LeetCode 3537 - Fill a Special Grid

The problem asks us to generate a special 2ⁿ x 2ⁿ grid filled with integers from 0 to 4ⁿ - 1 (note that 22n - 1 in the problem is likely a typo; it should be 4^n - 1) such that certain order constraints between quadrants are satisfied.

leetcodemediumarraydivide-and-conquermatrix

LeetCode Problem 3537

Difficulty: 🟡 Medium
Topics: Array, Divide and Conquer, Matrix

Solution

Problem Understanding

The problem asks us to generate a special 2ⁿ x 2ⁿ grid filled with integers from 0 to 4ⁿ - 1 (note that 22n - 1 in the problem is likely a typo; it should be 4^n - 1) such that certain order constraints between quadrants are satisfied. Specifically, the grid is divided recursively into four quadrants:

  • Top-right
  • Bottom-right
  • Bottom-left
  • Top-left

The numbers in these quadrants must satisfy a strict increasing order as you move clockwise starting from top-right: top-right < bottom-right < bottom-left < top-left. Additionally, each quadrant must itself be a special grid, which imposes a recursive structure.

The input n defines the size of the grid as 2ⁿ x 2ⁿ. For n = 0, the grid is a single cell containing 0. For n = 1, the grid is 2x2, and so on. The constraints guarantee that n <= 10, meaning the maximum grid size is 1024 x 1024, which is feasible for an algorithm with O(4^n) work because we can use recursion efficiently.

Edge cases include:

  • n = 0, which is a trivial single-cell grid.
  • Small n like 1 or 2 where it is easy to manually verify correctness.
  • Maximum n = 10 to ensure the algorithm scales without stack overflow or memory issues.

The key insight is recognizing that this is a recursive divide-and-conquer problem, where each quadrant is another smaller special grid, and we can systematically assign ranges of numbers to quadrants to maintain the ordering constraint.

Approaches

Brute Force Approach: One could try to generate all permutations of numbers from 0 to 4ⁿ - 1 and check each arrangement for the special property. While this is technically correct, it is clearly infeasible because the number of permutations grows factorially with the grid size. Even for n = 2, there are 16! permutations, which is prohibitively large.

Optimal Approach: The problem naturally suggests a recursive construction. The main observation is that the grid can be built recursively:

  • For a 1x1 grid, simply return [[start]].
  • For larger grids, split the range of numbers into four equal-sized chunks and assign them to quadrants in the order top-right, bottom-right, bottom-left, top-left. Recursively fill each quadrant with its respective subrange, incrementing numbers appropriately to maintain the ordering property.

This approach guarantees that the ordering between quadrants is maintained, and recursion ensures that each quadrant itself is a special grid.

Approach Time Complexity Space Complexity Notes
Brute Force O((4^n)!) O(4^n) Generate all permutations and check special property; infeasible
Optimal O(4^n) O(4^n) Recursive divide-and-conquer filling of quadrants; efficient and scales to n=10

Algorithm Walkthrough

  1. Define a recursive function fill_grid(size, start) that fills a size x size grid starting from the number start.
  2. If size == 1, return a 1x1 grid containing start. This is the base case.
  3. Otherwise, divide the grid into four quadrants of size half x half where half = size // 2.
  4. Compute the number of elements in one quadrant: quad_count = half * half.
  5. Recursively fill the top-right quadrant starting from start.
  6. Recursively fill the bottom-right quadrant starting from start + quad_count.
  7. Recursively fill the bottom-left quadrant starting from start + 2 * quad_count.
  8. Recursively fill the top-left quadrant starting from start + 3 * quad_count.
  9. Merge the four quadrants into a single size x size grid using standard 2D list concatenation, respecting their positions.
  10. Return the merged grid to the caller.

Why it works: Each quadrant receives a distinct, non-overlapping range of numbers. Since the ranges assigned to quadrants increase in the order top-right < bottom-right < bottom-left < top-left, the quadrant ordering condition is guaranteed. Recursion ensures that within each quadrant, the property holds as well.

Python Solution

from typing import List

class Solution:
    def specialGrid(self, n: int) -> List[List[int]]:
        def fill_grid(size: int, start: int) -> List[List[int]]:
            if size == 1:
                return [[start]]
            half = size // 2
            quad_count = half * half
            tr = fill_grid(half, start)
            br = fill_grid(half, start + quad_count)
            bl = fill_grid(half, start + 2 * quad_count)
            tl = fill_grid(half, start + 3 * quad_count)
            
            grid = []
            for top_row_tr, top_row_tl in zip(tr, tl):
                grid.append(top_row_tr + top_row_tl)
            for bottom_row_br, bottom_row_bl in zip(br, bl):
                grid.append(bottom_row_br + bottom_row_bl)
            return grid
        
        size = 2 ** n
        return fill_grid(size, 0)

This implementation first calculates the size of the grid as 2^n. It then recursively constructs the grid by dividing it into quadrants and assigning sequential ranges of numbers. The zip function merges the corresponding rows from quadrants into the correct positions.

Go Solution

func specialGrid(n int) [][]int {
    var fillGrid func(size int, start int) [][]int
    fillGrid = func(size int, start int) [][]int {
        if size == 1 {
            return [][]int{{start}}
        }
        half := size / 2
        quadCount := half * half
        tr := fillGrid(half, start)
        br := fillGrid(half, start + quadCount)
        bl := fillGrid(half, start + 2 * quadCount)
        tl := fillGrid(half, start + 3 * quadCount)
        
        grid := make([][]int, 0, size)
        for i := 0; i < half; i++ {
            row := append(tr[i], tl[i]...)
            grid = append(grid, row)
        }
        for i := 0; i < half; i++ {
            row := append(br[i], bl[i]...)
            grid = append(grid, row)
        }
        return grid
    }
    
    size := 1 << n
    return fillGrid(size, 0)
}

In Go, slices are used to construct and merge the quadrants. The append function is used to concatenate rows. The logic is otherwise identical to Python.

Worked Examples

Example n=1

Start with size = 2. The base case for 1x1 grids produces four numbers:

  • top-right: 0
  • bottom-right: 1
  • bottom-left: 2
  • top-left: 3

Merged:

[3,0]
[2,1]

This satisfies the special grid properties.

Example n=2

Start with size = 4. Each quadrant is filled recursively with 2x2 grids:

  • top-right: numbers 0-3
  • bottom-right: numbers 4-7
  • bottom-left: numbers 8-11
  • top-left: numbers 12-15

Merged 4x4 grid:

[[15,12,3,0],
 [14,13,2,1],
 [11,8,7,4],
 [10,9,6,5]]

Each quadrant is itself a 2x2 special grid, and the quadrant ordering is maintained.

Complexity Analysis

Measure Complexity Explanation
Time O(4^n) Each recursive call processes all cells once; there are 4^n cells in a 2^n x 2^n grid.
Space O(4^n) Stores the entire grid; recursion stack is O(log n), negligible compared to grid storage.

The recursion ensures we touch each cell exactly once, and constructing merged rows takes linear time per row.

Test Cases

sol = Solution()

# Provided examples
assert sol.specialGrid(0) == [[0]]  # single cell
assert sol.specialGrid(1) == [[3,0],[2,1]]  # 2x2 grid
assert sol.specialGrid(2) == [[15,12,3,0],[14,13,2,1],[11,8,7,4],[10,9,6,5]]  # 4x4 grid

# Edge cases
assert sol.specialGrid(3)[0][0] == 63  # top-left cell for n=3 should be the largest
assert sol.specialGrid(3)[7][7] == 0   # bottom-right cell should be the smallest

# Maximum size stress test (does not verify exact content, just ensures no crash)
res = sol.specialGrid(5)
assert len(res) == 32 and len(res[0]) == 32
Test Why
n=0 Base case, trivial 1x1 grid
n=1 Smallest non-tr