A Boolean Matrix Question (original) (raw)
Given a boolean matrix **mat where each cell contains either 0 or 1, the task is to modify it such that if a matrix cell matrix[i][j] is 1 then all the cells in its **i th row and **j th column will **become 1.
**Examples:
**Input: [[1, 0],
[0, 0]]
**Output: [[1, 1],
[1, 0]]**Input: [[1, 0, 0, 1],
[0, 0, 1, 0],
[0, 0, 0, 0]]
**Output: [[1, 1, 1, 1],
[1, 1, 1, 1],
[1, 0, 1, 1]]
**[Naive Approach] By Marking the Matrix
Assuming all the elements in the matrix are non-negative. Traverse through the matrix and if you find an element with value 1, then change all the zeros in its row and column to -1. The reason for not changing other elements to 1, but -1, is because that might affect other columns and rows.
Now traverse through the matrix again and if an element is -1 change it to 1, which will be the answer.
C++ `
// C++ Code For Boolean Matrix Question // By Marking the Matrix
#include #include using namespace std;
void booleanMatrix(vector<vector>& mat) { int rows = mat.size(), cols = mat[0].size();
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (mat[i][j] == 1) {
// Replace all the zeros in jth column by -1
for (int idx = 0; idx < rows; idx++) {
if (mat[idx][j] == 0) {
mat[idx][j] = -1;
}
}
// Replace all the zeros in ith row by -1
for (int idx = 0; idx < cols; idx++) {
if (mat[i][idx] == 0) {
mat[i][idx] = -1;
}
}
}
}
}
// Replace all the -1 by 1
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (mat[i][j] == -1) {
mat[i][j] = 1;
}
}
}}
int main() { vector<vector> mat = {{1, 0, 0, 1}, {0, 0, 1, 0}, {0, 0, 0, 0}}; booleanMatrix(mat); for (const vector& row : mat) { for (int val : row) { cout << val << " "; } cout << endl; } return 0; }
Java
// Java Code For Boolean Matrix Question // By Marking the Matrix class GfG { static void booleanMatrix(int[][] mat) { int rows = mat.length, cols = mat[0].length;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (mat[i][j] == 1) {
// Replace all the zeros in jth column by -1
for (int idx = 0; idx < rows; idx++) {
if (mat[idx][j] == 0) {
mat[idx][j] = -1;
}
}
// Replace all the zeros in ith row by -1
for (int idx = 0; idx < cols; idx++) {
if (mat[i][idx] == 0) {
mat[i][idx] = -1;
}
}
}
}
}
// Replace all the -1 by 1
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (mat[i][j] == -1) {
mat[i][j] = 1;
}
}
}
}
public static void main(String[] args) {
int[][] mat = {{1, 0, 0, 1}, {0, 0, 1, 0}, {0, 0, 0, 0}};
booleanMatrix(mat);
for (int[] row : mat) {
for (int val : row) {
System.out.print(val + " ");
}
System.out.println();
}
}}
Python
python3 Code For Boolean Matrix Question
By Marking the Matrix
def booleanMatrix(mat): rows, cols = len(mat), len(mat[0])
for i in range(rows):
for j in range(cols):
if mat[i][j] == 1:
# Replace all the zeros in jth column by -1
for idx in range(rows):
if mat[idx][j] == 0:
mat[idx][j] = -1
# Replace all the zeros in ith row by -1
for idx in range(cols):
if mat[i][idx] == 0:
mat[i][idx] = -1
# Replace all the -1 by 1
for i in range(rows):
for j in range(cols):
if mat[i][j] == -1:
mat[i][j] = 1if name == "main": mat = [[1, 0, 0, 1], [0, 0, 1, 0], [0, 0, 0, 0]] booleanMatrix(mat) for row in mat: print(" ".join(map(str, row)))
C#
// C# Code For Boolean Matrix Question // By Marking the Matrix using System;
class GfG { static void booleanMatrix(int[, ] mat) { int rows = mat.GetLength(0), cols = mat.GetLength(1);
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (mat[i, j] == 1) {
// Replace all the zeros in jth column
// by -1
for (int idx = 0; idx < rows; idx++) {
if (mat[idx, j] == 0) {
mat[idx, j] = -1;
}
}
// Replace all the zeros in ith row by
// -1
for (int idx = 0; idx < cols; idx++) {
if (mat[i, idx] == 0) {
mat[i, idx] = -1;
}
}
}
}
}
// Replace all the -1 by 1
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (mat[i, j] == -1) {
mat[i, j] = 1;
}
}
}
}
static void Main() {
int[, ] mat = { { 1, 0, 0, 1 },
{ 0, 0, 1, 0 },
{ 0, 0, 0, 0 } };
booleanMatrix(mat);
for (int i = 0; i < mat.GetLength(0); i++) {
for (int j = 0; j < mat.GetLength(1); j++) {
Console.Write(mat[i, j] + " ");
}
Console.WriteLine();
}
}}
JavaScript
// Javascript Code For Boolean Matrix Question // By Marking the Matrix function booleanMatrix(mat) { let rows = mat.length, cols = mat[0].length;
for (let i = 0; i < rows; i++) {
for (let j = 0; j < cols; j++) {
if (mat[i][j] === 1) {
// Replace all the zeros in jth column by -1
for (let idx = 0; idx < rows; idx++) {
if (mat[idx][j] === 0) {
mat[idx][j] = -1;
}
}
// Replace all the zeros in ith row by -1
for (let idx = 0; idx < cols; idx++) {
if (mat[i][idx] === 0) {
mat[i][idx] = -1;
}
}
}
}
}
// Replace all the -1 by 1
for (let i = 0; i < rows; i++) {
for (let j = 0; j < cols; j++) {
if (mat[i][j] === -1) {
mat[i][j] = 1;
}
}
}}
// Driver Code let mat = [[1, 0, 0, 1], [0, 0, 1, 0], [0, 0, 0, 0]]; booleanMatrix(mat); mat.forEach(row => { console.log(row.join(' ')); });
`
Output
1 1 1 1 1 1 1 1 1 0 1 1
**Time Complexity: O((n * m)*(n + m)) where n is number of rows and m is number of columns in given matrix.
O(n * m) for traversing through each element and (n+m) for traversing to row and column of matrix elements having value 1.
**Space Complexity: O(1)
**[Better Approach] Using Extra Space
The idea is to use **two temporary arrays,
rowMarker[]and **colMarker[], to keep track of which rows and columns need to be updated.
Following are the steps for this approach
- Create two temporary arrays rowMarker[] and colMarker[]. Initialize all values of rowMarker[] and colMarker[] as 0.
- Traverse the input matrix mat[][]. If you finds mat[i][j] as 1, then mark rowMarker[i] and colMarker[j] as true.
- Traverse the input matrix mat[][] again. For each entry mat[i][j], check the values of rowMarker[i] and colMarker[j]. If any of the two values (rowMarker[i] or colMarker[j]) is true, then mark mat[i][j] as 1. C++ `
// C++ Code For Boolean Matrix Question // Using two extra arrays #include #include using namespace std;
void booleanMatrix(vector<vector>& mat) { int rows = mat.size(); int cols = mat[0].size();
// Arrays to keep track of rows and columns to be updated
vector<bool> rowMarker(rows, false);
vector<bool> colMarker(cols, false);
// Mark the rows and columns that need to be updated
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (mat[i][j] == 1) {
rowMarker[i] = true;
colMarker[j] = true;
}
}
}
// Update the matrix
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (rowMarker[i] || colMarker[j]) {
mat[i][j] = 1;
}
}
}}
int main() { vector<vector> mat = { {1, 0, 0, 1}, {0, 0, 1, 0}, {0, 0, 0, 0} };
booleanMatrix(mat);
for (const vector<int>& row : mat) {
for (int val : row) {
cout << val << " ";
}
cout << endl;
}
return 0;}
Java
// Java Code For Boolean Matrix Question // Using two extra arrays import java.util.*;
class GfG { static void booleanMatrix(int mat[][]) { int rows = mat.length; int cols = mat[0].length;
// Arrays to keep track of rows and columns to be marked
boolean[] rowMarker = new boolean[rows];
boolean[] colMarker = new boolean[cols];
// First pass: Mark the rows and columns to be updated
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (mat[i][j] == 1) {
rowMarker[i] = true;
colMarker[j] = true;
}
}
}
// Second pass: Update the matrix
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (rowMarker[i] || colMarker[j]) {
mat[i][j] = 1;
}
}
}
}
public static void main(String[] args) {
int[][] arr = {
{ 1, 0, 0, 1 },
{ 0, 0, 1, 0 },
{ 0, 0, 0, 0 }
};
booleanMatrix(arr);
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[0].length; j++) {
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
}}
Python
python3 Code For Boolean Matrix Question
Using two extra arrays
def booleanMatrix(mat): rows = len(mat) cols = len(mat[0])
# Arrays to keep track of rows and columns to be updated
rowMarker = [False] * rows
colMarker = [False] * cols
# First pass: Mark the rows and columns to be updated
for i in range(rows):
for j in range(cols):
if mat[i][j] == 1:
rowMarker[i] = True
colMarker[j] = True
# Second pass: Update the matrix
for i in range(rows):
for j in range(cols):
if rowMarker[i] or colMarker[j]:
mat[i][j] = 1if name == "main": arr = [[1, 0, 0, 1], [0, 0, 1, 0], [0, 0, 0, 0]] booleanMatrix(arr) for row in arr: print(" ".join(map(str, row)))
C#
// C# Code For Boolean Matrix Question // Using two extra arrays using System; using System.Collections.Generic;
class GfG { static void booleanMatrix(List<List> mat) { int rows = mat.Count; int cols = mat[0].Count;
// Arrays to keep track of rows and columns to be updated
bool[] rowMarker = new bool[rows];
bool[] colMarker = new bool[cols];
// First pass: Mark the rows and columns to be updated
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (mat[i][j] == 1) {
rowMarker[i] = true;
colMarker[j] = true;
}
}
}
// Second pass: Update the matrix
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (rowMarker[i] || colMarker[j]) {
mat[i][j] = 1;
}
}
}
}
// Test the BooleanMatrix function with a sample input
static void Main(string[] args) {
List<List<int>> arr = new List<List<int>> {
new List<int>{ 1, 0, 0, 1 },
new List<int>{ 0, 0, 1, 0 },
new List<int>{ 0, 0, 0, 0 }
};
booleanMatrix(arr);
foreach (var row in arr) {
Console.WriteLine(string.Join(" ", row));
}
}}
JavaScript
// Javascript Code For Boolean Matrix Question // Using two extra arrays function booleanMatrix(mat) {
const rows = mat.length;
const cols = mat[0].length;
// Arrays to keep track of rows and columns to be
// updated
const rowMarker = new Array(rows).fill(false);
const colMarker = new Array(cols).fill(false);
// First pass: Mark the rows and columns to be updated
for (let i = 0; i < rows; i++) {
for (let j = 0; j < cols; j++) {
if (mat[i][j] === 1) {
rowMarker[i] = true;
colMarker[j] = true;
}
}
}
// Second pass: Update the matrix
for (let i = 0; i < rows; i++) {
for (let j = 0; j < cols; j++) {
if (rowMarker[i] || colMarker[j]) {
mat[i][j] = 1;
}
}
}}
// Driver Code const arr = [ [ 1, 0, 0, 1 ], [ 0, 0, 1, 0 ], [ 0, 0, 0, 0 ] ]; booleanMatrix(arr); arr.forEach(row => console.log(row.join(" ")));
`
Output
1 1 1 1 1 1 1 1 1 0 1 1
**Time Complexity: O(n * m) where n is number of rows and m is number of columns in given matrix.
**Auxiliary Space: O(n + m) as we are taking two arrays one of size m and another of size n.
**[Expected Approach] Without Using Extra Space
Instead of taking two dummy arrays we can **use the first row and first column of the matrix for the same work. This will help to reduce the space complexity of the problem. While traversing for the second time, the first row and column will be computed first, which will affect the values of further elements. So we traverse in the reverse direction.
Since **matrix[0][0] are overlapping in first row and first column. Therefore take separate variable col0 (say) to check if the 0th column has 1 or not and use matrix[0][0] to check if the 0th row has 1 or not. Now traverse from the last element to the first element and check if matrix[i][0]==1 || matrix[0][j]==1 and if true set matrix[i][j]=1, else continue.
C++ `
// C++ Code For Boolean Matrix Question // Most Optimised Approach #include #include using namespace std;
void booleanMatrix(vector<vector>& mat) { int col0 = 0, rows = mat.size(), cols = mat[0].size();
// Step 1: Mark the first row and
// column if there is a 1 in the matrix
for (int i = 0; i < rows; i++) {
// Check if 0th column contains 1
if (mat[i][0] == 1) col0 = 1;
for (int j = 1; j < cols; j++) {
if (mat[i][j] == 1) {
mat[i][0] = 1;
mat[0][j] = 1;
}
}
}
// Step 2: Traverse the matrix in
// reverse direction and update values
for (int i = rows - 1; i >= 0; i--) {
for (int j = cols - 1; j >= 1; j--) {
if (mat[i][0] == 1 || mat[0][j] == 1) {
mat[i][j] = 1;
}
}
if (col0 == 1) {
mat[i][0] = 1;
}
}}
int main() { vector<vector> arr = {{1, 0, 0, 1}, {0, 0, 1, 0}, {0, 0, 0, 0}};
booleanMatrix(arr);
for (int i = 0; i < arr.size(); i++) {
for (int j = 0; j < arr[0].size(); j++) {
cout << arr[i][j] << " ";
}
cout << endl;
}
return 0;}
Java
// Java Code For Boolean Matrix Question // Most Optimised Approach import java.util.*;
class GfG {
static void booleanMatrix(int mat[][]) {
int col0 = 0;
int rows = mat.length, cols = mat[0].length;
// Step 1: Mark the first row and
// column if there is a 1 in the matrix
for (int i = 0; i < rows; i++) {
// Check if 0th column contains 1
if (mat[i][0] == 1) col0 = 1;
for (int j = 1; j < cols; j++) {
if (mat[i][j] == 1) {
mat[i][0] = 1;
mat[0][j] = 1;
}
}
}
// Step 2: Traverse the matrix in reverse
// direction and update values
for (int i = rows - 1; i >= 0; i--) {
for (int j = cols - 1; j >= 1; j--) {
if (mat[i][0] == 1 || mat[0][j] == 1) {
mat[i][j] = 1;
}
}
if (col0 == 1) {
mat[i][0] = 1;
}
}
}
public static void main(String[] args) {
int[][] arr = {
{1, 0, 0, 1},
{0, 0, 1, 0},
{0, 0, 0, 0}
};
booleanMatrix(arr);
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[0].length; j++) {
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
}}
Python
python3 Code For Boolean Matrix Question
Most Optimised Approach
def booleanMatrix(mat): col0 = 0 rows = len(mat) cols = len(mat[0])
# Step 1: Mark the first row and column
# if there is a 1 in the matrix
for i in range(rows):
# Check if 0th column contains 1
if mat[i][0] == 1:
col0 = 1
for j in range(1, cols):
if mat[i][j] == 1:
mat[i][0] = 1
mat[0][j] = 1
# Step 2: Traverse the matrix in reverse
# direction and update values
for i in range(rows - 1, -1, -1):
for j in range(cols - 1, 0, -1):
if mat[i][0] == 1 or mat[0][j] == 1:
mat[i][j] = 1
if col0 == 1:
mat[i][0] = 1 if name == "main": arr = [ [1, 0, 0, 1], [0, 0, 1, 0], [0, 0, 0, 0] ]
booleanMatrix(arr)
for row in arr:
print(" ".join(map(str, row)))C#
// C# Code For Boolean Matrix Question // Most Optimised Approach using System;
class GfG {
static void booleanMatrix(int[,] mat) {
int col0 = 0, rows = mat.GetLength(0),
cols = mat.GetLength(1);
// Step 1: Mark the first row and column
// if there is a 1 in the matrix
for (int i = 0; i < rows; i++) {
// Check if 0th column contains 1
if (mat[i, 0] == 1) col0 = 1;
for (int j = 1; j < cols; j++) {
if (mat[i, j] == 1) {
mat[i, 0] = 1;
mat[0, j] = 1;
}
}
}
// Step 2: Traverse the matrix in reverse
// direction and update values
for (int i = rows - 1; i >= 0; i--) {
for (int j = cols - 1; j >= 1; j--) {
if (mat[i, 0] == 1 || mat[0, j] == 1) {
mat[i, j] = 1;
}
}
if (col0 == 1) {
mat[i, 0] = 1;
}
}
}
static void Main() {
int[,] arr = {
{1, 0, 0, 1},
{0, 0, 1, 0},
{0, 0, 0, 0}
};
booleanMatrix(arr);
for (int i = 0; i < arr.GetLength(0); i++) {
for (int j = 0; j < arr.GetLength(1); j++) {
Console.Write(arr[i, j] + " ");
}
Console.WriteLine();
}
}}
JavaScript
// Javascript Code For Boolean Matrix Question // Most Optimised Approach function booleanMatrix(mat) { let col0 = 0; const rows = mat.length; const cols = mat[0].length;
// Step 1: Mark the first row and column
// if there is a 1 in the matrix
for (let i = 0; i < rows; i++) {
// Check if 0th column contains 1
if (mat[i][0] === 1) col0 = 1;
for (let j = 1; j < cols; j++) {
if (mat[i][j] === 1) {
mat[i][0] = 1;
mat[0][j] = 1;
}
}
}
// Step 2: Traverse the matrix in reverse direction and update values
for (let i = rows - 1; i >= 0; i--) {
for (let j = cols - 1; j >= 1; j--) {
if (mat[i][0] === 1 || mat[0][j] === 1) {
mat[i][j] = 1;
}
}
if (col0 === 1) {
mat[i][0] = 1;
}
}}
// Driver Code const arr = [ [1, 0, 0, 1], [0, 0, 1, 0], [0, 0, 0, 0] ];
booleanMatrix(arr);
for (let i = 0; i < arr.length; i++) { console.log(arr[i].join(" ")); }
`
Output
1 1 1 1 1 1 1 1 1 0 1 1
**Time Complexity: O(n * m) where n is number of rows and m is number of columns in given matrix.
**Auxiliary Space: O(1)