Array Traversal Patterns in DSA Rust - Build from Scratch

Array Traversal Patterns

📖 Scenario: You are helping a small shop owner who keeps a list of daily sales numbers. They want to analyze these numbers to understand their business better.

🎯 Goal: You will create a Rust program that stores daily sales in an array, then uses a loop to go through each sale and calculate the total sales.

📋 What You'll Learn

Create an array with exact daily sales numbers

Create a variable to hold the total sales

Use a for loop with variable sale to traverse the array

Add each sale to the total

Print the total sales at the end

💡 Why This Matters

🌍 Real World

Summing daily sales is a common task in business to understand total revenue over a period.

💼 Career

Knowing how to traverse arrays and sum values is a basic skill for software developers working with data.

1

Create the daily sales array

Create an array called daily_sales with these exact values: 100, 200, 150, 175, 125

DSA Rust

# Create the daily_sales array with the given values
# Your code here

Hint

Use let daily_sales = [100, 200, 150, 175, 125]; to create the array.

2

Create a total sales variable

Add a variable called total and set it to 0 to hold the sum of sales

DSA Rust

fn main() {
    let daily_sales = [100, 200, 150, 175, 125];
    // Create total variable to hold sum
    // Your code here
}

Hint

Use let mut total = 0; to create a variable that can change.

3

Traverse the array and sum sales

Use a for loop with variable sale to go through daily_sales and add each sale to total

DSA Rust

fn main() {
    let daily_sales = [100, 200, 150, 175, 125];
    let mut total = 0;
    // Use for loop to add each sale to total
    // Your code here
}

Hint

Use for sale in daily_sales { total += sale; } to add each sale.

4

Print the total sales

Write a println! statement to print the text Total sales: followed by the value of total

DSA Rust

fn main() {
    let daily_sales = [100, 200, 150, 175, 125];
    let mut total = 0;
    for sale in daily_sales {
        total += sale;
    }
    // Print the total sales
    // Your code here
}

Hint

Use println!("Total sales: {}", total); to show the result.

Practice

1. What does the following Rust code snippet do?
for i in 0..5 { println!("{}", i); }

easy

A. Prints numbers from 1 to 4

B. Prints numbers from 1 to 5

C. Prints numbers from 0 to 5

D. Prints numbers from 0 to 4

Solution

1. Step 1: Understand the range syntax 0..5

   The range 0..5 in Rust includes numbers starting at 0 up to but not including 5.

2. Step 2: Trace the loop iterations

   The loop variable i takes values 0, 1, 2, 3, and 4, printing each number.

3. Final Answer:

   Prints numbers from 0 to 4 -> Option D

4. Quick Check:

   Range 0..5 excludes 5, so output is 0 to 4 [OK]

Hint: Remember Rust ranges exclude the end number [OK]

Common Mistakes:

• Thinking 0..5 includes 5
• Confusing 0..5 with 0..=5
• Assuming loop starts at 1

2. Which of the following is the correct syntax to iterate over the indices of an array arr in Rust?

easy

A. for i in 0..arr.len() { println!("{}", arr[i]); }

B. for i in 0..arr { println!("{}", i); }

C. for i in arr { println!("{}", i); }

D. for i in arr.len() { println!("{}", i); }

Solution

1. Step 1: Check valid iteration over array indices

   Rust arrays have a length method .len(), so 0..arr.len() gives valid indices.

2. Step 2: Access elements by index

   Using arr[i] inside the loop accesses each element correctly.

3. Final Answer:

   for i in 0..arr.len() { println!("{}", arr[i]); } -> Option A

4. Quick Check:

   Use 0..arr.len() for index-based iteration [OK]

Hint: Use arr.len() for array length, not arr alone [OK]

Common Mistakes:

• Using 0..arr instead of 0..arr.len()
• Trying to iterate directly over arr without reference
• Using arr.len() without range syntax

3. What is the output of this Rust code?

let arr = [10, 20, 30];
for i in (0..arr.len()).rev() {
    print!("{} ", arr[i]);
}

medium

A. 10 20 30

B. 0 1 2

C. 30 20 10

D. Compilation error

Solution

1. Step 1: Understand the range and reverse

   The range 0..arr.len() is 0..3, which is 0,1,2. Using .rev() reverses it to 2,1,0.

2. Step 2: Print elements in reverse order

   arr[2] = 30, arr[1] = 20, arr[0] = 10, so output is "30 20 10 "

3. Final Answer:

   30 20 10 -> Option C

4. Quick Check:

   Reverse iteration prints elements backward [OK]

Hint: Use .rev() to reverse range iteration [OK]

Common Mistakes:

• Forgetting .rev() reverses the order
• Printing indices instead of array values
• Confusing range end with inclusive

4. Identify the error in this Rust code snippet that tries to print all elements of an array arr:

let arr = [1, 2, 3];
for i in 0..arr.len() {
    println!("{}", arr[i+1]);
}

medium

A. Syntax error due to missing semicolon

B. Index out of bounds error at runtime

C. No error, prints all elements correctly

D. Compilation error due to type mismatch

Solution

1. Step 1: Analyze the loop indices and access

   The loop runs i from 0 to 2 (arr.len()=3). Accessing arr[i+1] means indices 1,2,3.

2. Step 2: Identify out of bounds access

   arr[3] does not exist, causing runtime panic (index out of bounds).

3. Final Answer:

   Index out of bounds error at runtime -> Option B

4. Quick Check:

   Accessing arr[i+1] exceeds array length [OK]

Hint: Check array indices carefully to avoid out of bounds [OK]

Common Mistakes:

• Assuming arr[i+1] is safe for all i
• Ignoring runtime panics
• Confusing compile-time and runtime errors

5. Given a Rust array arr of integers, which code snippet correctly prints only the even-indexed elements (0, 2, 4, ...) of the array?

hard

A. for i in (0..arr.len()).step_by(2) { println!("{}", arr[i]); }

B. for i in 0..arr.len() { if i % 2 == 1 { println!("{}", arr[i]); } }

C. for i in 0..arr.len() { println!("{}", arr[i * 2]); }

D. for i in (0..arr.len()).rev().step_by(2) { println!("{}", arr[i]); }

Solution

1. Step 1: Understand step_by usage

   The method step_by(2) skips every second index, so 0, 2, 4, ...

2. Step 2: Confirm correct element access

   Using arr[i] inside the loop prints elements at even indices correctly.

3. Final Answer:

   for i in (0..arr.len()).step_by(2) { println!("{}", arr[i]); } -> Option A

4. Quick Check:

   step_by(2) iterates over even indices [OK]

Hint: Use step_by(2) to jump every two elements [OK]

Common Mistakes:

• Using i % 2 == 1 instead of 0 for even indices
• Multiplying i by 2 causing out of bounds
• Using rev() which reverses order unnecessarily