Problem Description

You need to implement a class that manages a 3D binary matrix (a cube) of size n x n x n where each cell can contain either 0 or 1.

The class Matrix3D should support the following operations:

1. Constructor Matrix3D(int n): Creates a 3D binary matrix of dimensions n x n x n with all cells initially set to 0.

2. setCell(int x, int y, int z): Sets the cell at position (x, y, z) to 1. The position is specified by:

   • x: the layer index (0 to n-1)
   • y: the row index within layer x (0 to n-1)
   • z: the column index within row y (0 to n-1)

3. unsetCell(int x, int y, int z): Sets the cell at position (x, y, z) to 0.

4. largestMatrix(): Returns the layer index x that contains the maximum number of 1s. If multiple layers have the same maximum count of 1s, return the largest layer index among them.

For example, in a 3x3x3 matrix, you have 3 layers (x = 0, 1, 2), and each layer is a 3x3 2D matrix. The largestMatrix() function counts how many 1s are in each layer and returns the layer with the most 1s. If layer 0 has 5 ones and layer 2 has 5 ones, it would return 2 (the larger index).

Intuition

The key insight is that we need to efficiently track which layer (2D plane at index x) has the most 1s, and handle ties by returning the largest index.

Since we're frequently updating individual cells and querying for the layer with maximum 1s, we need to avoid recounting all cells every time largestMatrix() is called. Instead, we can maintain a running count of 1s for each layer.

We maintain:

• A 3D array g to store the actual matrix values
• An array cnt where cnt[x] tracks how many 1s are in layer x
• When setting a cell from 0 to 1, we increment cnt[x] by 1
• When unsetting a cell from 1 to 0, we decrement cnt[x] by 1

However, finding the maximum in cnt array each time largestMatrix() is called would still be O(n). To optimize this further, we need a data structure that maintains the layers sorted by their count of 1s.

This leads us to use a SortedList (ordered set) that stores tuples (count, layer_index). By using a custom sorting key (-count, -layer_index), we ensure:

• Layers with more 1s appear first (negative count for descending order)
• Among layers with equal counts, larger indices appear first (negative index for descending order)

With this structure, largestMatrix() becomes O(1) - just return the layer index from the first element in the sorted list. The trade-off is that setCell and unsetCell operations now have O(log n) complexity for updating the sorted list, but this is acceptable since we optimize the frequent query operation.

The solution also handles edge cases by checking if a cell is already set/unset before modifying it, preventing incorrect count updates.

Learn more about Heap (Priority Queue) patterns.

Solution Approach

The implementation uses a combination of a 3D array for storage and a sorted list for efficient querying:

Data Structures:

1. 3D Array g: A n x n x n array where g[x][y][z] stores the binary value (0 or 1) at position (x, y, z).
2. Count Array cnt: An array of length n where cnt[x] tracks the total number of 1s in layer x.
3. Sorted List sl: An ordered set that maintains tuples (cnt[x], x) sorted by:
   • First by count in descending order (more 1s come first)
   • Then by layer index in descending order (larger indices come first for ties)
   This is achieved using the key function: lambda x: (-x[0], -x[1])

Implementation Details:

__init__(n):

• Initialize the 3D array g with all zeros: [[[0] * n for _ in range(n)] for _ in range(n)]
• Initialize count array cnt with all zeros: [0] * n
• Create an empty SortedList with the custom sorting key

setCell(x, y, z):

1. Check if g[x][y][z] is already 1; if so, return immediately (avoid duplicate counting)
2. Set g[x][y][z] = 1
3. Remove the old tuple (cnt[x], x) from the sorted list using discard()
4. Increment cnt[x] by 1
5. Add the updated tuple (cnt[x], x) back to the sorted list

unsetCell(x, y, z):

1. Check if g[x][y][z] is already 0; if so, return immediately
2. Set g[x][y][z] = 0
3. Remove the old tuple (cnt[x], x) from the sorted list
4. Decrement cnt[x] by 1
5. Only add (cnt[x], x) back to the sorted list if cnt[x] > 0 (layers with no 1s are excluded)

largestMatrix():

• If the sorted list is not empty, return sl[0][1] (the layer index from the first element)
• If the sorted list is empty (all layers have 0 ones), return n - 1 (the largest possible layer index)

Time Complexity:

• setCell and unsetCell: O(log n) due to sorted list operations
• largestMatrix: O(1) since we just access the first element
• Space Complexity: O(n³) for the 3D array plus O(n) for auxiliary structures

Example Walkthrough

Let's walk through a small example with a 3x3x3 matrix to illustrate how the solution works.

Initial State:

• Create Matrix3D(3) - a 3x3x3 cube with all zeros
• g = 3D array of all zeros
• cnt = [0, 0, 0] (no 1s in any layer)
• sl = [] (empty sorted list)

Step 1: setCell(0, 1, 1)

• Check: g[0][1][1] == 0? Yes, proceed
• Set g[0][1][1] = 1
• Remove (0, 0) from sorted list (using discard, no error if not present)
• Update cnt[0] = 0 + 1 = 1
• Add (1, 0) to sorted list
• State: cnt = [1, 0, 0], sl = [(1, 0)]

Step 2: setCell(2, 0, 0)

• Check: g[2][0][0] == 0? Yes, proceed
• Set g[2][0][0] = 1
• Remove (0, 2) from sorted list (not present, no effect)
• Update cnt[2] = 0 + 1 = 1
• Add (1, 2) to sorted list
• State: cnt = [1, 0, 1], sl = [(1, 2), (1, 0)] (sorted by (-count, -index))

Step 3: largestMatrix()

• Check first element: sl[0] = (1, 2)
• Return layer index: 2 (layer 2 chosen over layer 0 due to larger index)

Step 4: setCell(0, 2, 2)

• Check: g[0][2][2] == 0? Yes, proceed
• Set g[0][2][2] = 1
• Remove (1, 0) from sorted list
• Update cnt[0] = 1 + 1 = 2
• Add (2, 0) to sorted list
• State: cnt = [2, 0, 1], sl = [(2, 0), (1, 2)] (layer 0 now has most 1s)

Step 5: largestMatrix()

• Check first element: sl[0] = (2, 0)
• Return layer index: 0 (layer 0 has 2 ones, more than layer 2's 1 one)

Step 6: unsetCell(0, 1, 1)

• Check: g[0][1][1] == 1? Yes, proceed
• Set g[0][1][1] = 0
• Remove (2, 0) from sorted list
• Update cnt[0] = 2 - 1 = 1
• Add (1, 0) back to sorted list (since cnt[0] > 0)
• State: cnt = [1, 0, 1], sl = [(1, 2), (1, 0)] (back to tie situation)

Step 7: largestMatrix()

• Check first element: sl[0] = (1, 2)
• Return layer index: 2 (tie broken by larger index)

This example demonstrates:

1. How the sorted list maintains layers ordered by count (descending) and index (descending for ties)
2. How setCell/unsetCell operations update both the matrix and the sorted list
3. How largestMatrix() returns the correct layer in O(1) time by checking the first element
4. How ties are handled by returning the largest layer index

Solution Implementation

Time and Space Complexity

Time Complexity:

• __init__: O(n^3) for initializing the 3D matrix, plus O(1) for initializing the count array and SortedList, resulting in O(n^3) overall.
• setCell: O(log n) - The method checks if a cell is already set in O(1), then performs discard and add operations on the SortedList, each taking O(log n) time.
• unsetCell: O(log n) - Similar to setCell, it checks the cell value in O(1), then performs discard operation in O(log n), and conditionally adds to the SortedList in O(log n).
• largestMatrix: O(1) - Simply accesses the first element of the SortedList and returns a value.

Space Complexity:

• The 3D matrix self.g requires O(n^3) space to store n × n × n elements.
• The count array self.cnt requires O(n) space for storing counts of each x-plane.
• The SortedList self.sl stores at most n tuples, requiring O(n) space.
• Overall space complexity: O(n^3) dominated by the 3D matrix storage.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Forgetting to Check if Cell State Has Already Changed

Pitfall: Not checking whether a cell is already set before setting it (or already unset before unsetting it) leads to incorrect counting. For example:

• Calling setCell(0, 0, 0) twice would incorrectly increment cnt[0] twice
• This breaks the invariant that cnt[x] should equal the actual number of 1s in layer x

Solution: Always check the current state before modifying:

def setCell(self, x: int, y: int, z: int) -> None:
    if self.grid[x][y][z] == 1:  # Already set, do nothing
        return
    # ... proceed with setting

2. Adding Zero-Count Entries to Sorted List

Pitfall: After unsetting cells, if a layer's count drops to 0, keeping (0, x) in the sorted list causes issues:

• When all layers have 0 ones, the sorted list would contain multiple (0, x) entries
• largestMatrix() would return an arbitrary layer index instead of the expected n-1

Solution: Only add non-zero counts to the sorted list:

def unsetCell(self, x: int, y: int, z: int) -> None:
    # ... after decrementing count
    if self.cell_count[x] > 0:  # Only add if count is positive
        self.sorted_planes.add((self.cell_count[x], x))

3. Incorrect Sorting Order for Tie-Breaking

Pitfall: Using the wrong sorting key leads to incorrect results when multiple layers have the same count. The problem requires returning the largest index among tied layers, but default sorting would return the smallest.

Solution: Use negative values in the sort key to reverse both orderings:

self.sorted_planes = SortedList(key=lambda item: (-item[0], -item[1]))
# First element: -count (larger counts first)
# Second element: -index (larger indices first for ties)

4. Not Handling Empty Sorted List in largestMatrix()

Pitfall: When all cells are 0 (sorted list is empty), directly accessing sl[0] throws an IndexError.

Solution: Check if the sorted list is non-empty before accessing:

def largestMatrix(self) -> int:
    if self.sorted_planes:
        return self.sorted_planes[0][1]
    else:
        return len(self.grid) - 1  # Return largest valid index

5. Using remove() Instead of discard() for SortedList Updates

Pitfall: Using remove() when the element might not exist (e.g., first time setting a cell in a layer) raises a ValueError.

Solution: Use discard() which silently does nothing if the element doesn't exist:

# Safe - won't raise error if (cnt[x], x) doesn't exist
self.sorted_planes.discard((self.cell_count[x], x))