Spiral Matrix Leetcode Problem 54 [Python Solution]

In this tutorial, we dismantle the Spiral Matrix Leetcode problem, a Math and Geometry related quetion and is categorized as a medium-level problem.

The challenge is set by Microsoft, and it’s a great exercise to strengthen your coding skills.

We will walk through the problem statement, provide an efficient Python solution, analyze the time and space complexity, and explain the reasoning behind our approach.

Problem Overview
The problem statement is quite simple: given an m x n matrix, we need to return all its elements in spiral order.

In other words, we need to traverse the matrix in a spiral pattern, starting from the top-left corner and moving to the right, down, left, and up, until all elements have been visited.

Let’s look at a couple of examples to understand this better:

Example 1:

Input:
matrix = [   [1, 2, 3],
   [4, 5, 6],
   [7, 8, 9]
 ]
Output: [1, 2, 3, 6, 9, 8, 7, 4, 5]

Example 2:

Input:
 matrix = [
   [1, 2, 3, 4],
   [5, 6, 7, 8],
   [9, 10, 11, 12]
 ]
 
 Output: [1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7]

Constraints
Before we dive into the solution, let’s understand the constraints:

The matrix has dimensions m x n.
The values of m and n are such that 1 <= m, n <= 10.
The matrix elements range from -100 to 100. Now that we have a clear picture of the problem, let’s proceed with our solution.

Efficient Python Code Solution

To solve this problem efficiently, we’ll use a systematic approach.

We’ll maintain four pointers: left, right, top, and bottom, which represent the boundaries of the current sub-matrix we’re working on.

We’ll iterate through the matrix in a spiral order, updating these pointers as we go.

Here’s the Python code for our solution:

View on GitHub ➹

def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
    if not matrix:
        return []

    result = []
    left, right = 0, len(matrix[0]) - 1
    top, bottom = 0, len(matrix) - 1

    while left &lt;= right and top &lt;= bottom:
        # Traverse the top row
        for i in range(left, right + 1):
            result.append(matrix[top][i])
        top += 1

        # Traverse the right column
        for i in range(top, bottom + 1):
            result.append(matrix[i][right])
        right -= 1

        if not (left &lt;= right and top &lt;= bottom):
            break

        # Traverse the bottom row
        for i in range(right, left - 1, -1):
            result.append(matrix[bottom][i])
        bottom -= 1

        # Traverse the left column
        for i in range(bottom, top - 1, -1):
            result.append(matrix[i][left])
        left += 1

    return result

This code starts by initializing the result list and the four boundary pointers.

Then, it iterates through the matrix in the spiral order by updating these pointers and adding the elements to the result list.

Let’s break down the code step by step:

We check if the matrix is empty.
If it is, we return an empty list because there are no elements to process.
We initialize left and right to represent the leftmost and rightmost columns, top and bottom to represent the top and bottom rows of the sub-matrix.
We enter a while loop that continues as long as left is less than or equal to right and top is less than or equal to bottom.
This condition ensures that we haven’t covered all elements.
Inside the loop, we traverse the top row from left to right, adding each element to the result.
We then increment top to indicate that this row has been visited.
Next, we traverse the right column from top to bottom, adding the elements to the result.
We decrement right to indicate that this column has been visited.
We check if the condition for continuing the loop is still met.
If not, we break out of the loop since we’ve visited all elements.
If the condition is still valid, we proceed to traverse the bottom row from right to left, adding elements and decrementing bottom.
Finally, we traverse the left column from bottom to top, adding elements and incrementing left.
The loop continues until all elements have been visited.
We return the result list, which contains the matrix elements in spiral order.

The code efficiently handles the problem and produces the desired output.

Time and Space Complexity

Now, let’s analyze the time and space complexity of our solution:

Time Complexity: The time complexity of this solution is O(m * n), where m is the number of rows and n is the number of columns in the matrix.

We visit each element exactly once, and our operations inside the loop are constant time.

Space Complexity: The space complexity is O(1) since we only use a constant amount of extra space to store the result list.

We don’t use additional data structures that grow with the input size.

Reasoning Behind Our Approach

The reasoning behind our approach is straightforward.

We tackle the problem by iteratively processing the outermost layer of the matrix while shrinking the boundaries inward.

This ensures that we visit all elements in the desired spiral order.

By maintaining four pointers that represent the boundaries of the current sub-matrix, we can efficiently traverse the matrix without the need for additional data structures or excessive memory usage.

This approach is intuitive and clear, making it suitable for solving the problem.

Conclusion

In this blog post, we’ve explored the Spiral Matrix problem from LeetCode.

We provided a Python solution that efficiently returns all elements of the matrix in spiral order.

The code is accompanied by a detailed explanation, and we analyzed the time and space complexity of the solution.

We hope this guide has been helpful for understanding how to tackle this problem.

If you have any questions, comments, or suggestions, please feel free to ask in the comments section below.

Coding challenges like this one are excellent for improving your problem-solving skills, and we encourage you to keep practicing and exploring similar problems.

Question Link – LeetCode Spiral Matrix

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Happy coding!

Spiral Matrix Leetcode Problem 54 [Python Solution]

Example 1:

Example 2:

Efficient Python Code Solution

Time and Space Complexity

Reasoning Behind Our Approach

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