Array traversal in Java - Mini Project: Build & Apply

Array traversal

📖 Scenario: You are working on a simple program to process a list of daily temperatures recorded in a week. You want to look at each temperature one by one to understand the weather pattern.

🎯 Goal: Build a Java program that stores temperatures in an array and then uses a loop to visit each temperature and print it out.

📋 What You'll Learn

Create an array of integers with exactly these values: 23, 25, 22, 20, 24, 26, 21

Create a variable to hold the length of the array

Use a for loop with an index variable i to traverse the array

Print each temperature value inside the loop

💡 Why This Matters

🌍 Real World

Traversing arrays is a basic skill used in many programs that handle lists of data, like weather apps, shopping lists, or scores.

💼 Career

Understanding how to loop through arrays is essential for any programming job, as it helps process collections of information efficiently.

1

Create the temperature array

Create an integer array called temperatures with these exact values: 23, 25, 22, 20, 24, 26, 21.

Java

// Your code here

Hint

Use curly braces {} to list the values inside the array.

2

Create a variable for array length

Create an integer variable called length and set it to the length of the temperatures array using temperatures.length.

Java

int[] temperatures = {23, 25, 22, 20, 24, 26, 21};
// Your code here

Hint

The length property gives the number of elements in the array.

3

Traverse the array with a for loop

Use a for loop with an integer index variable i starting at 0 and running while i < length. Inside the loop, access the current temperature with temperatures[i].

Java

int[] temperatures = {23, 25, 22, 20, 24, 26, 21};
int length = temperatures.length;
// Your code here

Hint

Use for (int i = 0; i < length; i++) to loop through the array indexes.

4

Print each temperature

Inside the for loop, use System.out.println(temperatures[i]) to print each temperature value.

Java

int[] temperatures = {23, 25, 22, 20, 24, 26, 21};
int length = temperatures.length;
for (int i = 0; i < length; i++) {
    // Your code here
}

Hint

Use System.out.println(temperatures[i]) to print each value on its own line.

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