Game of Thrones - I Discussions | Algorithms

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1084 Discussions

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23 hours ago+ 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";
}

2 weeks ago+ 0 comments

Approach to solve:

1 Count Character Frequencies Use a hash map (unordered_map) or array (int freq[26]) to store how many times each character appears in the given string.

2 Count Odd Frequency Characters Traverse through the frequency table and count how many characters have an odd frequency.

3 Check Conditions for a Palindrome If odd frequency count <= 1, the string can be rearranged into a palindrome --> Return "YES". Otherwise, return "NO".

2 weeks ago+ 0 comments

Here is my Python solution!

def gameOfThrones(s):
    s = list(s)
    if len(s) % 2 == 0:
        for string in set(s):
            if s.count(string) % 2 == 1:
                return "NO"
        return "YES"
    chances = 1
    for string in set(s):
        if s.count(string) % 2 == 1:
            chances -= 1
    return "YES" if chances >= 0 else "NO"

1 month ago+ 0 comments

string gameOfThrones(string s) {
 map<char,int> mp;
 for(char i:s){
    mp[i]++;
 }
 int count =0;
 for(int i=0;i<mp.size();i++){
    if(mp[i]%2!=0)count++;
 }
 if(count>1)return "NO";
 else return "YES";
}

1 month ago+ 0 comments

string gameOfThrones(string s) { map mp; for(char i:s){ mp[i]++; } int count =0; for(int i=0;i1)return "NO"; else return "YES"; }

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