Zero Matrix

Zero Matrix: Write an algorithm such that if an element in an MxN matrix is 0, its entire row and column are set to 0.

👉 Link here to the repo to solve the problem

👌 Tips

If you just cleared the rows and columns as you found Os, you'd likely wind up clearing the whole matrix.Try finding the cells with zeros first before making any changes to the matrix.

Can you use 0 (N) additional space instead of 0 (N2 )? What information do you really need from the list of cells that are zero?

You probably need some data storage to maintain a list of the rows and columns that need to be zeroed. Can you reduce the additional space usage to a(1) by using the matrix itself for data storage?

👊 Solution 1

At first glance, this problem seems easy: just iterate through the matrix and every time we see a cell with value zero, set its row and column to 0. There's one problem with that solution though: when we come across other cells in that row or column, we'll see the zeros and change their row and column to zero. Pretty soon, our entire matrix will be set to zeros.

One way around this is to keep a second matrix which flags the zero locations. We would then do a second pass through the matrix to set the zeros. This would ta ke 0 (MN) space.

Do we really need O(MN) space? No. Since we're going to set the entire row and column to zero, we don't need to track that it was exactly ce11[2] [4] (row 2, column 4). We only need to know that row 2 has a zero somewhere, and column 4 has a zero somewhere.We'll set the entire row and column to zero anyway, so why would we care to keep track of the exact location of the zero?

The code below implements this algorithm.We use two arrays to keep track ofall the rows with zeros and all the columns with zeros.We then nullify rows and columns based on the values in these arrays.

void solution(int[][] matrix) {
    int length = matrix.length;
    boolean[] row = new boolean[length];
    boolean[] column = new boolean[length];

    // Store the row and column index with value 0
    for (int i = 0; i < length; i++) {
        for (int j = 0; j < matrix[0].length; j++) {
            if (matrix[i][j] == 0) {
                row[i] = true;
                column[j] = true;
            }
        }
    }

    // nullify rows
    for (int i = 0; i < row.length; i++) {
        if (row[i]) nullifyRow(matrix, i);
    }

    // nullify columns
    for (int i = 0; i < column.length; i++) {
        if (column[i]) nullifyColumn(matrix, i);
    }
}

void nullifyRow(int[][] matrix, int row) {
    for (int i = 0; i < matrix[0].length; i++) {
        matrix[row][i] = 0;
    }
}

void nullifyColumn(int[][] matrix, int col) {
    for (int i = 0; i < matrix.length; i++) {
        matrix[i][col] = 0;
    }
}

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Question borrowed from “Cracking the coding interview”