Stack Implementation Using Array in DSA Rust - Time & Space Complexity

Time Complexity: Stack Implementation Using Array

O(1)

Understanding Time Complexity

We want to understand how fast operations like push and pop run in a stack built with an array.

How does the time taken change as we add more items?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct Stack {
    data: Vec<i32>,
}

impl Stack {
    fn push(&mut self, val: i32) {
        self.data.push(val);
    }

    fn pop(&mut self) -> Option<i32> {
        self.data.pop()
    }
}
    

This code shows a stack using a Rust vector (array). Push adds an item to the end, pop removes the last item.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding or removing one element at the end of the array.
• How many times: Each push or pop runs once per call, no loops inside.

How Execution Grows With Input

Each push or pop does a fixed amount of work, no matter how many items are in the stack.

Input Size (n)	Approx. Operations per push/pop
10	1
100	1
1000	1

Pattern observation: The time stays the same even as the stack grows larger.

Final Time Complexity

Time Complexity: O(1)

This means each push or pop takes the same short time, no matter how many items are in the stack.

Common Mistake

[X] Wrong: "Push or pop takes longer as the stack grows because the array is bigger."

[OK] Correct: The operations only add or remove at the end, so they do not need to look at all items.

Interview Connect

Knowing that stack operations run in constant time shows you understand efficient data structures, a key skill in coding interviews.

Self-Check

"What if the stack was implemented using a linked list instead of an array? How would the time complexity change?"

Practice

1. What is the main purpose of using an array to implement a stack?

easy

A. To store elements in a first-in-first-out order

B. To store elements in a fixed-size, last-in-first-out order

C. To sort elements automatically

D. To allow random access to elements anywhere

Solution

1. Step 1: Understand stack behavior

   A stack stores elements in last-in-first-out order, meaning the last element added is the first to be removed.

2. Step 2: Role of array in stack

   An array provides a fixed-size container to hold elements in order, supporting push and pop operations at one end.

3. Final Answer:

   To store elements in a fixed-size, last-in-first-out order -> Option B

4. Quick Check:

   Stack = LIFO with fixed size [OK]

Hint: Stack means last-in-first-out, arrays hold fixed-size data [OK]

Common Mistakes:

• Confusing stack with queue (FIFO)
• Thinking array sorts elements automatically
• Assuming random access is main stack feature

2. Which of the following is the correct Rust syntax to declare a fixed-size array of 5 integers initialized to zero for stack storage?

easy

A. let stack: [i32; 5] = [0; 5];

B. let stack = [0, 5];

C. let stack: Vec<i32> = Vec::new();

D. let stack = (0; 5);

Solution

1. Step 1: Identify fixed-size array syntax in Rust

   Rust fixed-size arrays use syntax: let name: [type; size] = [initial_value; size];

2. Step 2: Match options to syntax

   let stack: [i32; 5] = [0; 5]; matches this exactly: let stack: [i32; 5] = [0; 5]; initializes 5 integers to zero.

3. Final Answer:

   let stack: [i32; 5] = [0; 5]; -> Option A

4. Quick Check:

   Fixed-size array syntax = let stack: [i32; 5] = [0; 5]; [OK]

Hint: Fixed-size array uses [type; size] and [value; size] syntax [OK]

Common Mistakes:

• Using Vec instead of fixed-size array
• Incorrect array initialization syntax
• Confusing tuple syntax with array

3. What will be the output of the following Rust code implementing a stack push operation?

let mut stack: [i32; 3] = [0; 3];
let mut top = -1;

// Push function
fn push(stack: &mut [i32; 3], top: &mut i32, value: i32) {
    if *top < 2 {
        *top += 1;
        stack[*top as usize] = value;
    }
}

push(&mut stack, &mut top, 10);
push(&mut stack, &mut top, 20);
println!("{:?}", stack);

medium

A. [20, 10, 0]

B. [0, 10, 20]

C. [10, 20, 0]

D. [0, 0, 0]

Solution

1. Step 1: Trace push calls and stack changes

   Initially, stack = [0, 0, 0], top = -1. First push: top becomes 0, stack[0] = 10 -> [10, 0, 0]. Second push: top becomes 1, stack[1] = 20 -> [10, 20, 0].

2. Step 2: Print final stack state

   Printing stack shows [10, 20, 0].

3. Final Answer:

   [10, 20, 0] -> Option C

4. Quick Check:

   Push adds at top index, stack = [10, 20, 0] [OK]

Hint: Track top index and update array at that position [OK]

Common Mistakes:

• Not updating top index before assignment
• Confusing array indices order
• Assuming push adds at end regardless of top

4. Identify the error in this Rust stack pop function and select the correct fix:

fn pop(stack: &[i32; 3], top: &mut i32) -> Option<i32> {
    if *top >= 0 {
        let val = stack[*top as usize];
        *top -= 1;
        Some(val)
    } else {
        None
    }
}

medium

A. Change stack parameter to mutable: &mut [i32; 3]

B. Remove top parameter and use global variable

C. Change return type to i32 instead of Option<i32>

D. No error, function is correct

Solution

1. Step 1: Analyze the pop function code

   The function takes an immutable reference to the stack (&[i32; 3]) and a mutable reference to top. It only reads from the stack at index *top and decrements top.

2. Step 2: Verify compilation and logic

   No writes to the stack array occur, so an immutable reference is sufficient and correct. The function handles empty stack with None and returns the popped value properly.

3. Final Answer:

   No error, function is correct -> Option D

4. Quick Check:

   Immutable stack reference OK for read-only access in pop [OK]

Hint: Pop reads stack immutably and mutates top separately; no stack mutation needed [OK]

Common Mistakes:

• Assuming stack must be mutable for pop operations
• Suggesting unnecessary global variables
• Incorrectly changing return type from Option<i32>

5. Given a fixed-size stack implemented using an array of size 4, which sequence of operations will cause a stack overflow error?

let mut stack: [i32; 4] = [0; 4];
let mut top = -1;

// Push function
fn push(stack: &mut [i32; 4], top: &mut i32, value: i32) {
    if *top < 3 {
        *top += 1;
        stack[*top as usize] = value;
    } else {
        println!("Stack Overflow");
    }
}

// Operations:
push(&mut stack, &mut top, 5);
push(&mut stack, &mut top, 10);
push(&mut stack, &mut top, 15);
push(&mut stack, &mut top, 20);
push(&mut stack, &mut top, 25);

hard

A. The last push call with value 25 causes overflow

B. The first push call with value 5 causes overflow

C. No overflow occurs with these operations

D. Overflow occurs at the third push call with value 15

Solution

1. Step 1: Understand stack size and top limit

   Stack size is 4, so top can go from -1 to 3 (indices 0 to 3).

2. Step 2: Trace push calls and top value

   Push 5: top=0 Push 10: top=1 Push 15: top=2 Push 20: top=3 Push 25: top=3 already, so condition *top < 3 fails, triggers overflow.

3. Final Answer:

   The last push call with value 25 causes overflow -> Option A

4. Quick Check:

   Stack size 4, 5th push overflows [OK]

Hint: Stack max index = size-1; pushing beyond causes overflow [OK]

Common Mistakes:

• Counting top starting at 0 instead of -1
• Assuming overflow at fewer pushes
• Ignoring condition check in push