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  1. Prepare
  2. Algorithms
  3. Implementation
  4. ACM ICPC Team
  5. Discussions

ACM ICPC Team

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1239 Discussions

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/fCUW_MhWEW8

    vector<int> acmTeam(vector<string> topic) {
    int h = 0, cp = 0;
    for(int i = 0; i < topic.size() - 1; i++){
        bitset<500> a = bitset<500>(topic[i]);
        for(int j = i + 1; j < topic.size(); j++){
         bitset<500> b = bitset<500>(topic[j]);
         bitset<500> c = a | b;
         int x = c.count();
         if( x > h){h = x; cp = 1;}
         else if( x == h) cp++;
        }
    }
    return {h, cp};
}

  • farukcan
     + 0 comments
    public static List<int> acmTeam(List<string> topic)
{
        int m = topic.Count; // member count
        int t = topic[0].Length; // topic count
        List<int> teams = new(); // team known topic counts
        for(int i=0; i < m-1; i++)
        {
                for(int j=i+1; j < m; j++)
                {
                    int known = 0;
                    for(int k=0; k < t; k++)
                    {
                        if(topic[i][k]=='1' || topic[j][k]=='1')
                        {
                                known++;
                                continue;
                        }
                    }
                    teams.Add(known);   
                }
        }
        int max = teams.Max();
        int count = teams.Count(t=> t==max);
        return new List<int>{ max, count };
}

  • pub_ashishk28
     + 0 comments

    python3:

    from itertools import combinations

def acmTeam(topic):
    # Write your code here
    
    r = [ {i+1 for i,j in enumerate(s) if j == "1"} for s in topic]
         
    x = [ list(u[0].union(u[1])) for u in combinations(r, r=2) ]
    x = list(map(lambda i: len(i), x))
    
    return [max(x), x.count(max(x))]

  • patrickgao2020
     + 0 comments

    Here is my Python solution using itertools!

    def acmTeam(topic):
    teams = list(itertools.combinations(topic, 2))
    subjects = []
    for team in teams:
        known = 0
        for topic in range(len(team[0])):
            if team[0][topic] == "1" or team[1][topic] == "1":
                known += 1
        subjects.append(known)
    return [max(subjects), subjects.count(max(subjects))]

  • ali313xA
     + 0 comments
    // i have chnages the given array from topics -> persons. 
// now understand, it like if person[i][k] == 1, that means ith person knows the topic k. 

function acmTeam(persons) {
    let max = 0; // max topic that the team know 
    let count = 0; // the team which knows the max topics 
    
    for(let i = 0; i < persons.length; i++) { 
        for(let j = i + 1; j < persons.length; j++) {
            let know = 0; 
            for(let k = 0; k < persons[0].length; k++) {
                if(persons[i][k] === '1' || persons[j][k] === '1'){
                    know += 1; 
                }
            }
            
            if(know > max) {
                max = know; 
                count = 1; // reset the count for the nex max
            }
            else if(know === max){
                count += 1; // otherwise increase the count, we found another max 
            }
        }
    }
    return [max, count]; 
		// should i tell you a secret, >> you are g___r___e__a__t 😍!!
   //  Happy Coding 
}