Design a 3D Binary Matrix with Efficient Layer Tracking Python/Java solution

Most implementations maintain an array of layer counts and a TreeSet (Java) or sorted structure such as heap/ordered container in Python. Each update modifies the count and reinserts the layer into the ordered structure. The same logic translates cleanly to C++, Go, Python, and Java.

What is the best approach for Design a 3D Binary Matrix with Efficient Layer Tracking?

The most efficient design uses a per-layer counter combined with an ordered set (balanced tree). Each update adjusts the layer's count and reinserts it into the ordered structure in O(log L) time. The best layer can then be retrieved instantly in O(1). This avoids scanning the entire matrix after each modification.

Is Design a 3D Binary Matrix with Efficient Layer Tracking asked at Google/Amazon/Meta?

Problems that combine matrix updates with ordered data structures are common in interviews at companies like Google, Amazon, and Meta. Variants frequently appear in system design-style coding rounds where candidates must maintain dynamic statistics efficiently.

What data structure is used in Design a 3D Binary Matrix with Efficient Layer Tracking?

The core structures are arrays or matrices for storing the grid, counters for each layer, and an ordered set or priority queue to keep layers sorted by their metric. Hash tables may also be used to map layers to their current counts during updates.

What is the time complexity of Design a 3D Binary Matrix with Efficient Layer Tracking?

The optimized solution runs updates in O(log L) time where L is the number of layers. Retrieving the current best layer is O(1) because the ordered set keeps layers sorted by their counts. Space complexity is O(L) for storing layer statistics and the ordering structure.

Description Hints 2 Solution Video 13:48

3391. Design a 3D Binary Matrix with Efficient Layer Tracking

Q: Is Design a 3D Binary Matrix with Efficient Layer Tracking easy or hard?

The problem is typically rated Medium because the logic is straightforward once the right data structure is chosen. The challenge lies in recognizing that recomputing layer statistics is inefficient and that maintaining them with an ordered set reduces operations to O(log L).

Q: How to solve Design a 3D Binary Matrix with Efficient Layer Tracking in O(log n)?

Maintain a counter for each layer that tracks how many cells contain 1. Store layers in an ordered set keyed by this count. When a cell changes value, update the counter and adjust its position in the set, which costs O(log L). The highest or lowest layer metric can then be accessed directly.

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You are given a n x n x n binary 3D array matrix.

Implement the Matrix3D class:

Matrix3D(int n) Initializes the object with the 3D binary array matrix, where all elements are initially set to 0.
void setCell(int x, int y, int z) Sets the value at matrix[x][y][z] to 1.
void unsetCell(int x, int y, int z) Sets the value at matrix[x][y][z] to 0.
int largestMatrix() Returns the index x where matrix[x] contains the most number of 1's. If there are multiple such indices, return the largest x.

Example 1:

Input:
["Matrix3D", "setCell", "largestMatrix", "setCell", "largestMatrix", "setCell", "largestMatrix"]
[[3], [0, 0, 0], [], [1, 1, 2], [], [0, 0, 1], []]

Output:
[null, null, 0, null, 1, null, 0]

Explanation

Matrix3D matrix3D = new Matrix3D(3); // Initializes a 3 x 3 x 3 3D array matrix, filled with all 0's.
matrix3D.setCell(0, 0, 0); // Sets matrix[0][0][0] to 1.
matrix3D.largestMatrix(); // Returns 0. matrix[0] has the most number of 1's.
matrix3D.setCell(1, 1, 2); // Sets matrix[1][1][2] to 1.
matrix3D.largestMatrix(); // Returns 1. matrix[0] and matrix[1] tie with the most number of 1's, but index 1 is bigger.
matrix3D.setCell(0, 0, 1); // Sets matrix[0][0][1] to 1.
matrix3D.largestMatrix(); // Returns 0. matrix[0] has the most number of 1's.

Example 2:

Input:
["Matrix3D", "setCell", "largestMatrix", "unsetCell", "largestMatrix"]
[[4], [2, 1, 1], [], [2, 1, 1], []]

Output:
[null, null, 2, null, 3]

Explanation

Matrix3D matrix3D = new Matrix3D(4); // Initializes a 4 x 4 x 4 3D array matrix, filled with all 0's.
matrix3D.setCell(2, 1, 1); // Sets matrix[2][1][1] to 1.
matrix3D.largestMatrix(); // Returns 2. matrix[2] has the most number of 1's.
matrix3D.unsetCell(2, 1, 1); // Sets matrix[2][1][1] to 0.
matrix3D.largestMatrix(); // Returns 3. All indices from 0 to 3 tie with the same number of 1's, but index 3 is the biggest.

Constraints:

1 <= n <= 100
0 <= x, y, z < n
At most 10⁵ calls are made in total to setCell and unsetCell.
At most 10⁴ calls are made to largestMatrix.

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Topics

Hints open one at a time

Hint 1

Keep track of the number of 1’s on each matrix.

Hint 2

Use a data structure that allows to efficiently insert, delete or get the biggest pair on a list of pairs.

13:48

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Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

["Matrix3D","setCell","largestMatrix","setCell","largestMatrix","setCell","largestMatrix"]
[[3],[0,0,0],[],[1,1,2],[],[0,0,1],[]]

Input

["Matrix3D","setCell","largestMatrix","unsetCell","largestMatrix"]
[[4],[2,1,1],[],[2,1,1],[]]

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