100626: Fill a Special Grid

Difficulty: Medium

Topics: Recursion, Matrix Manipulation

Companies: Microsoft, Meta, NVIDIA

Pro Tip: This problem requires understanding of recursive quadrant filling and mathematical patterns. Optimal solutions run in O(4^N) time complexity.

Problem Statement

Construct a 2^N x 2^N grid where each quadrant follows strict ordering rules recursively. Numbers must range from 0 to 2^2N - 1.

Example 1

Input: N = 1
Output: [[3,0],[2,1]]
Explanation: Quadrants ordered as 0 < 1 < 2 < 3

Example 2

Input: N = 2
Output: [[15,12,3,0],[14,13,2,1],[11,8,7,4],[10,9,6,5]]
Explanation: Each quadrant maintains required ordering recursively

Problem Link: View on LeetCode ↗

Approach 1: Recursive Quadrant Filling

Divide grid into quadrants and fill recursively following the ordering rules.

Algorithm Steps

Base case: 1x1 grid contains 0
For larger grids:
- Divide into four 2^N-1 x 2^N-1 quadrants
- Fill quadrants in reverse order (TR → BR → BL → TL)
- Offset values based on quadrant position

Complexity Analysis

Time Complexity	Space Complexity
O(4^N)	O(4^N)

Recursive Solution

class Solution {
public:
    vector<vector<int>> specialGrid(int N) {
        int size = pow(2, N);
        vector<vector<int>> grid(size, vector<int>(size));
        fill(grid, 0, 0, size, 0);
        return grid;
    }
    
    void fill(vector<vector<int>>& grid, int x, int y, int size, int start) {
        if(size == 1) {
            grid[x][y] = start;
            return;
        }
        
        int half = size / 2;
        int quarter = (size * size) / 4;
        
        // Fill quadrants in TR → BR → BL → TL order
        fill(grid, x, y + half, half, start);                // Top-right
        fill(grid, x + half, y + half, half, start + quarter);   // Bottom-right
        fill(grid, x + half, y, half, start + 2*quarter);  // Bottom-left
        fill(grid, x, y, half, start + 3*quarter);          // Top-left
    }
};

class Solution {
    public int[][] specialGrid(int N) {
        int size = (int) Math.pow(2, N);
        int[][] grid = new int[size][size];
        fill(grid, 0, 0, size, 0);
        return grid;
    }
    
    private void fill(int[][] grid, int x, int y, int size, int start) {
        if(size == 1) {
            grid[x][y] = start;
            return;
        }
        
        int half = size / 2;
        int quarter = (size * size) / 4;
        
        fill(grid, x, y + half, half, start);              // Top-right
        fill(grid, x + half, y + half, half, start + quarter); // Bottom-right
        fill(grid, x + half, y, half, start + 2*quarter);    // Bottom-left
        fill(grid, x, y, half, start + 3*quarter);          // Top-left
    }
}

class Solution:
    def specialGrid(self, N: int) -> List[List[int]]:
        size = 2 ** N
        grid = [[0] * size for _ in range(size)]
        self.fill(grid, 0, 0, size, 0)
        return grid
    
    def fill(self, grid, x, y, size, start):
        if size == 1:
            grid[x][y] = start
            return
        
        half = size // 2
        quarter = (size * size) // 4
        
        self.fill(grid, x, y + half, half, start)          # Top-right
        self.fill(grid, x + half, y + half, half, start + quarter) # Bottom-right
        self.fill(grid, x + half, y, half, start + 2*quarter)  # Bottom-left
        self.fill(grid, x, y, half, start + 3*quarter)      # Top-left

Implementation Note: This recursive approach fills quadrants in reverse order (TR → BR → BL → TL) with appropriate value offsets to maintain ordering.

Approach 2: Mathematical Pattern Construction

Use bit manipulation and coordinate patterns to calculate values directly.

Algorithm Steps

For each cell (i,j):
- Determine quadrant level
- Calculate value based on quadrant position
- Use bitwise operations to mirror coordinates

Complexity Analysis

Time Complexity	Space Complexity
O(4^N)	O(4^N)

Mathematical Solution

class Solution {
public:
    vector<vector<int>> specialGrid(int N) {
        int size = pow(2, N);
        vector<vector<int>> grid(size, vector<int>(size));
        
        for(int i = 0; i < size; ++i) {
            for(int j = 0; j < size; ++j) {
                int val = 0;
                for(int k = N-1; k >= 0; --k) {
                    int mask = 1 << k;
                    int x = (i & mask) ? 1 : 0;
                    int y = (j & mask) ? 1 : 0;
                    val = (val << 2) | ((3 - 2*x - y) & 3);
                }
                grid[i][j] = val;
            }
        }
        return grid;
    }
};

class Solution {
    public int[][] specialGrid(int N) {
        int size = (int) Math.pow(2, N);
        int[][] grid = new int[size][size];
        
        for(int i = 0; i < size; ++i) {
            for(int j = 0; j < size; ++j) {
                int val = 0;
                for(int k = N-1; k >= 0; --k) {
                    int mask = 1 << k;
                    int x = (i & mask) != 0 ? 1 : 0;
                    int y = (j & mask) != 0 ? 1 : 0;
                    val = (val << 2) | ((3 - 2*x - y) & 3);
                }
                grid[i][j] = val;
            }
        }
        return grid;
    }
}

class Solution:
    def specialGrid(self, N: int) -> List[List[int]]:
        size = 2 ** N
        grid = [[0] * size for _ in range(size)]
        
        for i in range(size):
            for j in range(size):
                val = 0
                for k in range(N-1, -1, -1):
                    mask = 1 << k
                    x = (i & mask) != 0
                    y = (j & mask) != 0
                    val = (val << 2) | ((3 - 2*x - y) & 3)
                grid[i][j] = val
        return grid

Implementation Note: This mathematical approach uses bitwise operations to calculate cell values directly based on their coordinates, avoiding recursion.

Edge Cases and Special Considerations

Key edge cases to consider when implementing your solution:

1. Base Case (N=0)

Input: 0 → Output: [[0]]
Explanation: Single cell grid with value 0

2. N=1 Verification

Input: 1 → Output: [[3,0],[2,1]]
Explanation: Verify quadrant ordering TR=0, BR=1, BL=2, TL=3

3. Large N Values

Input: 3 → Output: 8x8 grid
Explanation: Ensure recursive pattern holds for deep nesting

Important: Test all quadrant boundaries and ensure recursive calls maintain proper offset calculations.

Frequently Asked Questions

1. Why does the recursive approach work?

The recursive approach maintains the quadrant ordering by assigning value ranges that ensure each quadrant's maximum is less than the next quadrant's minimum.

2. How does the mathematical solution work?

It encodes quadrant positions at each level using bit patterns, calculating the value through bitwise operations that mirror the recursive structure.

3. Can we optimize further?

Both approaches are optimal for their paradigms. The mathematical solution avoids recursion stack overhead but may be harder to understand.

4. How to handle very large N?

Both approaches work up to N=15 (32768x32768 grid) within memory limits. For N>15, consider iterative approaches with memory optimization.

5. Why reverse order in recursion?

Filling TR first ensures smaller values are placed first, followed by larger values in subsequent quadrants to maintain the required ordering.

Final Thoughts: This problem demonstrates beautiful recursive patterns and mathematical relationships in grid construction. The recursive approach is more intuitive, while the mathematical solution offers deeper insights into bit manipulation patterns.