@shruti ...u have used 2 arrays row n col extra?????
On Mon, Oct 3, 2011 at 11:47 PM, Ankur Garg <ankurga...@gmail.com> wrote:
> Many times this problem has been discussed ..Please check the archives yaar
> :(
>
>
>
> On Mon, Oct 3, 2011 at 11:39 PM, Shruti Gupta <shruti.gupt...@gmail.com>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 <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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