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  1. Prepare
  2. Algorithms
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  4. Game of Thrones - I
  5. Discussions

Game of Thrones - I

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution : explanation here : https://youtu.be/yhSaL48IHds

    solution 1 : 

    string gameOfThrones(string s) {
    map<char, int> mp;
    for(int i = 0; i < s.size(); i++){
        if(mp[s[i]] != 0){
            mp[s[i]]--;
            if(mp[s[i]] == 0) mp.erase(s[i]);
        }
        else mp[s[i]]++;
    }
    if(mp.size() > 1) return "NO";
    return "YES";
}

    solution 2 : 

    string gameOfThrones(string s) {
    map<char, int> mp;
    for(int i = 0; i < s.size(); i++){
        if(mp[s[i]] == 1)  mp.erase(s[i]);
        else mp[s[i]]++;
    }
    if(mp.size() > 1) return "NO";
    return "YES";
}

  • illogicalsleepi1
     + 0 comments
    from collections import Counter
def gameOfThrones(s):
    freq=Counter(s)
    c=0
    for i in freq:
        if freq[i]%2==1:
            c+=1
    return 'NO' if c>1 else 'YES'

  • vuhuutuan0
     + 0 comments

    My answer with Typescript, noted

    function gameOfThrones(key: string): string {
    // 0. define a map, couting [char] in [key]
    let hash = new Map<string, number>()
    for (let c of key) hash.set(c, (hash.get(c) || 0) + 1)
    
    // 1. define some variable
    let key_is_odd: boolean = key.length % 2 != 0
    let flag: boolean = true

    // 2. loop hash and check [char_count] of each [char]...
    for (const char_count of hash.values()) {
        const char_count_is_odd = char_count % 2 != 0
        
        // 2.1 [char_count] that divided by 2, mean it can be split 2 side
        //  -> skip check
        // 2.2 [char_count] that not divided by 2, but [key] is 
        //  -> there is at least one extra char, return NO
        // 2.3 [char_count] and [key] is both odd, it can be have only 1 extra char
        //  -> if have more than 1, return NO
        //  -> set flag = false to know that already have 1
        if (!char_count_is_odd) continue
        if (!key_is_odd) return 'NO'
        if (!flag) return 'NO'

        flag = false
    }

    return 'YES'
}

  • highsoft_soi
     + 0 comments

    Perl:

    sub gameOfThrones {
    my @s = split("", shift);
    my %h;
    my $cnt = 0;

    foreach (@s) {
        $h{$_}++;
    }

    foreach my $v (values %h) {
        if ($v % 2 > 0) {
            $cnt++;
        }        
    }  

    return ($cnt == 1 || $cnt == 0) ? "YES" : "NO";
}

  • dorirgob
     + 0 comments

    The odd even check is actually unnecessary I've just figured. You can skip that part but the logic is pretty much the same. Java: 

    public static String gameOfThrones(String s) {
  int n = s.length();
  if (n == 0) {
    return "NO";
  } else if (n == 1) {
    return "YES";
  }
  int[] alphabet = new int[26];

  for (char c : s.toCharArray()) {
    alphabet[c - 'a']++;
  }
  if (n % 2 == 0) {
    for (int freq : alphabet) {
      if (freq % 2 != 0) {
        return "NO";
      }
    }
    return "YES";
  } else {
    int oddCount = 0;
    for (int freq : alphabet) {
      if (freq % 2 != 0) {
        oddCount++;
      }
    }
    if (oddCount == 1) {
      return "YES";
    } else {
      return "NO";
    }
  }
}
}