and process element at current index
- Increment index by 1
- Stop when index reaches array length

This pattern safely visits every element in order.

Full Transcript

Array traversal means visiting each element in an array one by one. We start at index zero because arrays in Java start counting from zero. We check if the current index is less than the array length to avoid errors. If true, 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 array length, the loop stops because there are no more elements to visit. This pattern ensures we safely and completely go through the array.

Practice

1. What does the following Java code snippet do?
for (int i = 0; i < arr.length; i++) { System.out.print(arr[i] + " "); }

easy

A. Prints all elements of the array from end to start

B. Prints only the first element of the array

C. Prints all elements of the array from start to end

D. Prints only the last element of the array

Solution

1. Step 1: Understand the loop range

   The loop starts at index 0 and runs while i is less than the array length, so it covers all valid indices from start to end.

2. Step 2: Analyze the print statement inside the loop

   Each element at index i is printed followed by a space, so all elements are printed in order.

3. Final Answer:

   Prints all elements of the array from start to end -> Option C

4. Quick Check:

   Loop from 0 to length-1 prints all elements [OK]

Hint: Loop from 0 to length-1 prints all elements in order [OK]

Common Mistakes:

• Confusing loop direction and thinking it prints backwards
• Assuming it prints only one element
• Ignoring the loop condition

2. Which of the following is the correct syntax to traverse an array arr backwards in Java?

easy

A. for (int i = arr.length; i > 0; i--)

B. for (int i = 0; i < arr.length; i++)

C. for (int i = 1; i <= arr.length; i++)

D. for (int i = arr.length - 1; i >= 0; i--)

Solution

1. Step 1: Identify correct start and end indices for backward traversal

   Backward traversal starts at last index (arr.length - 1) and ends at 0 inclusive.

2. Step 2: Check loop condition and decrement

   The loop continues while i is greater than or equal to 0, decrementing i each time, correctly covering all indices backwards.

3. Final Answer:

   for (int i = arr.length - 1; i >= 0; i--) -> Option D

4. Quick Check:

   Start at last index, go down to 0 [OK]

Hint: Start at length-1, loop down to 0 with i-- [OK]

Common Mistakes:

• Starting loop at arr.length (out of bounds)
• Using i > 0 instead of i >= 0 (misses index 0)
• Starting loop at 1 instead of length-1

3. What is the output of the following code?

int[] arr = {2, 4, 6, 8};
for (int i = 1; i < arr.length; i += 2) {
    System.out.print(arr[i] + " ");
}

medium

A. 4 8

B. 2 6

C. 4 6 8

D. 2 4 6 8

Solution

1. Step 1: Identify loop indices

   The loop starts at i=1 and increments by 2, so i takes values 1 and 3 (since arr.length=4).

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

   arr[1] = 4 and arr[3] = 8, so output is "4 8 "

3. Final Answer:

   4 8 -> Option A

4. Quick Check:

   Indices 1 and 3 printed: 4 and 8 [OK]

Hint: Loop i=1; i<length; i+=2 prints odd indices [OK]

Common Mistakes:

• Assuming loop prints all elements
• Confusing index increments
• Printing elements at even indices instead

4. Find the error in this code snippet that tries to print array elements backwards:

int[] arr = {1, 3, 5, 7};
for (int i = arr.length; i >= 0; i--) {
    System.out.print(arr[i] + " ");
}

medium

A. Loop starts at arr.length which is out of bounds

B. Loop condition should be i <= arr.length

C. Array index should start at 0

D. No error, code works fine

Solution

1. Step 1: Check starting index of loop

   arr.length is 4, but valid indices are 0 to 3. Starting at 4 causes ArrayIndexOutOfBoundsException.

2. Step 2: Correct starting index for backward traversal

   Loop should start at arr.length - 1 (3) to access last element safely.

3. Final Answer:

   Loop starts at arr.length which is out of bounds -> Option A

4. Quick Check:

   Array indices go from 0 to length-1 [OK]

Hint: Array indices max is length-1, not length [OK]

Common Mistakes:

• Starting loop at arr.length instead of arr.length-1
• Ignoring index out of bounds error
• Confusing loop condition signs

5. Given an integer array arr, which code snippet correctly prints all elements at even indices in reverse order?

hard

A. for (int i = arr.length - 1; i >= 0; i -= 2) { System.out.print(arr[i] + " "); }

B. for (int i = (arr.length % 2 == 0 ? arr.length - 2 : arr.length - 1); i >= 0; i -= 2) { System.out.print(arr[i] + " "); }

C. for (int i = 0; i < arr.length; i += 2) { System.out.print(arr[i] + " "); }

D. for (int i = 1; i < arr.length; i += 2) { System.out.print(arr[i] + " "); }

Solution

1. Step 1: Determine the highest even index to start from

   If array length is even, highest even index is length-2; if odd, it's length-1.

2. Step 2: Loop backwards by 2 steps to cover even indices

   Start from the highest even index and decrement by 2 to print all even indices in reverse order.

3. Final Answer:

   for (int i = (arr.length % 2 == 0 ? arr.length - 2 : arr.length - 1); i >= 0; i -= 2) { System.out.print(arr[i] + " "); } -> Option B

4. Quick Check:

   Start at highest even index, step down by 2 [OK]

Hint: Start at highest even index, decrement by 2 [OK]

Common Mistakes:

• Starting at length-1 without checking even index
• Using i -= 1 instead of i -= 2
• Printing odd indices instead of even