Frequencies of even and odd numbers in a matrix
Given a matrix of order m*n then the task is to find the frequency of even and odd numbers in matrix
Examples:
Input : m = 3, n = 3
{ 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 }
Output : Frequency of odd number = 5
Frequency of even number = 4
Input : m = 3, n = 3
{ 10, 11, 12 },
{ 13, 14, 15 },
{ 16, 17, 18 }
Output : Frequency of odd number = 4
Frequency of even number = 5
Implementation:
CPP
#include<bits/stdc++.h>
using namespace std;
#define MAX 100
void freq( int ar[][MAX], int m, int n)
{
int even = 0, odd = 0;
for ( int i = 0; i < m; ++i)
{
for ( int j = 0; j < n; ++j)
{
if ((ar[i][j] % 2) == 0)
++even;
else
++odd;
}
}
printf ( " Frequency of odd number = %d \n" , odd);
printf ( " Frequency of even number = %d \n" , even);
}
int main()
{
int m = 3, n = 3;
int array[][MAX] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
freq(array, m, n);
return 0;
}
|
Java
class GFG {
static final int MAX = 100 ;
static void freq( int ar[][], int m, int n) {
int even = 0 , odd = 0 ;
for ( int i = 0 ; i < m; ++i)
{
for ( int j = 0 ; j < n; ++j)
{
if ((ar[i][j] % 2 ) == 0 )
++even;
else
++odd;
}
}
System.out.print( " Frequency of odd number =" +
odd + " \n" );
System.out.print( " Frequency of even number = " +
even + " \n" );
}
public static void main(String[] args) {
int m = 3 , n = 3 ;
int array[][] = {{ 1 , 2 , 3 }, { 4 , 5 , 6 }, { 7 , 8 , 9 }};
freq(array, m, n);
}
}
|
Python3
MAX = 100
def freq(ar, m, n):
even = 0
odd = 0
for i in range (m):
for j in range (n):
if ((ar[i][j] % 2 ) = = 0 ):
even + = 1
else :
odd + = 1
print ( " Frequency of odd number =" , odd)
print ( " Frequency of even number =" , even)
m = 3
n = 3
array = [[ 1 , 2 , 3 ],
[ 4 , 5 , 6 ],
[ 7 , 8 , 9 ]]
freq(array, m, n)
|
C#
using System;
class GFG
{
static void freq( int [,]ar, int m, int n)
{
int even = 0, odd = 0;
for ( int i = 0; i < m; ++i)
{
for ( int j = 0; j < n; ++j)
{
if ((ar[i, j] % 2) == 0)
++even;
else
++odd;
}
}
Console.WriteLine( " Frequency of odd number =" +
odd );
Console.WriteLine( " Frequency of even number = " +
even );
}
public static void Main()
{
int m = 3, n = 3;
int [,]array = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
freq(array, m, n);
}
}
|
PHP
<?php
$MAX = 100;
function freq( $ar , $m , $n )
{
$even = 0; $odd = 0;
for ( $i = 0; $i < $m ; ++ $i )
{
for ( $j = 0; $j < $n ; ++ $j )
{
if (( $ar [ $i ][ $j ] % 2) == 0)
++ $even ;
else
++ $odd ;
}
}
echo " Frequency of odd number = "
, $odd , "\n" ;
echo " Frequency of even number = "
, $even ;
}
$m = 3; $n = 3;
$array = array ( array (1, 2, 3),
array (4, 5, 6),
array (7, 8, 9));
freq( $array , $m , $n );
?>
|
Javascript
<script>
let MAX = 100;
function freq(ar,m,n) {
let even = 0, odd = 0;
for (let i = 0; i < m; ++i)
{
for (let j = 0; j < n; ++j)
{
if ((ar[i][j] % 2) == 0)
++even;
else
++odd;
}
}
document.write( " Frequency of odd number =" +
odd + " <br>" );
document.write( " Frequency of even number = " +
even + "<br>" );
}
let m = 3, n = 3;
let array = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];
freq(array, m, n);
</script>
|
Output
Frequency of odd number = 5
Frequency of even number = 4
Time Complexity: O(n*m)
Auxiliary Space: O(1), as no extra space is used
Method: Using bitwise & operator
C++
#include <bits/stdc++.h>
using namespace std;
#define MAX 100
void freq( int ar[][MAX], int m, int n)
{
int even = 0, odd = 0;
for ( int i = 0; i < m; ++i) {
for ( int j = 0; j < n; ++j) {
if ((ar[i][j] & 1) == 0)
++even;
else
++odd;
}
}
cout << "Frequency of odd number = " << odd << endl;
cout << "Frequency of even number = " << even << endl;
}
int main()
{
int m = 3, n = 3;
int array[][MAX]
= { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
freq(array, m, n);
return 0;
}
|
Java
import java.util.*;
public class Main {
public static void freq( int [][] ar, int m, int n)
{
int even = 0 , odd = 0 ;
for ( int i = 0 ; i < m; i++) {
for ( int j = 0 ; j < n; j++) {
if ((ar[i][j] & 1 ) == 0 )
++even;
else
++odd;
}
}
System.out.println( "Frequency of odd number = " + odd);
System.out.println( "Frequency of even number = " + even);
}
public static void main(String[] args)
{
int m = 3 , n = 3 ;
int [][] array= { { 1 , 2 , 3 }, { 4 , 5 , 6 }, { 7 , 8 , 9 } };
freq(array, m, n);
}
}
|
Python3
def freq(ar, m, n):
even = 0
odd = 0
for i in range (m):
for j in range (n):
if ((ar[i][j] & 1 ) = = 0 ):
even + = 1
else :
odd + = 1
print ( "Frequency of odd numbers = " , end = "")
print (odd)
print ( "Frequency of even numbers = " , end = "")
print (even)
m = 3
n = 3
array = [[ 1 , 2 , 3 ], [ 4 , 5 , 6 ], [ 7 , 8 , 9 ]]
freq(array, m, n)
|
C#
using System;
public class Program
{
public static void Freq( int [,] ar, int m, int n)
{
int even = 0, odd = 0;
for ( int i = 0; i < m; i++)
{
for ( int j = 0; j < n; j++)
{
if ((ar[i, j] & 1) == 0)
++even;
else
++odd;
}
}
Console.WriteLine( "Frequency of odd number = " + odd);
Console.WriteLine( "Frequency of even number = " + even);
}
public static void Main( string [] args)
{
int m = 3, n = 3;
int [,] array = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
Freq(array, m, n);
}
}
|
Javascript
function freq(ar, m, n){
let even = 0;
let odd = 0;
for (let i = 0; i<m; i++)
{
for (let j = 0; j<n; j++)
{
if ((ar[i][j] & 1) == 0){
++even;
}
else {
++odd;
}
}
}
console.log( "Frequency of odd numbers = " + odd + "\n" );
console.log( "Frequency of even numbers = " + even + "\n" );
}
let m = 3, n = 3;
let array = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];
freq(array, m, n);
|
Output
Frequency of odd number = 5
Frequency of even number = 4
Time Complexity: O(n*m)
Auxiliary Space: O(1), as no extra space is used
Last Updated :
10 Apr, 2023
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