Given a matrix arr[][] of size N * M, the task is to print the matrix after removing all rows and columns from the matrix which consists of 0s only.
Examples:
Input: arr[][] ={ { 1, 1, 0, 1 }, { 0, 0, 0, 0 }, { 1, 1, 0, 1}, { 0, 1, 0, 1 } }
Output:
111
111
011
Explanation:
Initially, the matrix is as follows:
arr[][] = { { 1, 1, 0, 1 },
{ 0, 0, 0, 0 },
{ 1, 1, 0, 1 },
{ 0, 1, 0, 1 } }
Removing the 2nd row modifies the matrix to:
arr[][] = { { 1, 1, 0, 1 },
{ 1, 1, 0, 1 },
{ 0, 1, 0, 1 } }
Removing the 3rd column modifies the matrix to:
arr[][] = { { 1, 1, 1 },
{ 1, 1, 1 },
{ 0, 1, 1 } }Input: arr={{0, 1}, {0, 1}}
Output:
1
1
Approach: The idea is to count the number of 0s in all the rows and columns of the matrix and check if any rows or columns consist only of 0s or not. If found to be true, then remove those rows or the columns of the matrix. Follow the steps below to solve the problem:
- Traverse the matrix and count 1s in rows and columns.
- Now, traverse over the loop again and check for the following:
- If the count of 1s is found to be 0 for any row, skip that row.
- If the count of 1s is found to be greater than 0 for any column, print that element.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to remove the rows or columns from // the matrix which contains all 0s elements void removeZeroRowCol(vector<vector< int > >& arr)
{ // Stores count of rows
int n = arr.size();
// col[i]: Stores count of 0s
// in current column
int col[n + 1] = { 0 };
// row[i]: Stores count of 0s
// in current row
int row[n + 1] = { 0 };
// Traverse the matrix
for ( int i = 0; i < n; ++i) {
// Stores count of 0s
// in current row
int count = 0;
for ( int j = 0; j < n; ++j) {
// Update col[j]
col[j] += (arr[i][j] == 1);
// Update count
count += (arr[i][j] == 1);
}
// Update row[i]
row[i] = count;
}
// Traverse the matrix
for ( int i = 0; i < n; ++i) {
// If all elements of
// current row is 0
if (row[i] == 0) {
continue ;
}
for ( int j = 0; j < n; ++j) {
// If all elements of
// current column is 0
if (col[j] != 0)
cout << arr[i][j];
}
cout << "\n" ;
}
} // Driver Code int main()
{ vector<vector< int > > arr = { { 1, 1, 0, 1 },
{ 0, 0, 0, 0 },
{ 1, 1, 0, 1 },
{ 0, 1, 0, 1 } };
// Function Call
removeZeroRowCol(arr);
return 0;
} |
// Java program for the above approach import java.io.*;
class GFG{
// Function to remove the rows or columns from // the matrix which contains all 0s elements static void removeZeroRowCol( int arr[][])
{ // Stores count of rows
int n = arr.length;
// col[i]: Stores count of 0s
// in current column
int col[] = new int [n + 1 ];
// row[i]: Stores count of 0s
// in current row
int row[] = new int [n + 1 ];
// Traverse the matrix
for ( int i = 0 ; i < n; ++i)
{
// Stores count of 0s
// in current row
int count = 0 ;
for ( int j = 0 ; j < n; ++j)
{
if (arr[i][j] == 1 )
// Update col[j]
col[j] += 1 ;
else
col[j] += 0 ;
if (arr[i][j] == 1 )
// Update count
count += 1 ;
else
count += 0 ;
}
// Update row[i]
row[i] = count;
}
// Traverse the matrix
for ( int i = 0 ; i < n; ++i)
{
// If all elements of
// current row is 0
if (row[i] == 0 )
{
continue ;
}
for ( int j = 0 ; j < n; ++j)
{
// If all elements of
// current column is 0
if (col[j] != 0 )
System.out.print(arr[i][j]);
}
System.out.println();
}
} // Driver Code public static void main (String[] args)
{ int arr[][] = { { 1 , 1 , 0 , 1 },
{ 0 , 0 , 0 , 0 },
{ 1 , 1 , 0 , 1 },
{ 0 , 1 , 0 , 1 } };
// Function Call
removeZeroRowCol(arr);
} } // This code is contributed by AnkThon |
# Python3 program for the above approach # Function to remove the rows or columns from # the matrix which contains all 0s elements def removeZeroRowCol(arr) :
# Stores count of rows
n = len (arr)
# col[i]: Stores count of 0s
# in current column
col = [ 0 ] * (n + 1 )
# row[i]: Stores count of 0s
# in current row
row = [ 0 ] * (n + 1 )
# Traverse the matrix
for i in range (n) :
# Stores count of 0s
# in current row
count = 0 for j in range (n) :
# Update col[j]
col[j] + = (arr[i][j] = = 1 )
# Update count
count + = (arr[i][j] = = 1 )
# Update row[i]
row[i] = count
# Traverse the matrix
for i in range (n) :
# If all elements of
# current row is 0
if (row[i] = = 0 ) :
continue for j in range (n) :
# If all elements of
# current column is 0
if (col[j] ! = 0 ) :
print (arr[i][j], end = "")
print ()
arr = [ [ 1 , 1 , 0 , 1 ],
[ 0 , 0 , 0 , 0 ],
[ 1 , 1 , 0 , 1 ],
[ 0 , 1 , 0 , 1 ] ]
# Function Call removeZeroRowCol(arr) # This code is contributed by divyeshrabadiya07 |
// C# program for the above approach using System;
class GFG{
// Function to remove the rows or columns from // the matrix which contains all 0s elements static void removeZeroRowCol( int [,] arr)
{ // Stores count of rows
int n = arr.GetLength(0);
// col[i]: Stores count of 0s
// in current column
int [] col = new int [n + 1];
// row[i]: Stores count of 0s
// in current row
int [] row = new int [n + 1];
// Traverse the matrix
for ( int i = 0; i < n ; ++i)
{
// Stores count of 0s
// in current row
int count = 0;
for ( int j = 0; j < n ; ++j)
{
if (arr[i, j] == 1)
// Update col[j]
col[j] += 1;
else
col[j] += 0;
if (arr[i, j] == 1)
// Update count
count += 1;
else
count += 0;
}
// Update row[i]
row[i] = count;
}
// Traverse the matrix
for ( int i = 0; i < n; ++i)
{
// If all elements of
// current row is 0
if (row[i] == 0)
{
continue ;
}
for ( int j = 0; j < n; ++j)
{
// If all elements of
// current column is 0
if (col[j] != 0)
Console.Write(arr[i, j]);
}
Console.WriteLine();
}
} // Driver Code public static void Main (String[] args)
{ int [,] arr = { { 1, 1, 0, 1 },
{ 0, 0, 0, 0 },
{ 1, 1, 0, 1 },
{ 0, 1, 0, 1 } };
// Function Call
removeZeroRowCol(arr);
} } // This code is contributed by susmitakundugoaldanga |
<script> // Javascript program to implement // the above approach // Function to remove the rows or columns from // the matrix which contains all 0s elements function removeZeroRowCol(arr)
{ // Stores count of rows
let n = arr.length;
// col[i]: Stores count of 0s
// in current column
let col = Array.from({length: n+1}, (_, i) => 0);
// row[i]: Stores count of 0s
// in current row
let row = Array.from({length: n+1}, (_, i) => 0);
// Traverse the matrix
for (let i = 0; i < n; ++i)
{
// Stores count of 0s
// in current row
let count = 0;
for (let j = 0; j < n; ++j)
{
if (arr[i][j] == 1)
// Update col[j]
col[j] += 1;
else
col[j] += 0;
if (arr[i][j] == 1)
// Update count
count += 1;
else
count += 0;
}
// Update row[i]
row[i] = count;
}
// Traverse the matrix
for (let i = 0; i < n; ++i)
{
// If all elements of
// current row is 0
if (row[i] == 0)
{
continue ;
}
for (let j = 0; j < n; ++j)
{
// If all elements of
// current column is 0
if (col[j] != 0)
document.write(arr[i][j]);
}
document.write( "<br/>" );
}
} // Driver Code let arr = [[ 1, 1, 0, 1 ],
[ 0, 0, 0, 0 ],
[ 1, 1, 0, 1 ],
[ 0, 1, 0, 1 ]];
// Function Call
removeZeroRowCol(arr);
// This code is contributed by souravghosh0416.
</script> |
111 111 011
Time complexity: O(N*M)
Auxiliary Space: O(N+M)
Another efficient approach: The idea is to mark all rows and columns that contain all zeros with some other special integer (a value that is not present in the matrix) to identify that we do not need that particular row or column and whenever we reach While iterating over the matrix on that particular integer we skip that cell because we assume that the cell has been removed while deletion of the row or column which has all zeros.
Follow the below steps to implement the idea:
- Iterate over all rows one by one and check if that particular row contains all zeros or not.
If, particular row contains all zeros then fill special integer to mark that row containing all zeros. - Iterate over all columns one by one and check if that particular column contains all zeros or not.
If, particular column contains all zeros then fill special integer to mark that column containing all zeros. - Iterate over the matrix and check if any particular cell is marked with special integer.
If yes, then skip that cell.
Otherwise, print that cell.
Below is the implementation of above approach:
// C++ code to implement the above approach #include <bits/stdc++.h> using namespace std;
// Function to mark all row and column // to special integer which contains all zeros void removeZeroRowCol(vector<vector< int > >& A)
{ int i, j, m, n;
m = A.size();
if (m != 0)
n = A[0].size();
// Traversing by row
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
// If found that the complete row do
// not contain only zeros
if (A[i][j] == 1)
break ;
}
// If found that the complete row
// contain zero
if (j == n) {
for (j = 0; j < n; j++)
A[i][j] = -1;
}
}
// Traversing by column
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
// If found that the complete column
// do not contain all zeros
if (A[j][i] == 1)
break ;
}
// If found that the complete column
// contain only zeros
if (j == n) {
for (j = 0; j < n; j++)
A[j][i] = -1;
}
}
} // Function to print the matrix void print(vector<vector< int > >& A)
{ int i, j, m, n;
m = A.size();
if (m != 0)
n = A[0].size();
// Taking a flag which helps us in printing
// the matrix because if we found that the above
// row is contain only zero then we won't go to next
// line otherwise there will be space between
// two consecutive rows
bool flag = false ;
// Iterating over matrix
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
if (A[i][j] != -1) {
cout << A[i][j];
// Making flag true if row get
// printed
flag = true ;
}
}
// If row got printed then moving to the
// next line
if (flag) {
cout << endl;
flag = !flag;
}
}
} // Driver code int main()
{ // Initializing matrix
vector<vector< int > > arr{ { 1, 1, 0, 1 },
{ 0, 0, 0, 0 },
{ 1, 1, 0, 1 },
{ 0, 1, 0, 1 } };
// Function calling
removeZeroRowCol(arr);
// Function to print the matrix
print(arr);
return 0;
} |
/*package whatever //do not write package name here */ import java.io.*;
class GFG {
// Java code to implement the above approach
// Function to mark all row and column
// to special integer which contains all zeros
static void removeZeroRowCol( int [][] A)
{
int i = 0 , j = 0 , m = A.length, n = 0 ;
if (m != 0 )
n = A[ 0 ].length;
// Traversing by row
for (i = 0 ; i < m; i++) {
for (j = 0 ; j < n; j++) {
// If found that the complete row do
// not contain only zeros
if (A[i][j] == 1 )
break ;
}
// If found that the complete row
// contain zero
if (j == n) {
for (j = 0 ; j < n; j++)
A[i][j] = - 1 ;
}
}
// Traversing by column
for (i = 0 ; i < m; i++) {
for (j = 0 ; j < n; j++) {
// If found that the complete column
// do not contain all zeros
if (A[j][i] == 1 )
break ;
}
// If found that the complete column
// contain only zeros
if (j == n) {
for (j = 0 ; j < n; j++)
A[j][i] = - 1 ;
}
}
}
private static void e() {
}
// Function to print the matrix
static void print( int [][] A)
{
int i = 0 , j = 0 , m = 0 , n = 0 ;
m = A.length;
if (m != 0 )
n = A[ 0 ].length;
// Taking a flag which helps us in printing
// the matrix because if we found that the above
// row is contain only zero then we won't go to next
// line otherwise there will be space between
// two consecutive rows
boolean flag = false ;
// Iterating over matrix
for (i = 0 ; i < m; i++) {
for (j = 0 ; j < n; j++) {
if (A[i][j] != - 1 ) {
System.out.print(A[i][j]);
// Making flag true if row get
// printed
flag = true ;
}
}
// If row got printed then moving to the
// next line
if (flag) {
System.out.println();
flag = !flag;
}
}
}
// Driver Code
public static void main(String args[])
{
// Initializing matrix
int [][] arr = { { 1 , 1 , 0 , 1 },
{ 0 , 0 , 0 , 0 },
{ 1 , 1 , 0 , 1 },
{ 0 , 1 , 0 , 1 } };
// Function calling
removeZeroRowCol(arr);
// Function to print the matrix
print(arr);
}
} // This code is contributed by shinjanpatra |
# Python code to implement the above approach # Function to mark all row and column # to special integer which contains all zeros def removeZeroRowCol(A):
m = len (A)
if (m ! = 0 ):
n = len (A[ 0 ])
i,j = 0 , 0
# Traversing by row
for i in range (m):
for j in range (n):
# If found that the complete row do
# not contain only zeros
if (A[i][j] = = 1 ):
break
# If found that the complete row
# contain zero
if (j = = n - 1 ):
for j in range (n):
A[i][j] = - 1
# Traversing by column
for i in range (m):
for j in range (n):
# If found that the complete column
# do not contain all zeros
if (A[j][i] = = 1 ):
break
# If found that the complete column
# contain only zeros
if (j = = n - 1 ):
for j in range (n):
A[j][i] = - 1
# Function to print the matrix def Print (A):
m = len (A)
if (m ! = 0 ):
n = len (A[ 0 ])
# Taking a flag which helps us in printing
# the matrix because if we found that the above
# row is contain only zero then we won't go to next
# line otherwise there will be space between
# two consecutive rows
flag = False
# Iterating over matrix
for i in range (m):
for j in range (n):
if (A[i][j] ! = - 1 ):
print (A[i][j],end = "")
# Making flag true if row get
# printed
flag = True
# If row got printed then moving to the
# next line
if (flag):
print ()
flag = False if (flag = = True ) else True
# Driver code # Initializing matrix arr = [ [ 1 , 1 , 0 , 1 ],
[ 0 , 0 , 0 , 0 ],
[ 1 , 1 , 0 , 1 ],
[ 0 , 1 , 0 , 1 ] ]
# Function calling removeZeroRowCol(arr) # Function to print the matrix Print (arr)
# This code is contributed by shinjanpatra |
using System;
using System.Collections.Generic;
class GFG {
// Java code to implement the above approach
// Function to mark all row and column
// to special integer which contains all zeros
static void removeZeroRowCol( int [,] A)
{
int i = 0, j = 0, m = A.GetLength(0), n = 0;
if (m != 0)
n = A.GetLength(1);
// Traversing by row
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
// If found that the complete row do
// not contain only zeros
if (A[i,j] == 1)
break ;
}
// If found that the complete row
// contain zero
if (j == n) {
for (j = 0; j < n; j++)
A[i,j] = -1;
}
}
// Traversing by column
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
// If found that the complete column
// do not contain all zeros
if (A[j,i] == 1)
break ;
}
// If found that the complete column
// contain only zeros
if (j == n) {
for (j = 0; j < n; j++)
A[j,i] = -1;
}
}
}
private static void e() {
}
// Function to print the matrix
static void print( int [,] A)
{
int i = 0, j = 0, m = 0, n = 0;
m = A.GetLength(0);
if (m != 0)
n = A.GetLength(1);
// Taking a flag which helps us in printing
// the matrix because if we found that the above
// row is contain only zero then we won't go to next
// line otherwise there will be space between
// two consecutive rows
bool flag = false ;
// Iterating over matrix
for (i = 0; i < m; i++) {
for (j = 0; j < n; j++) {
if (A[i,j] != -1) {
Console.Write(A[i,j]);
// Making flag true if row get
// printed
flag = true ;
}
}
// If row got printed then moving to the
// next line
if (flag) {
Console.WriteLine();
flag = !flag;
}
}
}
// Driver Code
public static void Main( string [] args)
{
// Initializing matrix
int [,] arr = { { 1, 1, 0, 1 },
{ 0, 0, 0, 0 },
{ 1, 1, 0, 1 },
{ 0, 1, 0, 1 } };
// Function calling
removeZeroRowCol(arr);
// Function to print the matrix
print(arr);
}
} // This code is contributed by phasing17 |
// JS code to implement the above approach // Function to mark all row and column // to special integer which contains all zeros function removeZeroRowCol(A)
{ let m = A.length
if (m != 0)
n = A[0].length
let i = 0, j = 0
// Traversing by row
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
// If found that the complete row do
// not contain only zeros
if (A[i][j] == 1)
break
}
// If found that the complete row
// contain zero
if (j == n-1)
for (j = 0; j < n; j++)
A[i][j] = -1
}
// Traversing by column
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
// If found that the complete column
// do not contain all zeros
if (A[j][i] == 1)
break
}
// If found that the complete column
// contain only zeros
if (j == n-1)
for (j = 0; j < n; j++)
A[j][i] = -1
}
} // Function to print the matrix function Print(A)
{ let m = A.length
if (m != 0)
n = A[0].length
// Taking a flag which helps us in printing
// the matrix because if we found that the above
// row is contain only zero then we won't go to next
// line otherwise there will be space between
// two consecutive rows
flag = false
// Iterating over matrix
for (i = 0; i < m; i++)
{
for (j = 0; j < n; j++)
{
if (A[i][j] != -1)
{
process.stdout.write(A[i][j] + "" )
// Making flag true if row get
// printed
flag = true
}
}
// If row got printed then moving to the
// next line
if (flag)
{
console.log()
flag = !flag
}
}
} // Driver code // Initializing matrix let arr = [ [ 1, 1, 0, 1 ], [ 0, 0, 0, 0 ],
[ 1, 1, 0, 1 ],
[ 0, 1, 0, 1 ] ]
// Function calling removeZeroRowCol(arr) // Function to print the matrix Print(arr) // This code is contributed by phasing17 |
111 111 011
Time complexity: O(N*M)
Auxiliary Space: O(1)