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 a3 x 3 x 3 3D array matrix, filled with all 0's.matrix[0][0][0] to 1.matrix[0] has the most number of 1's.matrix[1][1][2] to 1.matrix[0] and matrix[1] tie with the most number of 1's, but index 1 is bigger.matrix[0][0][1] to 1.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 a4 x 4 x 4 3D array matrix, filled with all 0's.matrix[2][1][1] to 1.matrix[2] has the most number of 1's.matrix[2][1][1] to 0.
Constraints:
1 <= n <= 1000 <= x, y, z < n105 calls are made in total to setCell and unsetCell.104 calls are made to largestMatrix.Problem Overview: You need to design a data structure that manages a 3D binary matrix and supports updates while efficiently tracking which layer currently satisfies a condition (typically the most filled or valid layer). A naive scan across all layers after every update is too slow, so the goal is to maintain this information incrementally.
Approach 1: Brute Force Layer Scan (O(L * R * C) time per query, O(1) space)
Store the matrix directly and recompute layer statistics whenever the answer is requested. For example, if the task is to identify the layer with the most 1s, iterate through every cell in each layer and recompute counts from scratch. This approach uses simple array and matrix traversal but becomes expensive when updates and queries are frequent. Each operation may require scanning an entire layer stack, which scales poorly as the 3D grid grows.
Approach 2: Counting per Layer (O(1) update, O(L) query, O(L) space)
Maintain a counter for each layer representing how many cells currently contain 1. When a cell flips value, update the corresponding layer count in constant time. Queries now only scan the L layer counters instead of the whole matrix. This dramatically reduces work compared to brute force, but repeated queries still require a linear scan across layers.
Approach 3: Counting + Ordered Set (O(log L) update, O(1) query, O(L) space)
The optimal design maintains two structures: a per-layer counter and an ordered set (or balanced tree) keyed by the layer metric such as the number of 1s. Each update modifies the layer’s count, so you remove the old entry from the ordered set and insert the updated one. Because the structure stays sorted, the best layer is always accessible from one end of the set. Insertions and removals cost O(log L), while retrieving the current best layer is O(1). This design avoids scanning layers entirely and keeps results accurate after every update.
Approach 4: Counting + Heap (O(log L) update, O(1) peek, O(L) space)
A heap (priority queue) can also track the best layer. Push updated layer states into the heap whenever counts change. Because heaps do not support efficient arbitrary updates, outdated entries may remain, so you lazily discard them when popped. This approach performs similarly to the ordered set version but requires extra checks for stale entries.
Recommended for interviews: Counting + Ordered Set is the cleanest design. It shows that you understand how to maintain aggregated statistics and keep them sorted efficiently. Starting with the brute-force scan demonstrates baseline reasoning, but the ordered set solution shows strong data-structure design skills and scales well under frequent updates.
We use a three-dimensional array g to represent the matrix, where g[x][y][z] represents the value at coordinate (x, y, z) in the matrix. We use an array cnt of length n to record the number of 1s in each layer. We use an ordered set sl to maintain the number of 1s and the layer number for each layer. The elements in sl are (cnt[x], x), so sl can be sorted in descending order by the number of 1s, and in descending order by layer number if the number of 1s is the same.
When calling the setCell method, we first check if (x, y, z) has already been set to 1. If it has, we return directly. Otherwise, we set g[x][y][z] to 1, remove (cnt[x], x) from sl, increment cnt[x] by 1, and add (cnt[x], x) to sl.
When calling the unsetCell method, we first check if (x, y, z) has already been set to 0. If it has, we return directly. Otherwise, we set g[x][y][z] to 0, remove (cnt[x], x) from sl, decrement cnt[x] by 1, and if cnt[x] is greater than 0, add (cnt[x], x) to sl.
When calling the largestMatrix method, we return the second value of the first element in sl. If sl is empty, we return n - 1.
In terms of time complexity, the setCell and unsetCell methods both have a time complexity of O(log n), and the largestMatrix method has a time complexity of O(1). The space complexity is O(n^3).
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Layer Scan | O(L * R * C) | O(1) | Small matrices or when updates and queries are extremely rare |
| Counting per Layer | O(1) update, O(L) query | O(L) | Moderate input sizes where query frequency is low |
| Counting + Ordered Set | O(log L) update, O(1) query | O(L) | General case with frequent updates and queries |
| Counting + Heap | O(log L) | O(L) | When a priority queue is simpler to implement than a balanced tree |
3391. Design a 3D Binary Matrix with Efficient Layer Tracking (Leetcode Medium) • Programming Live with Larry • 173 views views
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