Array Declaration and Initialization in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Rust array initialization?

Consider the following Rust code snippet. What will be printed after execution?

DSA Rust

fn main() {
    let arr = [5; 4];
    for i in 0..arr.len() {
        print!("{} ", arr[i]);
    }
}

A Compilation error

B 5 5 5 5

C 5 4 3 2

D 0 0 0 0

💡 Hint

The syntax [value; count] creates an array with 'count' elements, each initialized to 'value'.

The code creates an array of length 4 where each element is 5. The loop prints each element separated by space.

❓ Predict Output

intermediate

What is the output after modifying an array element?

What will be printed by this Rust program?

DSA Rust

fn main() {
    let mut arr = [1, 2, 3, 4];
    arr[2] = 10;
    for val in arr.iter() {
        print!("{} ", val);
    }
}

A 1 2 10 4

B 1 2 3 4

C 10 2 3 4

D Runtime error

💡 Hint

Arrays declared with 'mut' can have their elements changed.

The third element (index 2) is changed from 3 to 10 before printing all elements.

❓ Predict Output

advanced

What error does this Rust code produce?

What error will this Rust code cause when compiled?

DSA Rust

fn main() {
    let arr = [1, 2, 3];
    let x = arr[3];
    println!("{}", x);
}

A Index out of bounds panic at runtime

B Compilation error: index out of range

C Prints 3

D Prints 0

💡 Hint

Array indices start at 0 and go up to length-1.

Accessing arr[3] is out of bounds since arr has indices 0,1,2. Rust panics at runtime for this.

🧠 Conceptual

advanced

How many elements does this Rust array have?

Given the declaration: let arr = [0; 7]; How many elements are in arr?

A 1

B 0

C 7

D Compilation error

💡 Hint

The syntax [value; count] creates an array with 'count' elements.

The array has 7 elements, each initialized to 0.

❓ Predict Output

expert

What is the output of this Rust array slice and iteration?

What will this Rust program print?

DSA Rust

fn main() {
    let arr = [2, 4, 6, 8, 10];
    let slice = &arr[1..4];
    for val in slice.iter() {
        print!("{} ", val);
    }
}

A Compilation error

B 2 4 6 8 10

C 6 8 10

D 4 6 8

💡 Hint

Slices use start index inclusive and end index exclusive.

The slice &arr[1..4] includes elements at indices 1, 2, and 3, which are 4, 6, and 8.

Practice

1. Which of the following is the correct way to declare an array of 5 integers initialized to zero in Rust?

easy

A. let arr = [5; 0];

B. let arr: [i32; 5] = [0; 5];

C. let arr: [i32] = [0; 5];

D. let arr = [0, 5];

Solution

1. Step 1: Understand Rust array syntax

   Rust arrays require a fixed size and type, declared as [type; size].

2. Step 2: Check initialization correctness

   [0; 5] means 5 elements all set to 0, which matches the declaration.

3. Final Answer:

   let arr: [i32; 5] = [0; 5]; -> Option B

4. Quick Check:

   Correct Rust array declaration = let arr: [i32; 5] = [0; 5]; [OK]

Hint: Use [value; size] to initialize arrays quickly [OK]

Common Mistakes:

• Missing size in array type declaration
• Using wrong syntax like [0, 5] for repeated values
• Declaring array without fixed size

2. What is the correct syntax to declare a mutable array of 3 floating-point numbers initialized to 1.0 in Rust?

easy

A. let mut arr: [f64; 3] = [1.0; 3];

B. let arr mut: [f64; 3] = [1.0; 3];

C. mut let arr: [f64; 3] = [1.0; 3];

D. let arr: mut [f64; 3] = [1.0; 3];

Solution

1. Step 1: Identify mutable variable syntax in Rust

   Mutable variables use 'let mut' before the variable name.

2. Step 2: Confirm array declaration and initialization

   Array type is [f64; 3] and initialized with [1.0; 3], which is correct.

3. Final Answer:

   let mut arr: [f64; 3] = [1.0; 3]; -> Option A

4. Quick Check:

   Mutable array syntax = let mut arr: [f64; 3] = [1.0; 3]; [OK]

Hint: Place 'mut' right after 'let' for mutable arrays [OK]

Common Mistakes:

• Placing 'mut' after variable name
• Using 'mut let' instead of 'let mut'
• Incorrect array type syntax

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

let arr = [2; 4];
println!("{:?}", arr);

medium

A. [2; 4]

B. [4, 4, 4, 4]

C. [2, 2, 2, 2]

D. Compilation error

Solution

1. Step 1: Understand array initialization [2; 4]

   This creates an array with 4 elements, each set to 2.

2. Step 2: Print array with debug format

   Using {:?} prints the array elements as a list.

3. Final Answer:

   [2, 2, 2, 2] -> Option C

4. Quick Check:

   Array with repeated values prints as list [OK]

Hint: Remember [value; size] repeats value size times [OK]

Common Mistakes:

• Confusing size and value positions
• Expecting syntax [2;4] to print literally
• Assuming compilation error due to syntax

4. Identify the error in the following Rust code snippet:

let arr: [i32; 3] = [1, 2];

medium

A. Array size mismatch error

B. Type mismatch error

C. Mutable variable required

D. No error, code is correct

Solution

1. Step 1: Check declared array size

   Array is declared with size 3 but initialized with only 2 elements.

2. Step 2: Understand Rust array size rules

   Rust requires the number of elements to match the declared size exactly.

3. Final Answer:

   Array size mismatch error -> Option A

4. Quick Check:

   Declared size must match initializer length [OK]

Hint: Array size and initializer length must be equal [OK]

Common Mistakes:

• Ignoring size mismatch
• Assuming Rust allows shorter initializers
• Confusing type mismatch with size error

5. You want to create a 2D array (matrix) of size 3x3 filled with the value 7 in Rust. Which code correctly declares and initializes this array?

hard

A. let matrix: [i32; 3][3] = [[7; 3]; 3];

B. let matrix: [[i32; 3]; 3] = [7; 9];

C. let matrix: [i32; 9] = [[7; 3]; 3];

D. let matrix: [[i32; 3]; 3] = [[7; 3]; 3];

Solution

1. Step 1: Understand 2D array type declaration

   A 3x3 array of i32 is declared as [[i32; 3]; 3], meaning 3 arrays each of size 3.

2. Step 2: Check initialization syntax

   [[7; 3]; 3] creates 3 arrays each filled with 3 sevens, forming a 3x3 matrix.

3. Final Answer:

   let matrix: [[i32; 3]; 3] = [[7; 3]; 3]; -> Option D

4. Quick Check:

   Nested arrays use [[type; cols]; rows] [OK]

Hint: Use nested [value; size] for multi-dimensional arrays [OK]

Common Mistakes:

• Incorrect type syntax like [i32; 3][3]
• Initializing flat array instead of nested
• Mismatch between declared size and initializer