How to check sparse matrix in python

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A matrix is a two-dimensional data object having m rows and n columns, therefore a total of m*n values. If most of the values of a matrix are 0 then we say that the matrix is sparse. 
Consider a definition of Sparse where a matrix is considered sparse if the number of 0s is more than half of the elements in the matrix,

Examples: 

Input : 1 0 3
        0 0 4
        6 0 0
Output : Yes
There are 5 zeros. This count
is more than half of matrix
size.

Input : 1 2 3
        0 7 8
        5 0 7 
Output: No 

To check whether a matrix is a sparse matrix, we only need to check the total number of elements that are equal to zero. If this count is more than (m * n)/2, we return true. 

Implementation:

C++

#include

using namespace std;

const int MAX = 100;

bool isSparse(int array[][MAX], int m, int n)

{

    int counter = 0;

    for (int i = 0; i < m; ++i)

        for (int j = 0; j < n; ++j)

            if (array[i][j] == 0)

                ++counter;

    return (counter > ((m * n) / 2));

}

int main()

{

    int array[][MAX] = { { 1, 0, 3 },

                        { 0, 0, 4 },

                        { 6, 0, 0 } };

    int m = 3,

        n = 3;

    if (isSparse(array, m, n))

        cout << "Yes";

    else

        cout << "No";

}

Java

import java.io.*;

class GFG {

    static int MAX = 100;

    static boolean isSparse(int array[][], int m, int n)

    {

        int counter = 0;

        for (int i = 0; i < m; ++i)

            for (int j = 0; j < n; ++j)

                if (array[i][j] == 0)

                    ++counter;

        return (counter > ((m * n) / 2));

    }

    public static void main(String args[])

    {

        int array[][] = { { 1, 0, 3 },

                            { 0, 0, 4 },

                            { 6, 0, 0 } };

        int m = 3,

            n = 3;

        if (isSparse(array, m, n))

            System.out.println("Yes");

        else

            System.out.println("No");

    }

}

Python3

MAX = 100

def isSparse(array,m, n) :

    counter = 0

    for i in range(0,m) :

        for j in range(0,n) :

            if (array[i][j] == 0) :

                counter = counter + 1

    return (counter >

            ((m * n) // 2))

array = [ [ 1, 0, 3 ],

          [ 0, 0, 4 ],

          [ 6, 0, 0 ] ]

m = 3

n = 3

if (isSparse(array, m, n)) :

    print("Yes")

else :

    print("No")

C#

using System;

class GFG {

    static bool isSparse(int [,]array, int m,

                                       int n)

    {

        int counter = 0;

        for (int i = 0; i < m; ++i)

            for (int j = 0; j < n; ++j)

                if (array[i,j] == 0)

                    ++counter;

        return (counter > ((m * n) / 2));

    }

    public static void Main()

    {

        int [,]array = { { 1, 0, 3 },

                         { 0, 0, 4 },

                         { 6, 0, 0 } };

        int m = 3,

            n = 3;

        if (isSparse(array, m, n))

            Console.WriteLine("Yes");

        else

            Console.WriteLine("No");

    }

}

PHP

$MAX = 100;

function isSparse( $array, $m, $n)

{

    $counter = 0;

    for ($i = 0; $i < $m; ++$i)

        for ($j = 0; $j < $n; ++$j)

            if ($array[$i][$j] == 0)

                ++$counter;

    return ($counter > (($m * $n) / 2));

}

    $array = array(array(1, 0, 3),

                   array(0, 0, 4),

                   array(6, 0, 0));

    $m = 3;

    $n = 3;

    if (isSparse($array, $m, $n))

        echo "Yes";

    else

        echo "No";

?>

Javascript

Time Complexity: O(m*n) 
Auxiliary Space: O(1)

This article is contributed by Aarti_Rathi and Vineet Joshi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to . See your article appearing on the GeeksforGeeks main page and help other Geeks.