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  4. Game of Thrones - I
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Game of Thrones - I

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1074 Discussions

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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";
}

  • devin1423
     + 0 comments

    One liner.

    def gameOfThrones(s):

    return "NO" if sum([1 for i in [s.count(i) for i in set(s)] if i % 2 != 0]) > 1 else "YES"

  • nsav2820
     + 0 comments

    Python 3

    from collections import Counter

def gameOfThrones(s):     
    # Write your code here
    counts_s = Counter(s)

    vals = counts_s.values()

    uneven_count = 0
    for ii in vals:
        if ii % 2 != 0:
            uneven_count +=1 

    if uneven_count > 1:
        return "NO"
    else:
        return "YES"

  • vbthani
     + 0 comments

    for Python3 Platform

    import collections

def gameOfThrones(s):
    d = collections.Counter(s)
    multipleOddVal = False
    
    for val in d.values():
        if (val % 2 == 1):
            if (multipleOddVal):
                return "NO"
                break
            else:
                multipleOddVal = True
    else:
        return "YES"

s = input()

result = gameOfThrones(s)

print(result)

  • danielcristianv
     + 0 comments

    string gameOfThrones(string s) { const int range = 'z' - 'a' + 1; size_t pairTable[range] = {0}; for(auto & chr : s) { pairTable[chr - 'a'] ^= 1; } size_t sum = 0; for(size_t idx = 0; idx < range; idx++) { sum += pairTable[idx]; } if((s.length() % 2) == sum) { return "YES"; } return "NO"; }