Array Traversal Patterns in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print all elements of the array using a for loop.

DSA Java

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

Drag options to blanks, or click blank then click option'

Ai

Bindex

Ck

Dj

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name not matching the loop condition causes errors.

Starting the loop variable at 1 instead of 0 skips the first element.

The loop variable is commonly named 'i' to iterate over array indices from 0 to length-1.

2fill in blank

medium

Complete the code to traverse the array backwards and print elements.

DSA Java

int[] arr = {10, 20, 30, 40, 50};
for (int i = arr.length - 1; [1] >= 0; i--) {
    System.out.print(arr[i] + " ");
}

Drag options to blanks, or click blank then click option'

Ak

Bj

Ci

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable in the condition than in the initialization.

Incorrect loop condition causing infinite loops.

The loop variable 'i' is used to traverse from the last index down to 0.

3fill in blank

hard

Fix the error in the code to sum all elements of the array.

DSA Java

int[] numbers = {2, 4, 6, 8};
int sum = 0;
for (int i = 0; i < numbers.length; [1]++) {
    sum += numbers[i];
}
System.out.println(sum);

Drag options to blanks, or click blank then click option'

Aindex

Bj

Ck

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Incrementing a variable not used in the loop condition.

Forgetting to increment the loop variable causing infinite loops.

The loop variable 'i' must be incremented to avoid infinite loop and correctly sum elements.

4fill in blank

hard

Fill both blanks to create a loop that prints only even numbers from the array.

DSA Java

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

Drag options to blanks, or click blank then click option'

Ai

Bj

Ck

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in loop parts causing errors.

Mismatching loop variable and array index variable.

The loop variable 'i' is used consistently in initialization and condition to traverse the array.

5fill in blank

hard

Fill all three blanks to create a loop that prints elements at odd indices.

DSA Java

int[] data = {5, 10, 15, 20, 25, 30};
for (int [1] = 1; [2] < data.length; [3] += 2) {
    System.out.print(data[[1]] + " ");
}

Drag options to blanks, or click blank then click option'

Ai

Ci++

Di += 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in loop parts.

Incrementing by 1 instead of 2 causing wrong output.

The loop variable 'i' starts at 1, condition uses 'i', and increments by 2 to access odd indices.

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