Array Traversal Patterns in DSA Java - Build from Scratch

Array Traversal Patterns

📖 Scenario: Imagine you are managing a small store's daily sales. You have a list of sales amounts for each day of the week. You want to learn how to look at each sale one by one to understand the pattern.

🎯 Goal: You will create a Java program that stores daily sales in an array, then uses a loop to look at each sale and print it out. This will teach you how to go through arrays step-by-step.

📋 What You'll Learn

Create an integer array called dailySales with exact values for 7 days

Create an integer variable called totalSales and set it to 0

Use a for loop with variable i to traverse dailySales

Add each sale to totalSales inside the loop

Print the total sales after the loop

💡 Why This Matters

🌍 Real World

Stores and businesses often keep daily sales data in lists or arrays. Going through these lists helps understand how much was sold each day and the total sales.

💼 Career

Knowing how to traverse arrays is a basic skill for software developers. It helps in processing lists of data like sales, user inputs, or sensor readings.

1

Create the daily sales array

Create an integer array called dailySales with these exact values: 50, 75, 60, 90, 100, 80, 65

DSA Java

# Create the dailySales array with the given values
// Your code here

Hint

Use int[] dailySales = { ... }; to create the array with the exact numbers.

2

Set up total sales counter

Create an integer variable called totalSales and set it to 0

DSA Java

int[] dailySales = {50, 75, 60, 90, 100, 80, 65};
// Create totalSales variable and set it to 0
// Your code here

Hint

Use int totalSales = 0; to start counting from zero.

3

Traverse the array and sum sales

Use a for loop with variable i to go through dailySales from start to end. Inside the loop, add dailySales[i] to totalSales

DSA Java

int[] dailySales = {50, 75, 60, 90, 100, 80, 65};
int totalSales = 0;
// Use a for loop to add each sale to totalSales
// Your code here

Hint

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

4

Print the total sales

Print the value of totalSales using System.out.println

DSA Java

int[] dailySales = {50, 75, 60, 90, 100, 80, 65};
int totalSales = 0;
for (int i = 0; i < dailySales.length; i++) {
    totalSales += dailySales[i];
}
// Print totalSales
// Your code here

Hint

Use System.out.println(totalSales); to show the total sales.

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