Logical OR of Two Binary Grids Represented as

Tested Python solution for LeetCode 558 with 18 pytest cases. Generate a practice environment with lcpy.

LeetCode 558, Medium. Topics: Divide and Conquer, Tree. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 18 parametrized pytest cases, and a playground notebook:

lcpy gen -n 558   # by problem number
lcpy gen -s logical_or_of_two_binary_grids_represented_as_quad_trees   # by problem name

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Problem

A Binary Matrix is a matrix in which all the elements are either 0 or 1.

Given quadTree1 and quadTree2. quadTree1 represents a n * n binary matrix and quadTree2 represents another n * n binary matrix.

Return a Quad-Tree representing the n * n binary matrix which is the result of logical bitwise OR of the two binary matrixes represented by quadTree1 and quadTree2.

Notice that you can assign the value of a node to True or False when isLeaf is False, and both are accepted in the answer.

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.
• isLeaf: True if the node is leaf node on the tree or False if the node has the 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: The input/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 a 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.

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Examples

Input: quadTree1 = [[0,1],[1,1],[1,1],[1,0],[1,0]]
, quadTree2 = [[0,1],[1,1],[0,1],[1,1],[1,0],null,null,null,null,[1,0],[1,0],[1,1],[1,1]]
Output: [[0,0],[1,1],[1,1],[1,1],[1,0]]
Explanation: quadTree1 and quadTree2 are shown above. You can see the binary matrix which is represented by each Quad-Tree.
If we apply logical bitwise OR on the two binary matrices we get the binary matrix below which is represented by the result Quad-Tree.
Notice that the binary matrices shown are only for illustration, you don't have to construct the binary matrix to get the result tree.

Input: quadTree1 = [[1,0]], quadTree2 = [[1,0]]
Output: [[1,0]]
Explanation: Each tree represents a binary matrix of size 1*1. Each matrix contains only zero.
The resulting matrix is of size 1*1 with also zero.

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Constraints

• quadTree1 and quadTree2 are both valid Quad-Trees each representing a n * n grid.
• n == 2^x where 0 <= x <= 9.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

from __future__ import annotations


# ruff: noqa: N803
class Node:
    def __init__(
        self,
        val: bool,
        isLeaf: bool,
        topLeft: Node | None = None,
        topRight: Node | None = None,
        bottomLeft: Node | None = None,
        bottomRight: Node | None = None,
    ) -> None:
        self.val = val
        self.isLeaf = isLeaf
        self.topLeft = topLeft
        self.topRight = topRight
        self.bottomLeft = bottomLeft
        self.bottomRight = bottomRight


class Solution:
    # Time: O(n)
    # Space: O(log n) recursion depth
    def intersect(self, quad_tree1: Node, quad_tree2: Node) -> Node:
        if quad_tree1.isLeaf:
            return Node(True, True) if quad_tree1.val else quad_tree2
        if quad_tree2.isLeaf:
            return Node(True, True) if quad_tree2.val else quad_tree1
        quadrants = (
            (quad_tree1.topLeft, quad_tree2.topLeft),
            (quad_tree1.topRight, quad_tree2.topRight),
            (quad_tree1.bottomLeft, quad_tree2.bottomLeft),
            (quad_tree1.bottomRight, quad_tree2.bottomRight),
        )
        merged: list[Node] = []
        for left, right in quadrants:
            assert left is not None and right is not None
            merged.append(self.intersect(left, right))
        if all(child.isLeaf and child.val for child in merged):
            return Node(True, True)
        top_left, top_right, bottom_left, bottom_right = merged
        return Node(False, False, top_left, top_right, bottom_left, bottom_right)

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Complexity

Time	Space
O(n)	O(log n) recursion depth

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