Z Pattern in C++ with Time Complexity Analysis

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]
]

Problem Definition

The problem requires generating a square matrix of size N x N, where N is a power of 2, filled with numbers from 1 to N² in a specific "Z Pattern". The traversal order is upper-left, upper-right, lower-left, and lower-right quadrants recursively.

Input:

• An integer N (N is a power of 2).

Output:

• A 2D array of size N x N filled with numbers from 1 to N² in the "Z Pattern".

Constraints:

• 1 ≤ N ≤ 1024 (N is a power of 2).

Example:

Input: N = 2
Output: [
    [1, 2], 
    [3, 4]
]

Understanding the Problem

The core challenge is to fill the matrix in a specific recursive order. This problem is significant in understanding recursive patterns and matrix manipulations, which have applications in graphics, simulations, and more.

Potential pitfalls include misunderstanding the recursive traversal order and incorrectly handling the base cases.

Approach

To solve this problem, we can use a recursive function to fill the matrix. The naive approach would be to fill the matrix linearly, but this does not meet the "Z Pattern" requirement.

Naive Solution:

Fill the matrix linearly from 1 to N². This approach is not optimal as it does not follow the required pattern.

Optimized Solution:

We can use a recursive function to fill the matrix in the required "Z Pattern". The idea is to divide the matrix into four quadrants and fill them recursively in the order: upper-left, upper-right, lower-left, lower-right.

Thought Process:

• Start with the entire matrix and the initial number 1.
• Divide the matrix into four quadrants.
• Recursively fill each quadrant in the specified order.
• Base case: When the size of the quadrant is 1, fill it with the current number.

Algorithm

Here is a step-by-step breakdown of the algorithm:

1. Initialize the matrix with zeros.
2. Create a recursive function to fill the matrix.
3. In the recursive function, check if the size of the current quadrant is 1. If so, fill it with the current number and return.
4. Otherwise, divide the current quadrant into four smaller quadrants.
5. Recursively fill each smaller quadrant in the order: upper-left, upper-right, lower-left, lower-right.

Code Implementation

#include <iostream>
#include <vector>

using namespace std;

// Function to fill the matrix in Z pattern
void fillZPattern(vector<vector<int>> &matrix, int startRow, int startCol, int size, int &num) {
    if (size == 1) {
        matrix[startRow][startCol] = num++;
        return;
    }
    
    int halfSize = size / 2;
    
    // Upper-left quadrant
    fillZPattern(matrix, startRow, startCol, halfSize, num);
    // Upper-right quadrant
    fillZPattern(matrix, startRow, startCol + halfSize, halfSize, num);
    // Lower-left quadrant
    fillZPattern(matrix, startRow + halfSize, startCol, halfSize, num);
    // Lower-right quadrant
    fillZPattern(matrix, startRow + halfSize, startCol + halfSize, halfSize, num);
}

vector<vector<int>> generateZPattern(int N) {
    vector<vector<int>> matrix(N, vector<int>(N, 0));
    int num = 1;
    fillZPattern(matrix, 0, 0, N, num);
    return matrix;
}

int main() {
    int N = 4;
    vector<vector<int>> result = generateZPattern(N);
    
    for (const auto &row : result) {
        for (int val : row) {
            cout << val << " ";
        }
        cout << endl;
    }
    
    return 0;
}

Complexity Analysis

The time complexity of the recursive approach is O(N²) because we fill each cell of the N x N matrix exactly once. The space complexity is O(N²) for storing the matrix.

Edge Cases

Potential edge cases include:

• N = 1: The smallest possible matrix.
• N = 1024: The largest possible matrix within the given constraints.

Each algorithm handles these edge cases by correctly filling the matrix according to the "Z Pattern".

Testing

To test the solution comprehensively, consider the following test cases:

• Simple case: N = 2
• Medium case: N = 4
• Large case: N = 1024

Use a variety of test cases to ensure the solution works for all possible inputs.

Thinking and Problem-Solving Tips

When approaching such problems:

• Break down the problem into smaller sub-problems.
• Think recursively and identify base cases.
• Draw diagrams to visualize the problem and solution.
• Practice similar problems to improve problem-solving skills.

Conclusion

In this blog post, we discussed how to generate a matrix filled with numbers in a "Z Pattern" 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.

Additional Resources

For further reading and practice, consider the following resources:

• LeetCode - Practice coding problems.
• GeeksforGeeks - Tutorials and problem-solving tips.
• C++ Documentation - Official C++ documentation.