I had inserted 0 instead of 1 The corrected code will be:
public static void setZeros(int[][] matrix) {
int[] row = new int[matrix.length];
int[] column = new int[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] == 1) {
row[i] = 1;
column[j] = 1;
}
}
}
// Set arr[i][j] to 0 if either row i or column j has a 0
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
if ((row[i] == 1 || column[j] == 1)) {
matrix[i][j] = 1;
}
}
}
On Oct 3, 11:06 pm, Shruti Gupta <shruti.gupt...@gmail.com> wrote:
> Hi!
> A effecient way to solve the problem in O(1) space is by making use of
> the fact that instead of keeping track of which cell has a 0, we can
> just know which row or column has zero, as eventually that row/col
> will become 0. The code looks like this:
>
> public static void setZeros(int[][] matrix) {
> int[] row = new int[matrix.length];
> int[] column = new int[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] = 1;
> column[j] = 1;
> }
> }
>
> }
>
> // Set arr[i][j] to 0 if either row i or column j has a 0
> for (int i = 0; i < matrix.length; i++) {
> for (int j = 0; j < matrix[0].length; j++) {
> if ((row[i] == 1 || column[j] == 1)) {
> matrix[i][j] = 0;
> }
> }
> }
>
> Thus there is no extra space taken.
>
> Shruti
>
> On Oct 3, 12:27 am, rahul sharma <rahul23111...@gmail.com> wrote:
>
>
>
>
>
>
>
> > nput is a matrix of size n x m of 0s and 1s.
>
> > eg:
> > 1 0 0 1
> > 0 0 1 0
> > 0 0 0 0
>
> > If a location has 1; make all the elements of that row and column = 1. eg
>
> > 1 1 1 1
> > 1 1 1 1
> > 1 0 1 1
>
> > Solution should be with Time complexity = O(n*m) and O(1) extra space
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