A Boolean Matrix Question | GeeksforGeeks
Given an m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
Given a boolean matrix mat[M][N] of size M X N, modify it such that if a matrix cell mat[i][j] is 1 (or true) then make all the cells of ith row and jth column as 1.
Method 1 (Use two temporary arrays)
1) Create two temporary arrays row[M] and col[N]. Initialize all values of row[] and col[] as 0.
2) Traverse the input matrix mat[M][N]. If you see an entry mat[i][j] as true, then mark row[i] and col[j] as true.
3) Traverse the input matrix mat[M][N] again. For each entry mat[i][j], check the values of row[i] and col[j]. If any of the two values (row[i] or col[j]) is true, then mark mat[i][j] as true.
Code from http://www.programcreek.com/2012/12/leetcode-set-matrix-zeroes-java/
http://www.codebytes.in/2014/12/write-algorithm-such-that-if-element-in.html
void matrix_set_zero(int *input, int N){
https://github.com/careercup/CtCI-6th-Edition/blob/master/Java/Ch%2001.%20Arrays%20and%20Strings/Q1_08_Zero_Matrix/QuestionB.java
public static void nullifyRow(int[][] matrix, int row) {
for (int j = 0; j < matrix[0].length; j++) {
matrix[row][j] = 0;
}
}
public static void nullifyColumn(int[][] matrix, int col) {
for (int i = 0; i < matrix.length; i++) {
matrix[i][col] = 0;
}
}
public static 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[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][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);
}
}
https://github.com/careercup/CtCI-6th-Edition/blob/master/Java/Ch%2001.%20Arrays%20and%20Strings/Q1_08_Zero_Matrix/QuestionA.java
public static void nullifyRow(int[][] matrix, int row) {
for (int j = 0; j < matrix[0].length; j++) {
matrix[row][j] = 0;
}
}
public static void nullifyColumn(int[][] matrix, int col) {
for (int i = 0; i < matrix.length; i++) {
matrix[i][col] = 0;
}
}
public static 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 (int i = 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);
}
}
}
Read full article from A Boolean Matrix Question | GeeksforGeeks
Given an m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
Method 1 (Use two temporary arrays)
1) Create two temporary arrays row[M] and col[N]. Initialize all values of row[] and col[] as 0.
2) Traverse the input matrix mat[M][N]. If you see an entry mat[i][j] as true, then mark row[i] and col[j] as true.
3) Traverse the input matrix mat[M][N] again. For each entry mat[i][j], check the values of row[i] and col[j]. If any of the two values (row[i] or col[j]) is true, then mark mat[i][j] as true.
Method 2 (A Space Optimized Version of Method 1)
This method is a space optimized version of above method 1. This method uses the first row and first column of the input matrix in place of the auxiliary arrays row[] and col[] of method 1. So what we do is: first take care of first row and column and store the info about these two in two flag variables rowFlag and colFlag. Once we have this info, we can use first row and first column as auxiliary arrays and apply method 1 for submatrix (matrix excluding first row and first column) of size (M-1)*(N-1).
This method is a space optimized version of above method 1. This method uses the first row and first column of the input matrix in place of the auxiliary arrays row[] and col[] of method 1. So what we do is: first take care of first row and column and store the info about these two in two flag variables rowFlag and colFlag. Once we have this info, we can use first row and first column as auxiliary arrays and apply method 1 for submatrix (matrix excluding first row and first column) of size (M-1)*(N-1).
1) Scan the first row and set a variable rowFlag to indicate whether we need to set all 1s in first row or not.
2) Scan the first column and set a variable colFlag to indicate whether we need to set all 1s in first column or not.
3) Use first row and first column as the auxiliary arrays row[] and col[] respectively, consider the matrix as submatrix starting from second row and second column and apply method 1.
4) Finally, using rowFlag and colFlag, update first row and first column if needed.
2) Scan the first column and set a variable colFlag to indicate whether we need to set all 1s in first column or not.
3) Use first row and first column as the auxiliary arrays row[] and col[] respectively, consider the matrix as submatrix starting from second row and second column and apply method 1.
4) Finally, using rowFlag and colFlag, update first row and first column if needed.
Time Complexity: O(M*N)
Auxiliary Space: O(1)
http://ideone.com/v2Bec8Auxiliary Space: O(1)
Code from http://www.programcreek.com/2012/12/leetcode-set-matrix-zeroes-java/
public void setZeroes(int[][] matrix) { boolean firstRowZero = false; boolean firstColumnZero = false; //set first row and column zero or not for(int i=0; i<matrix.length; i++){ if(matrix[i][0] == 0){ firstColumnZero = true; break; } } for(int i=0; i<matrix[0].length; i++){ if(matrix[0][i] == 0){ firstRowZero = true; break; } } //mark zeros on first row and column for(int i=1; i<matrix.length; i++){ for(int j=1; j<matrix[0].length; j++){ if(matrix[i][j] == 0){ matrix[i][0] = 0; matrix[0][j] = 0; } } } //use mark to set elements for(int i=1; i<matrix.length; i++){ for(int j=1; j<matrix[0].length; j++){ if(matrix[i][0] == 0 || matrix[0][j] == 0){ matrix[i][j] = 0; } } } //set first column and row if(firstColumnZero){ for(int i=0; i<matrix.length; i++) matrix[i][0] = 0; } if(firstRowZero){ for(int i=0; i<matrix[0].length; i++) matrix[0][i] = 0; } }
void modifyMatrix(bool mat[R][C]){ bool row[R]; bool col[C]; int i, j; /* Initialize all values of row[] as 0 */ for (i = 0; i < R; i++) { row[i] = 0; } /* Initialize all values of col[] as 0 */ for (i = 0; i < C; i++) { col[i] = 0; } /* Store the rows and columns to be marked as 1 in row[] and col[] arrays respectively */ for (i = 0; i < R; i++) { for (j = 0; j < C; j++) { if (mat[i][j] == 1) { row[i] = 1; col[j] = 1; } } } /* Modify the input matrix mat[] using the above constructed row[] and col[] arrays */ for (i = 0; i < R; i++) { for (j = 0; j < C; j++) { if ( row[i] == 1 || col[j] == 1 ) { mat[i][j] = 1; } } }}http://www.codebytes.in/2014/12/write-algorithm-such-that-if-element-in.html
public static void deleteZeros(int[][] a){ boolean rows[] = new boolean[a.length]; boolean columns[] = new boolean[a[0].length]; for(int i = 0;i<a.length;++i){ for(int j = 0;j<a[0].length;++j){ if(a[i][j]==0){rows[i] = true;columns[j] = true;} } } for(int i=0;i<rows.length;++i){ if(rows[i]) for(int j=0;j<columns.length;++j)a[i][j]=0; } for(int j=0;j<columns.length;++j){ if(columns[j]) for(int i=0;i<rows.length;++i)a[i][j]=0; } }http://algorithmsforever.blogspot.com/2011/11/zero-matrix.html
void matrix_set_zero(int *input, int N){
int *row = new int[N];
int *col = new int[N];
for(int i=0; i<N; i++){
for(int i=0; i<N; i++){
for(int i=0; i<N; i++){
}int *col = new int[N];
for(int i=0; i<N; i++){
row[i] = col[i] = 1;
}for(int i=0; i<N; i++){
for(int j=0; j<N; j++){
}
row[i] &= input[i][j];
col[j] &= input[i][j];
}col[j] &= input[i][j];
for(int i=0; i<N; i++){
for(int j=0; j<N; j++){
}
input[i][j] = row[i] & col[j];
}https://github.com/careercup/CtCI-6th-Edition/blob/master/Java/Ch%2001.%20Arrays%20and%20Strings/Q1_08_Zero_Matrix/QuestionB.java
public static void nullifyRow(int[][] matrix, int row) {
for (int j = 0; j < matrix[0].length; j++) {
matrix[row][j] = 0;
}
}
public static void nullifyColumn(int[][] matrix, int col) {
for (int i = 0; i < matrix.length; i++) {
matrix[i][col] = 0;
}
}
public static 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[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][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);
}
}
https://github.com/careercup/CtCI-6th-Edition/blob/master/Java/Ch%2001.%20Arrays%20and%20Strings/Q1_08_Zero_Matrix/QuestionA.java
public static void nullifyRow(int[][] matrix, int row) {
for (int j = 0; j < matrix[0].length; j++) {
matrix[row][j] = 0;
}
}
public static void nullifyColumn(int[][] matrix, int col) {
for (int i = 0; i < matrix.length; i++) {
matrix[i][col] = 0;
}
}
public static 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 (int i = 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);
}
}
}
Read full article from A Boolean Matrix Question | GeeksforGeeks
