给定一个正数“N”，任务是找出“N”是否平衡。如果“N”是平衡数，则输出“YES”，否则输出“NO”。
A number is balanced if the frequency of all the digits in it is same i.e. all the digits appear the same number of times.
例子：
Input: N = 1234567890
Output: YES
The frequencies of all the digits are same.
i.e. every digit appears same number of times.
Input: N = 1337
Output: NO方法：
- 创建一个大小为 10 的数组 freq[] ，它会将每个数字的频率存储在“N”中。
- 然后，检查“N”的所有数字是否具有相同的频率。
- 如果是，则打印 ‘YES’ 或 ‘NO’ 否则。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// returns true if the number
// passed as the argument
// is a balanced number.
bool isNumBalanced(int N)
{
string st = to_string(N);
bool isBalanced = true;
// frequency array to store
// the frequencies of all
// the digits of the number
int freq[10] = { 0 };
int i = 0;
int n = st.size();
for (i = 0; i < n; i++)
// store the frequency of
// the current digit
freq[st[i] - '0']++;
sort(freq,freq+10);
int k=9; // last index of the array
while(freq[k-1]!=0){
if(freq[k]!=freq[k-1]){
isBalanced=false;
}
k--;
}
// return true if
// the string is balanced
if (isBalanced)
return true;
else
return false;
}
// Driver code
int main()
{
int N = 1234567890;
// Function call
bool flag = isNumBalanced(N);
if (flag)
cout << "YES";
else
cout << "NO";
}
// This code is contributed by ihritik Java
// JAVA implementation using Hashmap and Collection
import java.util.*;
public class Num_balanced {
public static boolean isNumBalanced(int n)
{
// Calling integer to char array convert function
char[] num = num_to_arr(n);
// HashMap is used to store key value pairs
HashMap hp
= new HashMap();
// traverse char array and store array elements as
// key and their frequency as their value
for (int i = 0; i < num.length; i++)
{
// if element already exists in the HashMap, so
// we increment its previous value
if (hp.containsKey(num[i]))
{
hp.put(num[i], hp.get(num[i]) + 1);
}
// element does'nt exist in the HashMap, so we
// initialize its value with 1
else
{
hp.put(num[i], 1);
}
}
// use Collection to store values of all the keys
Collection c = (Collection)hp.values();
// use iterator to iterate over each value
Iterator iterator = c.iterator();
int temp = (Integer) iterator.next();
while (iterator.hasNext())
{
// compare each value to be equal, if not return
// false
if ((int) iterator.next() != temp) {
return false;
}
}
// each value was equal so return true
return true;
}
// This function converts integer into char array
public static char[] num_to_arr(int num)
{
// Convert integer into String
String str = Integer.toString(num);
char[] arr = new char[str.length()];
// Insert characters of the string into char array
for (int i = 0; i < str.length(); i++) {
arr[i] = str.charAt(i);
}
return arr;
}
// Driver code
public static void main(String[] args)
{
int n_1 = 1234567890;
// Function call
if (isNumBalanced(n_1))
{
// if number is balanced
System.out.println("YES");
}
else
{
// if number is not balanced
System.out.println("NO");
}
}
}
// Code Contributed by YESHU GARG Java
// Implementation of JAVA to check the
// given number is Balanced or not
import java.util.*;
public class Main
{
// returns true if the number
// passed as the argument
// is a balanced number.
public static boolean isNumBalanced(int num)
{
// to get the absolute value of the number
num = Math.abs(num);
// to convert the int number into a String
String str = num + "";
// to convert the String into Character Array
char[] ch_arr = str.toCharArray();
// HashSet is used to remove the duplicates
// in the Character Array
HashSet hs = new HashSet();
for (char ch : ch_arr)
{
// Adding the Characters in the Array in the Set
hs.add(ch);
}
// getting the length of the String
int str_len = str.length();
// getting the numbers of elements in the HashSet
int hs_len = hs.size();
// return true if
// the number is balanced
// checks for the number is balanced or not by
// comparing length of String and HashSet
if (hs_len <= str_len / 2 || hs_len == str_len)
{
return true;
}
return false;
}
// Driver Code
public static void main(String[] args)
{
int N = 1234567890;
// Function call
boolean flag = isNumBalanced(N);
if (flag)
System.out.println("YES");
else
System.out.println("NO");
}
}
// This code is contributed by Mano Python3
# Python3 implementation of the above approach
# Returns true if the number passed as
# the argument is a balanced number.
def isNumBalanced(N):
st = str(N)
isBalanced = True
# Frequency array to store the frequencies
# of all the digits of the number
freq = [0] * 10
n = len(st)
for i in range(0, n):
# store the frequency of the
# current digit
freq[int(st[i])] += 1
for i in range(0, 9):
# if freq[i] is not equal to
# freq[i + 1] at any index 'i'
# then set isBalanced to false
if freq[i] != freq[i + 1]:
isBalanced = False
# Return true if the string
# is balanced
if isBalanced:
return True
else:
return False
# Driver code
if __name__ == "__main__":
N = 1234567890
# Function call
flag = isNumBalanced(N)
if flag:
print("YES")
else:
print("NO")
# This code is contributed by Rituraj JainC#
// CSHARP implementation of the above approach
using System;
class Program
{
// returns true if the number
// passed as the argument
// is a balanced number.
static bool isNumBalanced(int N)
{
String st = "" + N;
bool isBalanced = true;
// frequency array to store
// the frequencies of all
// the digits of the number
int[] freq = new int[10];
int i = 0;
int n = st.Length;
for (i = 0; i < n; i++)
// store the frequency of
// the current digit
freq[st[i] - '0']++;
for (i = 0; i < 9; i++)
{
// if freq[i] is not
// equal to freq[i + 1] at
// any index ‘i’ then set
// isBalanced to false
if (freq[i] != freq[i + 1])
isBalanced = false;
}
// return true if
// the string is balanced
if (isBalanced)
return true;
else
return false;
}
// Driver code
static void Main()
{
int N = 1234567890;
// Function call
bool flag = isNumBalanced(N);
if (flag)
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
// This code is contributed by ANKITRAI1
}PHP
Javascript
输出
YES