Re: [algogeeks] Re: Urgent sol needed
@shruti ...u have used 2 arrays row n col extra?
On Mon, Oct 3, 2011 at 11:47 PM, Ankur Garg wrote:
> Many times this problem has been discussed ..Please check the archives yaar
> :(
>
>
>
> On Mon, Oct 3, 2011 at 11:39 PM, Shruti Gupta wrote:
>
>> 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 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 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
>>
>> --
>> You received this message because you are subscribed to the Google Groups
>> "Algorithm Geeks" group.
>> To post to this group, send email to algogeeks@googlegroups.com.
>> To unsubscribe from this group, send email to
>> algogeeks+unsubscr...@googlegroups.com.
>> For more options, visit this group at
>> http://groups.google.com/group/algogeeks?hl=en.
>>
>>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to this group, send email to algogeeks@googlegroups.com.
> To unsubscribe from this group, send email to
> algogeeks+unsubscr...@googlegroups.com.
> For more options, visit this group at
> http://groups.google.com/group/algogeeks?hl=en.
>
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Urgent sol needed
Many times this problem has been discussed ..Please check the archives yaar
:(
On Mon, Oct 3, 2011 at 11:39 PM, Shruti Gupta wrote:
> 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 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 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
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to this group, send email to algogeeks@googlegroups.com.
> To unsubscribe from this group, send email to
> algogeeks+unsubscr...@googlegroups.com.
> For more options, visit this group at
> http://groups.google.com/group/algogeeks?hl=en.
>
>
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Urgent sol needed
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 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 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
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Urgent sol needed
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 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
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
