Dado un número N, la tarea es verificar si el valor xor de la frecuencia de los dígitos es cero o no.
Ejemplos:
Input: N = 122233 Output: Yes Frequencies of 1, 2 and 3 are 1, 3, 2 respectively. And Xor of 1, 3 and 2 is 0. Input: N = 123 Output: No
Enfoque: cuente la frecuencia de todos los dígitos y luego itere sobre todas las frecuencias y córtelas si la respuesta es cero, luego imprima Sí, de lo contrario, No.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the above approach
#include<bits/stdc++.h>
using namespace std;
bool check(int s)
{
// creating a frequency array
int freq[10] = {0},r;
while(s != 0)
{
// Finding the last digit of the number
r = s % 10;
// Dividing the number by 10 to
// eliminate last digit
s = int(s / 10);
// counting frequency of each digit
freq[r] += 1;
}
int xor__ = 0;
// checking if the xor of all frequency is zero or not
for (int i=0;i<10;i++)
{
xor__ = xor__ ^ freq[i];
if(xor__ == 0)
return true;
else
return false;
}
}
// Driver function
int main()
{
int s = 122233;
if(check(s))
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
}
// This code is contributed by
// Surendra_Gangwar
Java
// Java implementation of the above approach
class GFG
{
static boolean check(int s)
{
// creating a frequency array
int[] freq = new int[10];
int r,i;
for(i=0;i<10;i++)
{
freq[i]= 0;
}
while(s != 0)
{
// Finding the last digit of the number
r = s % 10;
// Dividing the number by 10 to
// eliminate last digit
s = (int)(s / 10);
// counting frequency of each digit
freq[r] += 1;
}
int xor__ = 0;
// checking if the xor of all frequency is zero or not
for ( i=0;i<10;i++)
{
xor__ = xor__ ^ freq[i];
if(xor__ == 0)
return true;
else
return false;
}
return true;
}
// Driver function
public static void main(String[] args) {
int s = 122233;
if(check(s))
System.out.println("Yes\n");
else
System.out.println("No\n");
}
// This code is contributed by
// Rajput-Ji
}
Python3
# Python implementation of the above approach
def check(s):
# creating a frequency array
freq =[0]*10
while(s != 0):
# Finding the last digit of the number
r = s % 10
# Dividing the number by 10 to
# eliminate last digit
s = s//10
# counting frequency of each digit
freq[r]+= 1
xor = 0
# checking if the xor of all frequency is zero or not
for i in range(10):
xor = xor ^ freq[i]
if(xor == 0):
return True
else:
return False
s = 122233
if(check(s)):
print("Yes")
else:
print("No")
C#
// C# implementation of the above approach
using System;
class GFG
{
static bool check(int s)
{
// creating a frequency array
int[] freq = new int[10];
int r, i;
for(i = 0; i < 10; i++)
{
freq[i]= 0;
}
while(s != 0)
{
// Finding the last digit of the number
r = s % 10;
// Dividing the number by 10 to
// eliminate last digit
s = (int)(s / 10);
// counting frequency of each digit
freq[r] += 1;
}
int xor__ = 0;
// checking if the xor of all frequency is zero or not
for ( i = 0; i < 10; i++)
{
xor__ = xor__ ^ freq[i];
if(xor__ == 0)
return true;
else
return false;
}
return true;
}
// Driver code
public static void Main()
{
int s = 122233;
if(check(s))
Console.Write("Yes\n");
else
Console.Write("No\n");
}
}
// This code is contributed by Ita_c.
PHP
<?php
// PHP implementation of the above approach
function check($s)
{
// creating a frequency array
$freq =array_fill(0,10,0);
while($s != 0)
{
// Finding the last digit of the number
$r = $s % 10;
// Dividing the number by 10 to
// eliminate last digit
$s = (int)($s/10);
// counting frequency of each digit
$freq[$r]+= 1;
}
$xor = 0;
// checking if the xor of all frequency is zero or not
for($i=0;$i<10;$i++)
$xor = $xor ^ $freq[$i];
if($xor == 0)
return true;
else
return false;
}
// Main Drive
$s = 122233;
if(check($s))
print("Yes");
else
print("No");
// This code is contributed by mits
?>
Javascript
<script>
// Javascript implementation of
// the above approach
function check(s)
{
// creating a frequency array
let freq = new Array(10).fill(0), r;
while(s != 0)
{
// Finding the last digit of the number
r = s % 10;
// Dividing the number by 10 to
// eliminate last digit
s = parseInt(s / 10);
// counting frequency of each digit
freq[r] += 1;
}
let xor__ = 0;
// checking if the xor of all
// frequency is zero or not
for (let i=0;i<10;i++)
{
xor__ = xor__ ^ freq[i];
if(xor__ == 0)
return true;
else
return false;
}
}
// Driver function
let s = 122233;
if(check(s))
document.write("Yes");
else
document.write("No");
</script>
Producción:
Yes
Complejidad de Tiempo: O(|s|), Espacio Auxiliar: O(10)