Scale Engineer

Construct Quad Tree

Med
#0414Time: O(N^2 * log N). For each node in the Quad Tree, we iterate over its corresponding sub-grid. The recurrence relation is `T(N) = 4 * T(N/2) + O(N^2)`. At each level of recursion, the total work for the homogeneity check across all nodes at that level is `O(N^2)`. Since the depth of the tree is `log N`, the total time complexity is `O(N^2 * log N)`.Space: O(N^2). The space is dominated by the storage for the Quad Tree itself, which can have `O(N^2)` nodes in the worst-case scenario (e.g., a checkerboard pattern). The recursion stack depth contributes `O(log N)`.1 company
Algorithms
Divide and Conquer
Data structures
ArrayMatrixTree
Companies
Palantir Technologies

Prompt

Given a n * n matrix grid of 0's and 1's only. We want to represent grid with a Quad-Tree.

Return the root of the Quad-Tree representing grid.

A Quad-Tree is a tree data structure in which each internal node has exactly four children. Besides, each node has two attributes:

  • val: True if the node represents a grid of 1's or False if the node represents a grid of 0's. Notice that you can assign the val to True or False when isLeaf is False, and both are accepted in the answer.
  • isLeaf: True if the node is a leaf node on the tree or False if the node has four children.
class Node {
    public boolean val;
    public boolean isLeaf;
    public Node topLeft;
    public Node topRight;
    public Node bottomLeft;
    public Node bottomRight;
}

We can construct a Quad-Tree from a two-dimensional area using the following steps:

  1. If the current grid has the same value (i.e all 1's or all 0's) set isLeaf True and set val to the value of the grid and set the four children to Null and stop.
  2. If the current grid has different values, set isLeaf to False and set val to any value and divide the current grid into four sub-grids as shown in the photo.
  3. Recurse for each of the children with the proper sub-grid.

If you want to know more about the Quad-Tree, you can refer to the wiki.

Quad-Tree format:

You don't need to read this section for solving the problem. This is only if you want to understand the output format here. The output represents the serialized format of a Quad-Tree using level order traversal, where null signifies a path terminator where no node exists below.

It is very similar to the serialization of the binary tree. The only difference is that the node is represented as a list [isLeaf, val].

If the value of isLeaf or val is True we represent it as 1 in the list [isLeaf, val] and if the value of isLeaf or val is False we represent it as 0.

 

Example 1:

Input: grid = [[0,1],[1,0]]
Output: [[0,1],[1,0],[1,1],[1,1],[1,0]]
Explanation: The explanation of this example is shown below:
Notice that 0 represents False and 1 represents True in the photo representing the Quad-Tree.

Example 2:

Input: grid = [[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,1,1,1,1],[1,1,1,1,1,1,1,1],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0]]
Output: [[0,1],[1,1],[0,1],[1,1],[1,0],null,null,null,null,[1,0],[1,0],[1,1],[1,1]]
Explanation: All values in the grid are not the same. We divide the grid into four sub-grids.
The topLeft, bottomLeft and bottomRight each has the same value.
The topRight have different values so we divide it into 4 sub-grids where each has the same value.
Explanation is shown in the photo below:

 

Constraints:

  • n == grid.length == grid[i].length
  • n == 2x where 0 <= x <= 6

Approaches

3 approaches with complexity analysis and trade-offs.

This approach directly implements the recursive definition of the Quad Tree construction. A helper function is defined that takes the coordinates and size of the current sub-grid. This function first checks if all values in the sub-grid are the same. If they are, it creates a leaf node. If not, it creates an internal node, divides the grid into four equal quadrants, and makes a recursive call for each quadrant to build the sub-trees.

Algorithm

  • Create a recursive helper function build(grid, row, col, size).
  • In the helper function, check if the sub-grid from (row, col) of size x size is homogeneous.
  • To check for homogeneity, iterate through all size * size cells and compare them to the first cell's value.
  • If the sub-grid is homogeneous, create and return a leaf node with isLeaf=true and val set to the common value.
  • If the sub-grid is not homogeneous, create an internal node with isLeaf=false.
  • Recursively call build for the four quadrants:
    • topLeft = build(grid, row, col, size/2)
    • topRight = build(grid, row, col + size/2, size/2)
    • bottomLeft = build(grid, row + size/2, col, size/2)
    • bottomRight = build(grid, row + size/2, col + size/2, size/2)
  • Assign the results to the children of the internal node and return it.
  • The initial call is construct(grid) which calls build(grid, 0, 0, grid.length).

Walkthrough

The core of this method is a recursive function, let's call it build(row, col, size). This function is responsible for constructing the Quad Tree for the size x size sub-grid starting at (row, col).

The process within build(row, col, size) is as follows:

  1. Homogeneity Check: First, we determine if the sub-grid is homogeneous (contains all 0s or all 1s). We can do this by taking the value of the top-left cell, grid[row][col], and then iterating through all the cells in the size x size sub-grid. If we find any cell with a different value, the grid is not homogeneous.
  2. Leaf Node Creation: If the check passes (the grid is homogeneous), we create a new Node. We set isLeaf to true, val to the common value (true for 1, false for 0), and all four children pointers to null. This node is then returned.
  3. Internal Node Creation: If the check fails (the grid has mixed values), we create an internal node. We set isLeaf to false and val to an arbitrary value (e.g., true). Then, we divide the current size x size grid into four (size/2) x (size/2) sub-grids and make recursive calls to build for each of these four sub-grids, assigning the returned nodes to the topLeft, topRight, bottomLeft, and bottomRight children of the current internal node. This node is then returned.

The initial call to start the process would be build(0, 0, n), where n is the dimension of the input grid.

class Solution {    public Node construct(int[][] grid) {        return build(grid, 0, 0, grid.length);    }     private Node build(int[][] grid, int row, int col, int size) {        // Check if the sub-grid is homogeneous        boolean isHomogeneous = true;        int firstVal = grid[row][col];        for (int i = row; i < row + size; i++) {            for (int j = col; j < col + size; j++) {                if (grid[i][j] != firstVal) {                    isHomogeneous = false;                    break;                }            }            if (!isHomogeneous) {                break;            }        }         if (isHomogeneous) {            return new Node(firstVal == 1, true, null, null, null, null);        } else {            int newSize = size / 2;            Node topLeft = build(grid, row, col, newSize);            Node topRight = build(grid, row, col + newSize, newSize);            Node bottomLeft = build(grid, row + newSize, col, newSize);            Node bottomRight = build(grid, row + newSize, col + newSize, newSize);            // The 'val' for an internal node can be arbitrary, e.g., true.            return new Node(true, false, topLeft, topRight, bottomLeft, bottomRight);        }    }}

Complexity

Time

O(N^2 * log N). For each node in the Quad Tree, we iterate over its corresponding sub-grid. The recurrence relation is `T(N) = 4 * T(N/2) + O(N^2)`. At each level of recursion, the total work for the homogeneity check across all nodes at that level is `O(N^2)`. Since the depth of the tree is `log N`, the total time complexity is `O(N^2 * log N)`.

Space

O(N^2). The space is dominated by the storage for the Quad Tree itself, which can have `O(N^2)` nodes in the worst-case scenario (e.g., a checkerboard pattern). The recursion stack depth contributes `O(log N)`.

Trade-offs

Pros

  • Simple to understand and implement as it directly follows the problem definition.

  • Does not require any extra data structures for pre-computation.

Cons

  • Inefficient due to redundant computations. The homogeneity check for a sub-grid re-scans cells that were already scanned by its parent's check.

Solutions

/* // Definition for a QuadTree node. class Node { public boolean val; public boolean isLeaf; public Node topLeft; public Node topRight; public Node bottomLeft; public Node bottomRight; public Node() { this.val = false; this.isLeaf = false; this.topLeft = null; this.topRight = null; this.bottomLeft = null; this.bottomRight = null; } public Node(boolean val, boolean isLeaf) { this.val = val; this.isLeaf = isLeaf; this.topLeft = null; this.topRight = null; this.bottomLeft = null; this.bottomRight = null; } public Node(boolean val, boolean isLeaf, Node topLeft, Node topRight, Node bottomLeft, Node bottomRight) { this.val = val; this.isLeaf = isLeaf; this.topLeft = topLeft; this.topRight = topRight; this.bottomLeft = bottomLeft; this.bottomRight = bottomRight; } }; */ class Solution { public Node construct ( int [][] grid ) { return dfs ( 0 , 0 , grid . length - 1 , grid [ 0 ]. length - 1 , grid ); } private Node dfs ( int a , int b , int c , int d , int [][] grid ) { int zero = 0 , one = 0 ; for ( int i = a ; i <= c ; ++ i ) { for ( int j = b ; j <= d ; ++ j ) { if ( grid [ i ][ j ] == 0 ) { zero = 1 ; } else { one = 1 ; } } } boolean isLeaf = zero + one == 1 ; boolean val = isLeaf && one == 1 ; Node node = new Node ( val , isLeaf ); if ( isLeaf ) { return node ; } node . topLeft = dfs ( a , b , ( a + c ) / 2 , ( b + d ) / 2 , grid ); node . topRight = dfs ( a , ( b + d ) / 2 + 1 , ( a + c ) / 2 , d , grid ); node . bottomLeft = dfs (( a + c ) / 2 + 1 , b , c , ( b + d ) / 2 , grid ); node . bottomRight = dfs (( a + c ) / 2 + 1 , ( b + d ) / 2 + 1 , c , d , grid ); return node ; } }

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