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Fill a Special Grid
Fill a Special Grid uses recursive divide-and-conquer to populate a 2^n x 2^n matrix with sequential integers uniquely in each quadrant.
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Practice Focus
Medium · Array plus Divide and Conquer
Answer-first summary
Fill a Special Grid uses recursive divide-and-conquer to populate a 2^n x 2^n matrix with sequential integers uniquely in each quadrant.
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Start by placing the smallest 2x2 special grid manually and expand recursively. Divide the grid into quadrants, assign ranges of integers, and combine them carefully. This ensures each subgrid remains special while maintaining sequential integer order across the entire 2^n x 2^n matrix.
Problem Statement
Given a non-negative integer n, generate a 2^n x 2^n grid filled with integers from 0 to 2^(2n) - 1. The grid is special if each quadrant is a special grid itself and integers are assigned uniquely in a divide-and-conquer fashion.
Return the resulting 2^n x 2^n special grid. Any 1x1 grid is trivially special. Implement a recursive solution that respects the integer placement and quadrant hierarchy.
Examples
Example 1
Input: n = 0
Output: [[0]]
The only number that can be placed is 0, and there is only one possible position in the grid.
Example 2
Input: n = 1
Output: [[3,0],[2,1]]
The numbers in each quadrant are: Since 0 < 1 < 2 < 3 , this satisfies the given constraints.
Example 3
Input: n = 2
Output: [[15,12,3,0],[14,13,2,1],[11,8,7,4],[10,9,6,5]]
The numbers in each quadrant are: This satisfies the first three requirements. Additionally, each quadrant is also a special grid. Thus, this is a special grid.
Constraints
- 0 <= n <= 10
Solution Approach
Recursive Quadrant Filling
Divide the grid into four quadrants, recursively fill each quadrant with a distinct integer range, and merge them. Base case occurs when n = 0 with a single cell.
Integer Range Management
Track the current integer range for each quadrant carefully to prevent overlaps. Increment ranges as you recurse down to ensure sequential placement from 0 to 2^(2n)-1.
Combine Quadrants
After filling sub-quadrants, merge them in the correct order: top-left, top-right, bottom-left, bottom-right. This maintains the special property while preserving recursive integrity.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time and space depend on the recursion depth, which is O(n). Each level handles four quadrants of half size, resulting in O(4^n) operations for a full 2^n x 2^n grid. Space grows with recursion stack and storage for subgrids.
What Interviewers Usually Probe
- Look for a divide-and-conquer approach instead of iterative filling.
- Expect careful tracking of integer ranges for each quadrant.
- Recursive correctness and merging order are key to a valid solution.
Common Pitfalls or Variants
Common pitfalls
- Failing to properly offset integer ranges leads to duplicate numbers.
- Incorrect quadrant merging breaks the special property.
- Attempting an iterative approach without recursion may cause confusion and mistakes.
Follow-up variants
- Fill a 2^n x 2^n grid in spiral order while maintaining uniqueness.
- Modify the grid so each quadrant sums to the same total.
- Generate a special grid with different base patterns for n=0 instead of starting at 0.
FAQ
What is the key pattern used in Fill a Special Grid?
The main pattern is array plus divide-and-conquer, recursively dividing the grid into quadrants and filling each with sequential integers.
What is the base case when n = 0?
A 1x1 grid containing 0 is trivially special and serves as the base case for recursion.
How do you prevent integer overlap in quadrants?
By assigning distinct ranges of integers to each quadrant before recursing, ensuring no duplicates occur across the grid.
Can this problem be solved iteratively?
Iterative solutions are possible but tend to be error-prone; recursion aligns naturally with the divide-and-conquer structure.
What is the time complexity of filling a 2^n x 2^n special grid?
The complexity grows exponentially with n, roughly O(4^n), due to recursively filling four sub-quadrants at each level.
Solution
Solution 1
#### Python3
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