Remove all zero-rows and all zero-columns from a Matrix

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++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output

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++

// 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;
}

Java

/*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

Python3

# 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

C#

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

Javascript

// 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

Output

111
111
011

Time complexity: O(N*M)
Auxiliary Space: O(1)