Array traversal in Java - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print each element of the array.

Java

int[] numbers = {1, 2, 3, 4, 5};
for (int [1] = 0; i < numbers.length; i++) {
    System.out.println(numbers[i]);
}

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

Aj

Bi

Ck

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name different from the one in the loop condition causes errors.

Forgetting to initialize the loop variable.

The variable i is commonly used as the loop counter to traverse arrays in Java.

2fill in blank

medium

Complete the code to sum all elements in the array.

Java

int[] values = {10, 20, 30};
int sum = 0;
for (int i = 0; i < [1]; i++) {
    sum += values[i];
}
System.out.println(sum);

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

Avalues.length

Bvalues.length()

Cvalues.size

Dvalues.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses after length causes a compilation error.

Using size or size() which are not valid for arrays.

Arrays in Java have a length property (without parentheses) to get their size.

3fill in blank

hard

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

Java

String[] fruits = {"apple", "banana", "cherry"};
for (int i = 0; i <= [1]; i++) {
    System.out.println(fruits[i]);
}

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Afruits.length - 1

Bfruits.size

Cfruits.length

Dfruits.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= fruits.length causes runtime errors.

Using size or size() which are not valid for arrays.

The loop should run while i is less than fruits.length. Using i <= fruits.length - 1 is correct to avoid going out of bounds.

4fill in blank

hard

Fill both blanks to create a map of words to their lengths for words longer than 3 characters.

Java

Map<String, Integer> wordLengths = new HashMap<>();
String[] words = {"cat", "elephant", "dog", "lion"};
for (String [1] : words) {
    if ([2].length() > 3) {
        wordLengths.put([2], [2].length());
    }
}

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

Aword

Bw

Citem

Delement

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inside the loop causes errors.

Using variable names not declared in the for-each loop.

The variable word is used as the loop variable and inside the if condition to check length and add to the map.

5fill in blank

hard

Fill all three blanks to create a map of uppercase words to their lengths for words longer than 4 characters.

Java

Map<String, Integer> result = new HashMap<>();
String[] words = {"tree", "mountain", "river", "sky"};
for (String [1] : words) {
    if ([2].length() > 4) {
        result.put([3], [2].length());
    }
}

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

Aw

Cw.toUpperCase()

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Forgetting to convert the word to uppercase before putting it in the map.

The loop variable w is used to check length and to create the uppercase key for the map.

Practice

1. What does array traversal mean in Java?
int[] arr = {1, 2, 3};

easy

A. Creating a new array with the same elements

B. Visiting each element of the array one by one using a loop

C. Sorting the array elements in ascending order

D. Deleting elements from the array

Solution

1. Step 1: Understand array traversal meaning

   Array traversal means going through each element in the array one by one.

2. Step 2: Match with given options

   Only Visiting each element of the array one by one using a loop describes visiting each element using a loop, which is traversal.

3. Final Answer:

   Visiting each element of the array one by one using a loop -> Option B

4. Quick Check:

   Array traversal = visiting elements one by one [OK]

Hint: Traversal means visiting all elements step-by-step [OK]

Common Mistakes:

• Confusing traversal with sorting
• Thinking traversal means deleting elements
• Mixing traversal with creating new arrays

2. Which of the following is the correct syntax to traverse an array int[] nums in Java?

easy

A. for(int i = 0; i < nums.length; i++) { System.out.println(nums[i]); }

B. for(int i = 1; i <= nums.length; i++) { System.out.println(nums[i]); }

C. for(int i = 0; i <= nums.length; i++) { System.out.println(nums[i]); }

D. for(int i = 0; i < nums.length - 1; i++) { System.out.println(nums[i]); }

Solution

1. Step 1: Check loop start and end conditions

   The loop should start at 0 and run while i is less than nums.length to cover all indices.

2. Step 2: Validate array access inside loop

   Accessing nums[i] is correct if i is within 0 to length-1. for(int i = 0; i < nums.length; i++) { System.out.println(nums[i]); } satisfies this.

3. Final Answer:

   for(int i = 0; i < nums.length; i++) { System.out.println(nums[i]); } -> Option A

4. Quick Check:

   Loop from 0 to length-1 = correct traversal [OK]

Hint: Loop from 0 to less than length for full traversal [OK]

Common Mistakes:

• Starting loop at 1 instead of 0
• Using <= length causing index out of bounds
• Stopping loop early at length - 1

3. What will be the output of the following code?

int[] arr = {5, 10, 15};
for(int i = 0; i < arr.length; i++) {
  System.out.print(arr[i] + " ");
}

medium

A. 5 10 15

B. 15 10 5

C. 5 10 15 0

D. Error: ArrayIndexOutOfBoundsException

Solution

1. Step 1: Understand the loop iteration

   The loop runs from i=0 to i=2 (arr.length=3), printing arr[i] each time.

2. Step 2: Check printed values

   Values printed are arr[0]=5, arr[1]=10, arr[2]=15, each followed by a space.

3. Final Answer:

   5 10 15 -> Option A

4. Quick Check:

   Print elements in order = 5 10 15 [OK]

Hint: Prints elements in order from index 0 to length-1 [OK]

Common Mistakes:

• Assuming reverse order output
• Expecting extra zero at end
• Thinking loop causes index error

4. Identify the error in this array traversal code:

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

medium

A. System.out.println should be outside the loop

B. Array declaration is incorrect

C. Loop condition should be i < data.length, not <= data.length

D. No error, code runs fine

Solution

1. Step 1: Analyze loop condition

   The loop runs while i <= data.length, which means i can be equal to data.length (3).

2. Step 2: Check array index access

   Accessing data[3] causes ArrayIndexOutOfBoundsException because valid indices are 0 to 2.

3. Final Answer:

   Loop condition should be i < data.length, not <= data.length -> Option C

4. Quick Check:

   Index must be less than length to avoid error [OK]

Hint: Use i < length, not i <= length in loops [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Ignoring array index limits
• Thinking loop runs without error

5. Given an array int[] nums = {3, 6, 9, 12, 15};, write code to print only the even numbers by traversing the array.

hard

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

B. for(int i = 1; i <= nums.length; i++) { if(nums[i] % 2 == 0) System.out.print(nums[i] + " "); }

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

D. for(int i = 0; i < nums.length; i++) { if(nums[i] % 2 == 0) System.out.print(nums[i] + " "); }

Solution

1. Step 1: Traverse the array correctly

   Loop from i=0 to i < nums.length to visit all elements safely.

2. Step 2: Check condition for even numbers

   Use nums[i] % 2 == 0 to find even numbers and print them.

3. Final Answer:

   for(int i = 0; i < nums.length; i++) { if(nums[i] % 2 == 0) System.out.print(nums[i] + " "); } -> Option D

4. Quick Check:

   Traverse all and print only even numbers [OK]

Hint: Check nums[i] % 2 == 0 inside loop to print evens [OK]

Common Mistakes:

• Using wrong loop bounds causing errors
• Printing odd numbers instead of even
• Missing condition inside loop