Array Traversal Patterns in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested loop traversal on 2D array

What is the output of the following Java code that traverses a 2D array row-wise and prints elements?

DSA Java

int[][] arr = {{1, 2}, {3, 4}, {5, 6}};
for (int i = 0; i < arr.length; i++) {
    for (int j = 0; j < arr[i].length; j++) {
        System.out.print(arr[i][j] + " ");
    }
}

A 1 3 5 2 4 6

B 2 1 4 3 6 5

C 6 5 4 3 2 1

D 1 2 3 4 5 6

💡 Hint

Think about how the outer loop goes through rows and inner loop goes through columns.

The outer loop goes through each row, and the inner loop prints each element in that row from left to right, so the output is all elements in order: 1 2 3 4 5 6.

❓ Predict Output

intermediate

Output of reverse traversal on 1D array

What is the output of this Java code that traverses an array backwards and prints elements?

DSA Java

int[] arr = {10, 20, 30, 40};
for (int i = arr.length - 1; i >= 0; i--) {
    System.out.print(arr[i] + " ");
}

A 40 20 30 10

B 10 20 30 40

C 40 30 20 10

D 30 40 20 10

💡 Hint

The loop starts from the last index and moves to the first.

The loop prints elements starting from the last index to the first, so the output is 40 30 20 10.

🔧 Debug

advanced

Identify the error in diagonal traversal of square matrix

What error will this Java code produce when trying to print the main diagonal of a 3x3 matrix?

DSA Java

int[][] matrix = {{1,2,3},{4,5,6},{7,8,9}};
for (int i = 0; i < matrix.length; i++) {
    System.out.print(matrix[i][i] + " ");
}

A ArrayIndexOutOfBoundsException

B NullPointerException

C SyntaxError

D Prints: 1 5 9

💡 Hint

Check the loop condition and array indexing carefully.

The loop runs from i = 0 to i = matrix.length (which is 3), but valid indices are 0 to 2. Accessing matrix[3][3] causes ArrayIndexOutOfBoundsException.

🧠 Conceptual

advanced

Number of elements visited in zigzag traversal of 2D array

Consider a 3x3 2D array. If you traverse it in a zigzag pattern starting from top-left, moving right, then diagonally down-left, then right again, and so on until all elements are visited, how many elements will be visited?

A 8

B 9

C 6

D 12

💡 Hint

Think about how many elements are in the array and if any are skipped.

A 3x3 array has 9 elements. Zigzag traversal visits each element exactly once, so total visited elements are 9.

🚀 Application

expert

Output after alternating forward and backward traversal on array

Given the Java code below, what is the printed output?

DSA Java

int[] arr = {1, 2, 3, 4, 5};
for (int i = 0; i < arr.length; i++) {
    if (i % 2 == 0) {
        System.out.print(arr[i] + " ");
    } else {
        System.out.print(arr[arr.length - 1 - i] + " ");
    }
}

A 1 4 3 2 5

B 1 3 3 3 5

C 1 3 3 2 5

D 1 4 3 4 5

💡 Hint

Check how the index changes for even and odd i values.

For i=0 (even), print arr[0]=1; i=1 (odd), print arr[3]=4; i=2 (even), print arr[2]=3; i=3 (odd), print arr[1]=2; i=4 (even), print arr[4]=5. Output: 1 4 3 2 5.

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