Se da un cierto número y la tarea es contar los dígitos pares y los dígitos impares del número y también los dígitos pares están presentes incluso un número de veces y, de manera similar, para los números impares.
Print Yes If: If number contains even digits even number of time Odd digits odd number of times Else Print No
Ejemplos:
Input : 22233
Output : NO
count_even_digits = 3
count_odd_digits = 2
In this number even digits occur odd number of times and odd
digits occur even number of times so its print NO.
Input : 44555
Output : YES
count_even_digits = 2
count_odd_digits = 3
In this number even digits occur even number of times and odd
digits occur odd number of times so its print YES.
Solución eficiente para calcular dígitos pares e impares en un número.
C++
// C++ program to count
// even and odd digits
// in a given number
#include <iostream>
using namespace std;
// Function to count digits
int countEvenOdd(int n)
{
int even_count = 0;
int odd_count = 0;
while (n > 0)
{
int rem = n % 10;
if (rem % 2 == 0)
even_count++;
else
odd_count++;
n = n / 10;
}
cout << "Even count : "
<< even_count;
cout << "\nOdd count : "
<< odd_count;
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return 1;
else
return 0;
}
// Driver Code
int main()
{
int n;
n = 2335453;
int t = countEvenOdd(n);
if (t == 1)
cout << "\nYES" << endl;
else
cout << "\nNO" << endl;
return 0;
}
Java
// Java program to count
// even and odd digits
// in a given number
import java.io.*;
class GFG
{
// Function to count digits
static int countEvenOdd(int n)
{
int even_count = 0;
int odd_count = 0;
while (n > 0)
{
int rem = n % 10;
if (rem % 2 == 0)
even_count++;
else
odd_count++;
n = n / 10;
}
System.out.println ( "Even count : " +
even_count);
System.out.println ( "Odd count : " +
odd_count);
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return 1;
else
return 0;
}
// Driver Code
public static void main (String[] args)
{
int n;
n = 2335453;
int t = countEvenOdd(n);
if (t == 1)
System.out.println ( "YES" );
else
System.out.println( "NO") ;
}
}
Python3
# python program to count even and
# odd digits in a given number
# Function to count digits
def countEvenOdd(n):
even_count = 0
odd_count = 0
while (n > 0):
rem = n % 10
if (rem % 2 == 0):
even_count += 1
else:
odd_count += 1
n = int(n / 10)
print( "Even count : " , even_count)
print("\nOdd count : " , odd_count)
if (even_count % 2 == 0 and
odd_count % 2 != 0):
return 1
else:
return 0
# Driver code
n = 2335453;
t = countEvenOdd(n);
if (t == 1):
print("YES")
else:
print("NO")
# This code is contributed by Sam007.
C#
// C# program to count even and
// odd digits in a given number
using System;
class GFG {
// Function to count digits
static int countEvenOdd(int n)
{
int even_count = 0;
int odd_count = 0;
while (n > 0) {
int rem = n % 10;
if (rem % 2 == 0)
even_count++;
else
odd_count++;
n = n / 10;
}
Console.WriteLine("Even count : " +
even_count);
Console.WriteLine("Odd count : " +
odd_count);
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return 1;
else
return 0;
}
// Driver Code
public static void Main ()
{
int n;
n = 2335453;
int t = countEvenOdd(n);
if (t == 1)
Console.WriteLine ("YES");
else
Console.WriteLine("NO") ;
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to count
// even and odd digits
// in a given number
// Function to count digits
function countEvenOdd($n)
{
$even_count = 0;
$odd_count = 0;
while ($n > 0)
{
$rem = $n % 10;
if ($rem % 2 == 0)
$even_count++;
else
$odd_count++;
$n = (int)($n / 10);
}
echo("Even count : " .
$even_count);
echo("\nOdd count : " .
$odd_count);
if ($even_count % 2 == 0 &&
$odd_count % 2 != 0)
return 1;
else
return 0;
}
// Driver code
$n = 2335453;
$t = countEvenOdd($n);
if ($t == 1)
echo("\nYES");
else
echo("\nNO");
// This code is contributed by Ajit.
?>
Javascript
<script>
/// Javascript program to count
// even and odd digits
// in a given number
// Function to count digits
function countEvenOdd(n)
{
let even_count = 0;
let odd_count = 0;
while (n > 0)
{
let rem = n % 10;
if (rem % 2 == 0)
even_count++;
else
odd_count++;
n = Math.floor(n / 10);
}
document.write("Even count : "
+ even_count);
document.write("<br>" + "Odd count : "
+ odd_count);
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return 1;
else
return 0;
}
// Driver Code
let n;
n = 2335453;
let t = countEvenOdd(n);
if (t == 1)
document.write("<br>" + "YES" + "<br>");
else
document.write("<br>"+"NO" + "<br>");
// This code is contributed by Mayank Tyagi
</script>
Producción
Even count : 2 Odd count : 5 YES
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(1)
Otra solución para resolver este problema es una array o string de caracteres.
C++
// C++ program to count
// even and odd digits
// in a given number
// using char array
#include <bits/stdc++.h>
using namespace std;
// Function to count digits
int countEvenOdd(char num[],
int n)
{
int even_count = 0;
int odd_count = 0;
for (int i = 0; i < n; i++)
{
int x = num[i] - 48;
if (x % 2 == 0)
even_count++;
else
odd_count++;
}
cout << "Even count : "
<< even_count;
cout << "\nOdd count : "
<< odd_count;
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return 1;
else
return 0;
}
// Driver Code
int main()
{
char num[18] = { 1, 2, 3 };
int n = strlen(num);
int t = countEvenOdd(num, n);
if (t == 1)
cout << "\nYES" << endl;
else
cout << "\nNO" << endl;
return 0;
}
Java
// Java program to count
// even and odd digits
// in a given number
// using char array
import java.io.*;
class GFG
{
// Function to count digits
static int countEvenOdd(char num[],
int n)
{
int even_count = 0;
int odd_count = 0;
for (int i = 0; i < n; i++)
{
int x = num[i] - 48;
if (x % 2 == 0)
even_count++;
else
odd_count++;
}
System.out.println ("Even count : " +
even_count);
System.out.println( "Odd count : " +
odd_count);
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return 1;
else
return 0;
}
// Driver Code
public static void main (String[] args)
{
char num[] = { 1, 2, 3 };
int n = num.length;
int t = countEvenOdd(num, n);
if (t == 1)
System.out.println("YES") ;
else
System.out.println("NO") ;
}
}
// This code is contributed by vt_m
Python3
# Python3 program to count
# even and odd digits
# in a given number
# using char array
# Function to count digits
def countEvenOdd(num, n):
even_count = 0;
odd_count = 0;
num=list(str(num))
for i in num:
if i in ('0','2','4','6','8'):
even_count+=1
else:
odd_count+=1
print("Even count : ",
even_count);
print("Odd count : ",
odd_count);
if (even_count % 2 == 0 and
odd_count % 2 != 0):
return 1;
else:
return 0;
# Driver Code
num = (1, 2, 3);
n = len(num);
t = countEvenOdd(num, n);
if t == 1:
print("YES");
else:
print("NO");
# This code is contributed by mits.
C#
// C# program to count
// even and odd digits
// in a given number
// using char array
using System;
class GFG
{
// Function to count digits
static int countEvenOdd(char []num,
int n)
{
int even_count = 0;
int odd_count = 0;
for (int i = 0; i < n; i++)
{
int x = num[i] - 48;
if (x % 2 == 0)
even_count++;
else
odd_count++;
}
Console.WriteLine("Even count : " +
even_count);
Console.WriteLine( "Odd count : " +
odd_count);
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return 1;
else
return 0;
}
// Driver code
public static void Main ()
{
char [] num = { '1', '2', '3' };
int n = num.Length;
int t = countEvenOdd(num, n);
if (t == 1)
Console.WriteLine("YES") ;
else
Console.WriteLine("NO") ;
}
}
// This code is contributed by Sam007.
PHP
<?php
// PHP program to count
// even and odd digits
// in a given number
// using char array
// Function to count digits
function countEvenOdd($num, $n)
{
$even_count = 0;
$odd_count = 0;
for ($i = 0; $i < $n; $i++)
{
$x = $num[$i] - 48;
if ($x % 2 == 0)
$even_count++;
else
$odd_count++;
}
echo "Even count : " .
$even_count;
echo "\nOdd count : " .
$odd_count;
if ($even_count % 2 == 0 &&
$odd_count % 2 != 0)
return 1;
else
return 0;
}
// Driver Code
$num = array( 1, 2, 3 );
$n = strlen(num);
$t = countEvenOdd($num, $n);
if ($t == 1)
echo "\nYES" ,"\n";
else
echo "\nNO" ,"\n";
// This code is contributed by ajit.
?>
Javascript
<script>
// Javascript program to count
// even and odd digits
// in a given number
// using char array
// Function to count digits
function countEvenOdd(num, n)
{
even_count = 0;
odd_count = 0;
for(var i = 0; i < n; i++)
{
x = num[i] - 48;
if (x % 2 == 0)
even_count++;
else
odd_count++;
}
document.write("Even count : " +
even_count);
document.write("<br>Odd count : " +
odd_count);
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return 1;
else
return 0;
}
// Driver code
var num = [ 1, 2, 3 ];
n = num.length;
t = countEvenOdd(num, n);
if (t == 1)
document.write("<br>YES <br>");
else
document.write("<br>NO <br>");
// This code is contributed by akshitsaxenaa09
</script>
Producción
Even count : 1 Odd count : 2 NO
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(1)
Método n.º 3: uso de encasillamiento (enfoque simplificado):
- Tenemos que convertir el número dado en una string tomando una nueva variable.
- Atraviese la string, convierta cada elemento en un número entero.
- Si el carácter (dígito) es par, entonces el conteo aumentado
- De lo contrario, incremente el conteo impar.
- Si el conteo par es par y el conteo impar es impar, entonces imprima Sí.
- De lo contrario, imprima no.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of above approach
#include<bits/stdc++.h>
using namespace std;
string getResult(int n)
{
// Converting integer to String
string st = to_string(n);
int even_count = 0;
int odd_count = 0;
// Looping till length of String
for(int i = 0; i < st.length(); i++)
{
if ((st[i] % 2) == 0)
// Digit is even so increment even count
even_count += 1;
else
odd_count += 1;
}
// Checking even count is even and
// odd count is odd
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return "Yes";
else
return "no";
}
// Driver Code
int main(){
int n = 77788;
// Passing this number to get result function
cout<<getResult(n)<<endl;
}
// This code is contributed by shinjanpatra.
Java
// Java implementation of above approach
class GFG{
static String getResult(int n)
{
// Converting integer to String
String st = String.valueOf(n);
int even_count = 0;
int odd_count = 0;
// Looping till length of String
for(int i = 0; i < st.length(); i++)
{
if ((st.charAt(i) % 2) == 0)
// Digit is even so increment even count
even_count += 1;
else
odd_count += 1;
}
// Checking even count is even and
// odd count is odd
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return "Yes";
else
return "no";
}
// Driver Code
public static void main(String[] args)
{
int n = 77788;
// Passing this number to get result function
System.out.println(getResult(n));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python implementation of above approach def getResult(n): # Converting integer to string st = str(n) even_count = 0 odd_count = 0 # Looping till length of string for i in range(len(st)): if((int(st[i]) % 2) == 0): # digit is even so increment even count even_count += 1 else: odd_count += 1 # Checking even count is even and odd count is odd if(even_count % 2 == 0 and odd_count % 2 != 0): return 'Yes' else: return 'no' # Driver Code n = 77788 # passing this number to get result function print(getResult(n)) # this code is contributed by vikkycirus
C#
// C# implementation of above approach
using System;
using System.Collections.Generic;
class GFG{
static String getResult(int n)
{
// Converting integer to String
String st = String.Join("",n);
int even_count = 0;
int odd_count = 0;
// Looping till length of String
for(int i = 0; i < st.Length; i++)
{
if ((st[i] % 2) == 0)
// Digit is even so increment even count
even_count += 1;
else
odd_count += 1;
}
// Checking even count is even and
// odd count is odd
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return "Yes";
else
return "no";
}
// Driver Code
public static void Main(String[] args)
{
int n = 77788;
// Passing this number to get result function
Console.WriteLine(getResult(n));
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// Javascript implementation of above approach
function getResult(n)
{
// Converting integer to String
var st = n.toString();
var even_count = 0;
var odd_count = 0;
// Looping till length of String
for(var i = 0; i < st.length; i++)
{
if ((st.charAt(i) % 2) == 0)
// Digit is even so increment even count
even_count += 1;
else
odd_count += 1;
}
// Checking even count is even and
// odd count is odd
if (even_count % 2 == 0 &&
odd_count % 2 != 0)
return "Yes";
else
return "no";
}
// Driver Code
var n = 77788;
// Passing this number to get result function
document.write(getResult(n));
</script>
Producción
Yes
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(1)
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA