**QUESTION**:

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

**SOLUTION:**

- 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 take O(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 cell [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 of all the rows with zeros and all the columns with zeros. We then nullify rows and columns based on the values in these arrays.

void setZeros(int[][] matrix) { boolean[] row = new boolean[matrix .length]; boolean[] column = new boolean[matrix[0].length]; //Store the row and column index with value 0 for (int i= 0; i < matrix.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 (inti= 0; i < row.length; i++) { if (row[i]) nullifyRow(matrix, i); } // Nullify columns for (int j = 0; j < column.length; j++) { if (column[j]) nullifyColumn(matrix, j); } } void nullifyRow(int[][] matrix, int row) { for (int j = 0; j < matrix[0].length; j++) { matrix[row][j] = 0; } } void nullifyColumn(int[][] matrix, int col) { for (int i= 0; i < matrix.length; i++) { matrix[i][col] = 0; } }

- To make this somewhat more space-efficient we could use a bit vector instead of a boolean array.
- It would still, be O(N) space.

We can reduce the space to 0(1) by using the first row as a replacement for the row array and the first column as a replacement for the column array. This works as follows:

- Check if the first row and first column have any zeros, and set variables rowHasZero and columnHasZero. (We’ll nullify the first row and first column later, if necessary.)
- Iterate through the rest of the matrix, setting matrix[ i] [ 0) and matrix [ 0) [ j] to zero whenever there’s a zero in matrix[i] [j].
- Iterate through rest of matrix, nullifying row i if there’s a zero in matrix [ i] [ 0].
- Iterate through rest of matrix, nullifying column j if there’s a zero in matrix [0] [j].
- Nullify the first row and first column, if necessary (based on values from Step 1).

**This code is below:**

void setzeros(int[][] matrix) { boolean rowHasZero = false; boolean colHasZero = false; // Check if first row has a zero for (int j = 0; j < matrix[0].length; j++) { if (matrix[0][j] == 0) { rowHasZero = true; break; } } // Check if first column has a zero for (int i = 0; i < matrix.length; i++) { if (matrix[i][0] == 0) { colHasZero = true; break; } } // Check for zeros in the rest of the array for (int i= 1; i < matrix.length; i++) { for (int j = 1; j < matrix[0].length;j++) { if (matrix[iJ[jJ == 0) { matrix[i][0] = 0; matrix[0J[j] = 0; } } } // Nullify rows based on values in first column for (int i= 1; i < matrix.length; i++) { if (matrix[i][0] == 0) { nullifyRow(matrix, i); } } // Nullify columns based on values in first row for (int j = 1; j < matrix[0].length; j++) { if (matrix[0][j] == 0) { nullifyColumn(matrix, j); } } // Nullify first row if (rowHasZero) { nullifyRow(matrix, 0); } // Nullify first column if (colHasZero) { nullifyColumn(matrix, 0); } }

- This code has a lot of “do this for the rows, then the equivalent action for the column:’ In an interview, you could abbreviate this code by adding comments and TODOs that explain that the next chunk of code looks the same as the earlier code, but using rows.
- This would allow you to focus on the most important parts of the algorithm.