Special Grid Quest: Rule of Four

Special Grid Generator – Divide and Rule in Action

Problem Statement

You're given a non-negative integer N, and you need to generate a 2^N × 2^N grid filled with integers from 0 to 4^N - 1.

The twist? The grid has to be special, satisfying all the following rules:

  • All numbers in the top-right quadrant must be smaller than those in the bottom-right quadrant.

  • All numbers in the bottom-right quadrant must be smaller than those in the bottom-left quadrant.

  • All numbers in the bottom-left quadrant must be smaller than those in the top-left quadrant.

  • Each of the four quadrants must recursively also be a special grid.

Note: A 1x1 grid (when N = 0) is trivially special.

Example

Let’s take N = 1 (so grid size = 2x2)

Output:

[[3, 0],
 [2, 1]]

Explanation:

  • Top-right quadrant: 0

  • Bottom-right quadrant: 1

  • Bottom-left quadrant: 2

  • Top-left quadrant: 3

Ordering: 0 < 1 < 2 < 3 — perfect.
Each quadrant is 1x1, which is trivially special.

Best Data Structure for the Job

A simple 2D array is all we need. This problem relies on recursive logic, not any advanced data structures.

Different Approaches

Approach 1: Brute Force (Educational but Slow)

For small values of N, try all permutations of the numbers 0 to 4^N - 1, convert each permutation into a grid, and validate whether the grid is special by checking all the rules.

Use this only when N = 0 or 1, because it becomes computationally impossible beyond that.

import java.util.*;

public class BruteForceSpecialGrid {
    public List<int[][]> generateAllSpecialGrids(int N) {
        int size = 1 << N;
        int total = size * size;
        List<int[][]> results = new ArrayList<>();

        List<Integer> nums = new ArrayList<>();
        for (int i = 0; i < total; i++) {
            nums.add(i);
        }

        permute(nums, 0, size, results);
        return results;
    }

    private void permute(List<Integer> nums, int start, int size, List<int[][]> results) {
        if (start == nums.size()) {
            int[][] grid = toGrid(nums, size);
            if (isSpecial(grid, 0, 0, size)) {
                results.add(grid);
            }
            return;
        }

        for (int i = start; i < nums.size(); i++) {
            Collections.swap(nums, start, i);
            permute(nums, start + 1, size, results);
            Collections.swap(nums, start, i);
        }
    }

    private int[][] toGrid(List<Integer> nums, int size) {
        int[][] grid = new int[size][size];
        for (int i = 0; i < nums.size(); i++) {
            grid[i / size][i % size] = nums.get(i);
        }
        return grid;
    }

    private boolean isSpecial(int[][] grid, int row, int col, int size) {
        if (size == 1) return true;

        int newSize = size / 2;

        int[] max = new int[4];
        int[] min = new int[4];
        Arrays.fill(max, Integer.MIN_VALUE);
        Arrays.fill(min, Integer.MAX_VALUE);

        // Fill max and min for each quadrant
        for (int r = 0; r < size; r++) {
            for (int c = 0; c < size; c++) {
                int quadrant = (r >= newSize ? 2 : 0) + (c >= newSize ? 1 : 0);
                int val = grid[row + r][col + c];
                max[quadrant] = Math.max(max[quadrant], val);
                min[quadrant] = Math.min(min[quadrant], val);
            }
        }

        if (!(max[1] < min[3] && max[3] < min[2] && max[2] < min[0])) return false;

        return isSpecial(grid, row, col, newSize) &&
               isSpecial(grid, row, col + newSize, newSize) &&
               isSpecial(grid, row + newSize, col, newSize) &&
               isSpecial(grid, row + newSize, col + newSize, newSize);
    }
}

Approach 2: Optimal Recursive Divide & Conquer

Instead of generating all permutations, use recursion to fill each quadrant based on the required rules. Each quadrant gets numbers in a fixed order with specific offsets, ensuring global ordering without validation.

public class SpecialGridGenerator {
    public int[][] generateGrid(int N) {
        int size = 1 << N; // 2^N
        int[][] grid = new int[size][size];
        fill(grid, 0, 0, size, 0);
        return grid;
    }

    private void fill(int[][] grid, int row, int col, int size, int base) {
        if (size == 1) {
            grid[row][col] = base;
            return;
        }

        int newSize = size / 2;
        int offset = newSize * newSize;

        // Recursively fill each quadrant in required order
        fill(grid, row, col + newSize, newSize, base);                  // Top-right
        fill(grid, row + newSize, col + newSize, newSize, base + offset); // Bottom-right
        fill(grid, row + newSize, col, newSize, base + 2 * offset);     // Bottom-left
        fill(grid, row, col, newSize, base + 3 * offset);               // Top-left
    }
}

Common Mistakes and Pitfalls

  • Assuming it’s a dynamic programming problem — it’s pure recursion.

  • Getting quadrant positions mixed up during recursion.

  • Failing to assign increasing offset values across recursive calls.

  • Brute-forcing for large N — it's only good for small-scale verification.

Time and Space Complexity Analysis

Approach    Time Complexity       Space Complexity
Brute Force               O((4^N)!) (factorial time)             O(4^N)
Optimal Recursive               O(4^N)             O(4^N)

Real-world Applications

  • Procedural content generation (games, terrain, levels)

  • Grid layout design with hierarchical priority zones

  • Recursive structures like quadtrees or matrix partitioning

  • Fractal/grid-based simulations and recursive puzzle generation


I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

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