Spiral Matrix Traversal in DSA Rust - Time & Space Complexity

Time Complexity: Spiral Matrix Traversal

O(n x m)

Understanding Time Complexity

We want to understand how the time needed to read all elements in a matrix in a spiral order grows as the matrix size increases.

How does the number of steps change when the matrix gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn spiral_order(matrix: Vec>) -> Vec {
    let mut result = Vec::new();
    if matrix.is_empty() { return result; }
    let (mut top, mut bottom) = (0, matrix.len() - 1);
    let (mut left, mut right) = (0, matrix[0].len() - 1);

    while left <= right && top <= bottom {
        for col in left..=right {
            result.push(matrix[top][col]);
        }
        top += 1;

        for row in top..=bottom {
            result.push(matrix[row][right]);
        }
        if right == 0 { break; } else { right -= 1; }

        if top <= bottom {
            for col in (left..=right).rev() {
                result.push(matrix[bottom][col]);
            }
            if bottom == 0 { break; } else { bottom -= 1; }
        }

        if left <= right {
            for row in (top..=bottom).rev() {
                result.push(matrix[row][left]);
            }
            left += 1;
        }
    }
    result
}
    

This code reads all elements of a matrix in a spiral order, starting from the top-left corner and moving inward.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing each element of the matrix exactly once in a spiral pattern.
• How many times: Each element is visited once, so total operations equal the number of elements (rows x columns).

How Execution Grows With Input

As the matrix size grows, the number of elements grows with the product of rows and columns.

Input Size (n x n)	Approx. Operations
10 x 10	100
100 x 100	10,000
1000 x 1000	1,000,000

Pattern observation: The operations grow roughly with the square of the matrix dimension, because we visit every element once.

Final Time Complexity

Time Complexity: O(n x m)

This means the time grows proportionally to the total number of elements in the matrix, visiting each element once.

Common Mistake

[X] Wrong: "The spiral traversal has nested loops causing it to be O(n² x m²) or worse."

[OK] Correct: Although there are loops inside loops, each element is visited only once, so total steps equal the number of elements, not their square.

Interview Connect

Understanding how to analyze traversal of 2D structures like matrices is a key skill. It shows you can reason about loops and data access clearly, which is valuable in many coding problems.

Self-Check

"What if the matrix was not rectangular but jagged (rows of different lengths)? How would the time complexity change?"

Practice

1. What is the main idea behind Spiral Matrix Traversal?

easy

A. Traverse the matrix column by column from left to right

B. Traverse the matrix row by row from top to bottom

C. Traverse the matrix in a circular pattern from outer to inner layers

D. Traverse only the diagonal elements of the matrix

Solution

1. Step 1: Understand traversal pattern

   Spiral traversal means moving around the edges of the matrix and gradually moving inward in a circular way.

2. Step 2: Compare options

   Only Traverse the matrix in a circular pattern from outer to inner layers describes this circular, layer-by-layer movement. Others describe linear or partial traversals.

3. Final Answer:

   Traverse the matrix in a circular pattern from outer to inner layers -> Option C

4. Quick Check:

   Spiral traversal = circular outer to inner [OK]

Hint: Think of peeling an onion layer by layer [OK]

Common Mistakes:

• Confusing spiral with row-wise or column-wise traversal
• Thinking only diagonals are involved
• Ignoring the inward movement after outer layer

2. Which Rust loop structure is best suited to iterate over matrix indices for spiral traversal?

easy

A. loop { ... } with break condition

B. for i in 0..n { ... }

C. while i <= n { ... }

D. match statement for index matching

Solution

1. Step 1: Identify loop types in Rust

   Rust commonly uses for i in 0..n for fixed range iteration, ideal for matrix indices.

2. Step 2: Evaluate options for spiral traversal

   for i in 0..n { ... } uses range-based for loop, perfect for controlled index traversal. while i <= n { ... } is less idiomatic. loop { ... } with break condition is infinite loop needing manual breaks. match statement for index matching is for pattern matching, not iteration.

3. Final Answer:

   for i in 0..n { ... } -> Option B

4. Quick Check:

   Range-based for loop = best for index iteration [OK]

Hint: Use for loops with ranges for matrix indices [OK]

Common Mistakes:

• Using infinite loops without proper break
• Using while loops without clear index control
• Misusing match for iteration

3. What is the output of the spiral traversal on this 3x3 matrix?

let matrix = vec![
  vec![1, 2, 3],
  vec![4, 5, 6],
  vec![7, 8, 9],
];
// Spiral traversal code prints elements in order

medium

A. [1, 2, 3, 6, 9, 8, 7, 4, 5]

B. [1, 4, 7, 8, 9, 6, 3, 2, 5]

C. [1, 2, 3, 4, 5, 6, 7, 8, 9]

D. [9, 8, 7, 6, 5, 4, 3, 2, 1]

Solution

1. Step 1: Trace spiral traversal order

   Start top row left to right: 1, 2, 3; then right column down: 6, 9; bottom row right to left: 8, 7; left column up: 4; center element: 5.

2. Step 2: Write elements in order

   Combined order is [1, 2, 3, 6, 9, 8, 7, 4, 5].

3. Final Answer:

   [1, 2, 3, 6, 9, 8, 7, 4, 5] -> Option A

4. Quick Check:

   Outer to inner spiral = [1,2,3,6,9,8,7,4,5] [OK]

Hint: Follow edges clockwise, then move inward [OK]

Common Mistakes:

• Listing elements row-wise instead of spiral
• Reversing order of bottom or left edges
• Skipping center element

4. Identify the bug in this Rust spiral traversal snippet:

let mut top = 0;
let mut bottom = matrix.len() - 1;
let mut left = 0;
let mut right = matrix[0].len() - 1;
while top <= bottom && left <= right {
  for i in left..=right {
    print!("{} ", matrix[top][i]);
  }
  top += 1;
  for i in top..=bottom {
    print!("{} ", matrix[i][right]);
  }
  right -= 1;
  for i in (left..=right).rev() {
    print!("{} ", matrix[bottom][i]);
  }
  bottom -= 1;
  for i in (top..=bottom).rev() {
    print!("{} ", matrix[i][left]);
  }
  left += 1;
}

medium

A. The loop may print elements twice when top > bottom or left > right

B. The ranges in for loops are incorrect and cause syntax errors

C. The variables top, bottom, left, right are not initialized properly

D. The print statements use wrong matrix indices causing runtime panic

Solution

1. Step 1: Analyze loop conditions and updates

   The while loop condition is correct, but inside the loop, after updating top, bottom, left, right, the for loops still run without checking if top <= bottom or left <= right again.

2. Step 2: Identify consequence of missing checks

   This can cause the inner loops to run when top > bottom or left > right, printing elements twice or causing invalid access.

3. Final Answer:

   The loop may print elements twice when top > bottom or left > right -> Option A

4. Quick Check:

   Missing inner boundary checks = duplicate prints [OK]

Hint: Check boundaries before each edge traversal [OK]

Common Mistakes:

• Not checking boundaries inside the loop after updates
• Assuming while condition covers all cases
• Ignoring reverse ranges correctness

5. Given a rectangular matrix with dimensions 3x5, how many elements will be printed in spiral traversal, and what is the order of traversal directions?

hard

A. 15 elements; order: left to right, bottom to top, right to left, top to bottom, repeat

B. 15 elements; order: top to bottom, left to right, bottom to top, right to left, repeat

C. 8 elements; order: left to right, top to bottom, right to left, bottom to top, repeat

D. 15 elements; order: left to right, top to bottom, right to left, bottom to top, repeat

Solution

1. Step 1: Calculate total elements in 3x5 matrix

   3 rows * 5 columns = 15 elements to print in spiral traversal.

2. Step 2: Identify traversal direction order

   Spiral traversal moves: left to right (top row), top to bottom (right column), right to left (bottom row), bottom to top (left column), then repeats inward.

3. Final Answer:

   15 elements; order: left to right, top to bottom, right to left, bottom to top, repeat -> Option D

4. Quick Check:

   3*5=15; spiral directions = LR, TB, RL, BT [OK]

Hint: Count total elements; remember spiral direction cycle [OK]

Common Mistakes:

• Miscounting total elements
• Mixing up traversal direction order
• Assuming fewer elements printed