Given an integer N, which is a power of 2, return a square of length N filled with the numbers {1, 2, 3, ... N2} following the "Z Pattern".
This consists of recursively traversing the 4 quadrants in the order: upper-left, upper-right, lower-left, lower-right.
Example 1:
Input: N = 2 Output: [ [1, 2], [3, 4] ]
Example 2:
Input: N = 4 Output:[ [1, 2, 5, 6], [3, 4, 7, 8], [9, 10, 13, 14], [11, 12, 15, 16] ]
The core challenge of this problem is to fill a matrix in a specific "Z Pattern" order. This pattern involves recursively dividing the matrix into four quadrants and filling them in a specific sequence: upper-left, upper-right, lower-left, and lower-right. This problem is significant in understanding recursive matrix traversal and has applications in image processing and computer graphics.
Potential pitfalls include misunderstanding the recursive nature of the problem and incorrectly filling the matrix in the wrong order.
To solve this problem, we need to think recursively. The naive approach would be to try and fill the matrix iteratively, but this would be complex and error-prone. Instead, we can use a recursive function to fill the matrix quadrant by quadrant.
Here is a step-by-step approach:
Let's break down the algorithm step-by-step:
fill_z_pattern(matrix, start_row, start_col, size, start_num)
:
matrix[start_row][start_col]
to start_num
.def fill_z_pattern(matrix, start_row, start_col, size, start_num):
# Base case: if the size is 1, fill the single cell with the start_num
if size == 1:
matrix[start_row][start_col] = start_num
return start_num + 1
# Calculate the mid-point of the current bounds
mid = size // 2
# Recursively fill the four quadrants
start_num = fill_z_pattern(matrix, start_row, start_col, mid, start_num) # upper-left
start_num = fill_z_pattern(matrix, start_row, start_col + mid, mid, start_num) # upper-right
start_num = fill_z_pattern(matrix, start_row + mid, start_col, mid, start_num) # lower-left
start_num = fill_z_pattern(matrix, start_row + mid, start_col + mid, mid, start_num) # lower-right
return start_num
def z_pattern(N):
# Initialize an N x N matrix with zeros
matrix = [[0 for _ in range(N)] for _ in range(N)]
# Start the recursive filling with the initial bounds and starting number 1
fill_z_pattern(matrix, 0, 0, N, 1)
return matrix
# Example usage
N = 4
result = z_pattern(N)
for row in result:
print(row)
The time complexity of this approach is O(N^2) because we are filling each cell of the N x N matrix exactly once. The space complexity is also O(N^2) due to the storage required for the matrix.
Potential edge cases include:
These edge cases are handled naturally by the recursive approach.
To test the solution comprehensively, we should include a variety of test cases:
We can use Python's built-in unittest
framework to automate these tests.
When approaching such problems, it's essential to:
In this blog post, we discussed how to solve the "Z Pattern" problem using a recursive approach. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills and a deep understanding of recursive algorithms.
For further reading and practice, consider the following resources: