HackerRank Game of Thrones I problem solution

In this HackerRank Game of Thrones - I problem solution we have given a string, determine if it can be rearranged into a palindrome. Return the string YES or NO.

Problem solution in Python.

```def can_be_palindrome(string):
string = sorted(string)
current_letter_count = 1
middle = False
for index, char in enumerate(list(string[1:])):
if string[index] != char:
if current_letter_count % 2:
if not middle:
middle = True
else:
return False
current_letter_count += 1
return True

print('YES' if can_be_palindrome(input()) else 'NO')```

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Problem solution in Java.

```import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner myScan = new Scanner(System.in);
String inputString = myScan.nextLine();

String ans;
int[]count=new int[26];
for(int i=0;i<inputString.length();i++)
count[inputString.charAt(i)-'a']++;
int odds=0;
for(int i=0;i<26;i++)
if(count[i]%2==1)
odds++;
if(((inputString.length()%2==0)&&odds==0)||((inputString.length()%2==1)&&odds==1))
ans="YES";
else
ans="NO";

System.out.println(ans);
myScan.close();
}
}
```

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Problem solution in C++.

```#include <vector>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main() {
vector<long long> stat(26,0);
string str;
getline(cin,str);
for(int i=0; i<str.size(); i++){
stat[str[i]-'a']++;
}
int odd=0;
for(int i=0; i<stat.size(); i++){
if(stat[i] % 2 == 1)
odd++;
if(odd > 1){
cout<<"NO";
break;
}
}
if(!(odd >1 ))
cout<<"YES";
return 0;
}```

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Problem solution in C.

```#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
int x = 0, i, y = 0;
char a[1000001], b[27]={0};
scanf ("%s", &a);
x = strlen (a);
for (i = 0; i < x; i++){
b[a[i]-'a']++;
}
for (i = 0; i < 26; i++){
if (b[i]%2 == 1) y++;
}
if (x % 2 == 1){
if (y == 1){
printf ("YES");
return 0;
}
}else{
if (y == 0){
printf ("YES");
return 0;
}
}
printf("NO");
return 0;
}```

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