2024-02-24 Daily Challenge

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In: DailyChallenge

Today I have done leetcode's February LeetCoding Challenge with cpp.

February LeetCoding Challenge 24

Description

Find All People With Secret

You are given an integer n indicating there are n people numbered from 0 to n - 1. You are also given a 0-indexed 2D integer array meetings where meetings[i] = [xi, yi, timei] indicates that person xi and person yi have a meeting at timei. A person may attend multiple meetings at the same time. Finally, you are given an integer firstPerson.

Person 0 has a secret and initially shares the secret with a person firstPerson at time 0. This secret is then shared every time a meeting takes place with a person that has the secret. More formally, for every meeting, if a person xi has the secret at timei, then they will share the secret with person yi, and vice versa.

The secrets are shared instantaneously. That is, a person may receive the secret and share it with people in other meetings within the same time frame.

Return a list of all the people that have the secret after all the meetings have taken place. You may return the answer in any order.

 

Example 1:

Input: n = 6, meetings = [[1,2,5],[2,3,8],[1,5,10]], firstPerson = 1
Output: [0,1,2,3,5]
Explanation:
At time 0, person 0 shares the secret with person 1.
At time 5, person 1 shares the secret with person 2.
At time 8, person 2 shares the secret with person 3.
At time 10, person 1 shares the secret with person 5.​​​​
Thus, people 0, 1, 2, 3, and 5 know the secret after all the meetings.

Example 2:

Input: n = 4, meetings = [[3,1,3],[1,2,2],[0,3,3]], firstPerson = 3
Output: [0,1,3]
Explanation:
At time 0, person 0 shares the secret with person 3.
At time 2, neither person 1 nor person 2 know the secret.
At time 3, person 3 shares the secret with person 0 and person 1.
Thus, people 0, 1, and 3 know the secret after all the meetings.

Example 3:

Input: n = 5, meetings = [[3,4,2],[1,2,1],[2,3,1]], firstPerson = 1
Output: [0,1,2,3,4]
Explanation:
At time 0, person 0 shares the secret with person 1.
At time 1, person 1 shares the secret with person 2, and person 2 shares the secret with person 3.
Note that person 2 can share the secret at the same time as receiving it.
At time 2, person 3 shares the secret with person 4.
Thus, people 0, 1, 2, 3, and 4 know the secret after all the meetings.

 

Constraints:

  • 2 <= n <= 105
  • 1 <= meetings.length <= 105
  • meetings[i].length == 3
  • 0 <= xi, yi <= n - 1
  • xi != yi
  • 1 <= timei <= 105
  • 1 <= firstPerson <= n - 1

Solution

BFS way

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
  using pi = pair<int, int>;
public:
  vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
    map<int, vector<pi>> simultaneousMeeting;
    for(const auto &meeting : meetings) {
      simultaneousMeeting[meeting[2]].push_back({meeting[0], meeting[1]});
    }

    set<int> answerSet{0, firstPerson};
    for(const auto &[_time, meetings] : simultaneousMeeting) {
      if(meetings.size() == 1) {
        if(answerSet.count(meetings[0].first) || answerSet.count(meetings[0].second)) {
          answerSet.insert(meetings[0].first);
          answerSet.insert(meetings[0].second);
        }
        continue;
      }
      map<int, vector<int>> graph;
      set<int> peopleKnown;

      for(const auto &[x, y] : meetings) {
        graph[x].push_back(y);
        graph[y].push_back(x);

        if(answerSet.count(x)) peopleKnown.insert(x);
        if(answerSet.count(y)) peopleKnown.insert(y);
      }
      queue<int> q;
      for(auto people : peopleKnown) q.push(people); 
      
      while(q.size()) {
        auto current = q.front();
        q.pop();
        for(auto next : graph[current]) {
          if(answerSet.count(next)) continue;
          answerSet.insert(next);
          q.push(next);
        }
      }
    }
    return vector<int>(answerSet.begin(), answerSet.end());
  }
};

// Accepted
// 55/55 cases passed (602 ms)
// Your runtime beats 46.31 % of cpp submissions
// Your memory usage beats 21.72 % of cpp submissions (243.7 MB)

dijkstra way

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
  using pi = pair<int, int>;
public:
  vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
    sort(meetings.begin(), meetings.end(), [](const vector<int> &a, const vector<int> &b) {
      return a[2] < b[2];
    });
    unordered_map<int, vector<pair<int, int>>> graph;
    for(const auto &meeting : meetings) {
      graph[meeting[0]].push_back({meeting[2], meeting[1]});
      graph[meeting[1]].push_back({meeting[2], meeting[0]});
    }
    vector<int> knownTime(n, -1);
    priority_queue<pi, vector<pi>, greater<pi>> pq;
    pq.push({0, 0});
    pq.push({0, firstPerson});
    while(pq.size()) {
      auto [time, person] = pq.top();
      pq.pop();
      if(knownTime[person] != -1) continue;
      knownTime[person] = time;
      for(const auto &[meetingTime, next] : graph[person]) {
        // cout << person << ' ' << time << ' ' << next << ' ' << meetingTime << endl;
        if(meetingTime < time || knownTime[next] != -1) continue;
        pq.push({meetingTime, next});
      }
    }
    vector<int> answer;
    for(int i = 0; i < n; ++i) {
      if(knownTime[i] != -1) answer.push_back(i);
    }
    return answer;
  }
};

// Accepted
// 55/55 cases passed (482 ms)
// Your runtime beats 65.98 % of cpp submissions
// Your memory usage beats 43.44 % of cpp submissions (186.8 MB)

union set way

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
struct UnionSet {
  vector<int> parent;
public:
  UnionSet(int size): parent(size) {
    iota(parent.begin(), parent.end(), 0);
  }

  int find(int x) {
    if(parent[x] != x) parent[x] = find(parent[x]);
    return parent[x];
  }

  void reset(int x) {
    parent[x] = x;
  }

  void merge(int x, int y) {
    x = find(x);
    y = find(y);
    parent[x] = y;
  }
};
class Solution {
  using pi = pair<int, int>;
public:
  vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
    sort(meetings.begin(), meetings.end(), [](const vector<int> &a, const vector<int> &b) {
      return a[2] < b[2];
    });
    UnionSet us(n);
    us.merge(0, firstPerson);
    for(int i = 0 ; i< meetings.size(); ++i) {
      int time = meetings[i][2];
      us.merge(meetings[i][0], meetings[i][1]);
      if(i != meetings.size() - 1 && time == meetings[i + 1][2]) continue;
      for(int j = i; j >= 0 && meetings[j][2] == time; --j) {
        if(us.find(meetings[j][0]) != us.find(0)) us.reset(meetings[j][0]);
        if(us.find(meetings[j][1]) != us.find(0)) us.reset(meetings[j][1]);
      }
    }
    vector<int> answer;
    for(int i = 0; i < n; ++i) {
      if(us.find(i) == us.find(0)) answer.push_back(i);
    }
    return answer;
  }
};

// Accepted
// 55/55 cases passed (330 ms)
// Your runtime beats 99.18 % of cpp submissions
// Your memory usage beats 97.54 % of cpp submissions (135.4 MB)
Algorithm