检查数字N的所有数字的频数的Xor是否为零

给定数字N，任务是检查数字频率的异或值是否为零。

例子：

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

方法：计算所有数字的频率，然后遍历所有频率，如果答案为零，则对它们进行异或运算，然后打印“是”，否则。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include
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"<

---

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.

输出：

Yes