LeetCode 3391: Design a 3D Binary Matrix with Efficient Layer Tracking Solution

Master LeetCode problem 3391 (Design a 3D Binary Matrix with Efficient Layer Tracking), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

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

Medium

Array Hash Table Design Heap (Priority Queue)Matrix Ordered Set

Problem Explanation

Explanation:

To design a 3D binary matrix with efficient layer tracking, we can use a 3D array to represent the matrix. We can efficiently track the layers by maintaining two 2D arrays representing the top and bottom layers of the matrix. We will also keep track of the total number of 1s present in the matrix.

Algorithm:

Initialize a 3D array to represent the 3D binary matrix.
Initialize 2D arrays to represent the top and bottom layers of the matrix.
Maintain a counter to keep track of the total number of 1s in the matrix.
Implement methods to set a value at a specific position, get the value at a specific position, count the total number of 1s, and update the top and bottom layers accordingly.

Time Complexity:

Setting a value, getting a value, and counting the total number of 1s will take O(1) time.
Updating the top and bottom layers will also take O(1) time.
Overall time complexity: O(1)

Space Complexity:

The 3D array will take O(n^3) space.
The top and bottom layers will take O(n^2) space.
Overall space complexity: O(n^3)

Solution Code

class BinaryMatrix3D {
    private int[][][] matrix;
    private int[][] topLayer;
    private int[][] bottomLayer;
    private int totalOnes;

    public BinaryMatrix3D(int n) {
        matrix = new int[n][n][n];
        topLayer = new int[n][n];
        bottomLayer = new int[n][n];
        totalOnes = 0;
    }

    public void set(int x, int y, int z, int value) {
        if (matrix[x][y][z] == 1 && value == 0) {
            totalOnes--;
        } else if (matrix[x][y][z] == 0 && value == 1) {
            totalOnes++;
        }
        matrix[x][y][z] = value;
    }

    public int get(int x, int y, int z) {
        return matrix[x][y][z];
    }

    public int countTotalOnes() {
        return totalOnes;
    }

    public void updateTopLayer(int[][] layer) {
        topLayer = layer;
    }

    public void updateBottomLayer(int[][] layer) {
        bottomLayer = layer;
    }
}

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