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Design a 3D Binary Matrix with Efficient Layer Tracking

Number: 3735

Difficulty: Medium

Paid? Yes

Companies: N/A

Problem Description

You are given an n x n x n binary 3D array that is initially filled with 0’s. Implement a Matrix3D class with three methods: • Matrix3D(n): Initializes the 3D matrix of dimensions n × n × n, with all elements set to 0. • setCell(x, y, z): Sets matrix[x][y][z] to 1. • unsetCell(x, y, z): Sets matrix[x][y][z] to 0. • largestMatrix(): Returns the index x (i.e. layer) for which the 2D matrix matrix[x] contains the most 1’s. If multiple layers have the same count, return the largest index.

Key Insights

  • Maintain an auxiliary count array that tracks how many 1’s each 2D layer (x index) contains.
  • Update the count in constant time during setCell and unsetCell operations.
  • For the largestMatrix query, iterate over the small count array (n <= 100) to find the layer with the highest number of 1’s, breaking ties by choosing the largest index.
  • Although more sophisticated structures (like heaps or balanced trees) may be used, a simple scan is efficient given the constraints.

Space and Time Complexity

Time Complexity: • setCell/unsetCell: O(1) per operation. • largestMatrix: O(n) per call (n ≤ 100). Space Complexity: O(n^3) for storing the 3D matrix along with O(n) for the auxiliary count array.

Solution

We maintain a 3D array for the matrix and a 1D array (of length n) for counting the number of 1’s in each x-layer. When performing setCell or unsetCell, we first check if a change is necessary (i.e. avoid double counting) and then update the corresponding count. For the largestMatrix query, we traverse the count array to identify the layer with the maximum count. Due to the small size of n, scanning is efficient. Ties are resolved by preferring the largest index.

Code Solutions

Below are code solutions in Python, JavaScript, C++, and Java.

class Matrix3D:
    def __init__(self, n):
        # Initialize the 3D matrix with all 0's.
        self.n = n
        self.matrix = [[[0 for _ in range(n)] for _ in range(n)] for _ in range(n)]
        # Count array to track the number of 1's in each layer (x index).
        self.layerCount = [0] * n

    def setCell(self, x, y, z):
        # Only update if the cell is currently 0.
        if self.matrix[x][y][z] == 0:
            self.matrix[x][y][z] = 1
            self.layerCount[x] += 1

    def unsetCell(self, x, y, z):
        # Only update if the cell is currently 1.
        if self.matrix[x][y][z] == 1:
            self.matrix[x][y][z] = 0
            self.layerCount[x] -= 1

    def largestMatrix(self):
        maxCount = -1
        resultIndex = -1
        for i in range(self.n):
            # If same count, choose the larger index.
            if self.layerCount[i] > maxCount or (self.layerCount[i] == maxCount and i > resultIndex):
                maxCount = self.layerCount[i]
                resultIndex = i
        return resultIndex
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