# Game of Thrones – I – HackerRank Solution

In this post, we will solve Game of Thrones – I HackerRank Solution. This problem (Game of Thrones – I) is a part of HackerRank Problem Solving series.

## Solution – Game of Thrones – I – HackerRank Solution

### C++

```#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main(int argc, char const *argv[])
{
string str;
cin >> str;
int A[26];
for (int i = 0; i < 26; ++i)
{
A[i] = 0;
}
for (int i = 0; i < str.length(); ++i)
{
A[str[i] - 'a'] ++;
}

int sum = 0;
for (int i = 0; i < 26; ++i)
{
sum = sum + (A[i] % 2);
}

if (sum >= 2)
{
cout << "NO" << endl;
}
else
{
cout << "YES" << endl;
}

return 0;
}
```

### Python

```s = input()
cnt = {}
for char in s:
if char in cnt:
cnt[char] += 1
else:
cnt[char] = 1
odd = False
for key in cnt:
if cnt[key] % 2 == 1:
if odd:
print("NO")
break
odd = True
else:
print("YES")
```

### Java

```import java.io.*;
import java.util.*;

public class Solution {

public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner input = new Scanner(System.in);
String s = input.nextLine();

Map<Character, Integer> letters = new HashMap<>();
for(char c : s.toCharArray())
{
if(letters.containsKey(c))
letters.put(c, letters.get(c) + 1);
else
letters.put(c, 1);
}

int odd = 0;
int even = 0;
for(Integer frequency : letters.values())
{
if(frequency % 2 == 1)
{
odd++;
continue;
}

if(frequency % 2 == 0)
even++;
}

if(odd > 1)
System.out.println("NO");
else
System.out.println("YES");

}
}
```

Note: This problem (Game of Thrones – I) is generated by HackerRank but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.