HackerRank - ACM ICPC Team, Taum and B'day, Organizing Containers of Balls, Encryption

[풀이 - ACM ICPC Team]

int CalcKnown(string s1, string s2, int count)
{
    int result = 0;
    //
    for(int i = 0; i < count; ++i)
    {
        if(s1[i] == '1' || s2[i] == '1')
            ++result;
    }
    //
    return result;
}

vector<int> acmTeam(vector<string> topic)
{
    vector<int> result;
    map<int, int> m;
    int attendeeCount = topic.size();
    int topicCount = topic[0].size();
    // make pair & calc Value
    for(int i = 0; i < attendeeCount; ++i)
    {
        for(int j = i+1; j < attendeeCount; ++j)
        {
            int value = CalcKnown(topic[i], topic[j], topicCount);
            m[value]++;
        }
    }
    //
    auto iter = m.rbegin();
    result.push_back(iter->first); // maxCount
    result.push_back(iter->second); // teamCountOfMax
    return result;
}

 

[풀이 - Taum and B'day]

long taumBday(long b, long w, long bc, long wc, long z)
{
    long result = 0;
    //
    long totalCount = b + w;
    long d = abs(bc - wc);
    // what is cheap?
    long lowPrice = 0, lowCount = 0;
    bc > wc ?
    (lowPrice = wc, lowCount = w) : 
    (lowPrice = bc, lowCount = b); 
    // Change is effective?
    if(d <= z) // Yes
        result = b * bc + w * wc;
    else // No ( buy at low price and pay 'change pee')
        result = lowPrice * totalCount + (totalCount - lowCount) * z;
    //
    return result;
}

 

[풀이 - Organizing Containers of Balls]

string organizingContainers(vector<vector<int>> container)
{
    string result = "";
    //
    int n = container.size();
    vector<int> containerSize(n, 0);
    vector<int> ballCount(n, 0);
    //
    for(int j = 0; j < n; ++j)
        for(int i = 0; i < n; ++i)
        {
            containerSize[j] += container[j][i];
            ballCount[i] += container[j][i];
        }
    //
    sort(containerSize.begin(), containerSize.end());
    sort(ballCount.begin(), ballCount.end());
    //
    for(int i = 0; i < n; ++i)
        if(containerSize[i] != ballCount[i])
            return "Impossible";
    //
    return "Possible";
}

1. [각 컨테이너가 가지고 있는 원소의 합]과 [Type별 Ball 갯수의 총합]을 각각 배열에 저장

2. 두 배열을 정렬해서 전부 일치하면 "Possible"

# Each container contains only balls of the same type.

위 문장을 컨테이너 번호와 Ball Type 번호가 일치해야 한다고 해석해서 문제가 안풀렸었다..

컨테이너 번호와 Ball Type 번호가 일치하지 않아도 가능..

 

[풀이 - Encryption ]

string encryption(string s)
{
    string result = "";
    string temp = s;
    // Remove Space(' ')
    remove(temp.begin(), temp.end(), ' ');
    //
    int L = temp.size();
    int f = floor(sqrt(L));
    int c = ceil(sqrt(L));
    if(f * c < L)
        f++;
    // cut string by column
    vector<string> v(f, "");
    int i = -1, j = 0;
    while(j < L)
    {
        if(j % c == 0)
            ++i;
        //
        if(j < L)
            v[i] += temp[j];
        //
        ++j;
    }
    // append string to result
    for(int i = 0 ; i < c; ++i)
    {
        for(int j = 0 ; j < f; ++j)
        {
            if(i < v[j].size())
                result += v[j][i];
        }
        result += " ";
    }
    //
    return result;
}