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2092. Find All People With Secret

Description

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

Solutions

Solution 1

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class Solution:
    def findAllPeople(
        self, n: int, meetings: List[List[int]], firstPerson: int
    ) -> List[int]:
        vis = [False] * n
        vis[0] = vis[firstPerson] = True
        meetings.sort(key=lambda x: x[2])
        i, m = 0, len(meetings)
        while i < m:
            j = i
            while j + 1 < m and meetings[j + 1][2] == meetings[i][2]:
                j += 1
            s = set()
            g = defaultdict(list)
            for x, y, _ in meetings[i : j + 1]:
                g[x].append(y)
                g[y].append(x)
                s.update([x, y])
            q = deque([u for u in s if vis[u]])
            while q:
                u = q.popleft()
                for v in g[u]:
                    if not vis[v]:
                        vis[v] = True
                        q.append(v)
            i = j + 1
        return [i for i, v in enumerate(vis) if v]

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class Solution {
    public List<Integer> findAllPeople(int n, int[][] meetings, int firstPerson) {
        boolean[] vis = new boolean[n];
        vis[0] = true;
        vis[firstPerson] = true;
        int m = meetings.length;
        Arrays.sort(meetings, Comparator.comparingInt(a -> a[2]));
        for (int i = 0; i < m;) {
            int j = i;
            for (; j + 1 < m && meetings[j + 1][2] == meetings[i][2]; ++j)
                ;
            Map<Integer, List<Integer>> g = new HashMap<>();
            Set<Integer> s = new HashSet<>();
            for (int k = i; k <= j; ++k) {
                int x = meetings[k][0], y = meetings[k][1];
                g.computeIfAbsent(x, key -> new ArrayList<>()).add(y);
                g.computeIfAbsent(y, key -> new ArrayList<>()).add(x);
                s.add(x);
                s.add(y);
            }
            Deque<Integer> q = new ArrayDeque<>();
            for (int u : s) {
                if (vis[u]) {
                    q.offer(u);
                }
            }
            while (!q.isEmpty()) {
                int u = q.poll();
                for (int v : g.getOrDefault(u, Collections.emptyList())) {
                    if (!vis[v]) {
                        vis[v] = true;
                        q.offer(v);
                    }
                }
            }
            i = j + 1;
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; ++i) {
            if (vis[i]) {
                ans.add(i);
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> findAllPeople(int n, vector<vector<int>>& meetings, int firstPerson) {
        vector<bool> vis(n);
        vis[0] = vis[firstPerson] = true;
        sort(meetings.begin(), meetings.end(), [&](const auto& x, const auto& y) {
            return x[2] < y[2];
        });
        for (int i = 0, m = meetings.size(); i < m;) {
            int j = i;
            for (; j + 1 < m && meetings[j + 1][2] == meetings[i][2]; ++j)
                ;
            unordered_map<int, vector<int>> g;
            unordered_set<int> s;
            for (int k = i; k <= j; ++k) {
                int x = meetings[k][0], y = meetings[k][1];
                g[x].push_back(y);
                g[y].push_back(x);
                s.insert(x);
                s.insert(y);
            }
            queue<int> q;
            for (int u : s)
                if (vis[u])
                    q.push(u);
            while (!q.empty()) {
                int u = q.front();
                q.pop();
                for (int v : g[u]) {
                    if (!vis[v]) {
                        vis[v] = true;
                        q.push(v);
                    }
                }
            }
            i = j + 1;
        }
        vector<int> ans;
        for (int i = 0; i < n; ++i)
            if (vis[i])
                ans.push_back(i);
        return ans;
    }
};

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func findAllPeople(n int, meetings [][]int, firstPerson int) []int {
    vis := make([]bool, n)
    vis[0], vis[firstPerson] = true, true
    sort.Slice(meetings, func(i, j int) bool {
        return meetings[i][2] < meetings[j][2]
    })
    for i, j, m := 0, 0, len(meetings); i < m; i = j + 1 {
        j = i
        for j+1 < m && meetings[j+1][2] == meetings[i][2] {
            j++
        }
        g := map[int][]int{}
        s := map[int]bool{}
        for _, e := range meetings[i : j+1] {
            x, y := e[0], e[1]
            g[x] = append(g[x], y)
            g[y] = append(g[y], x)
            s[x], s[y] = true, true
        }
        q := []int{}
        for u := range s {
            if vis[u] {
                q = append(q, u)
            }
        }
        for len(q) > 0 {
            u := q[0]
            q = q[1:]
            for _, v := range g[u] {
                if !vis[v] {
                    vis[v] = true
                    q = append(q, v)
                }
            }
        }
    }
    var ans []int
    for i, v := range vis {
        if v {
            ans = append(ans, i)
        }
    }
    return ans
}

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function findAllPeople(n: number, meetings: number[][], firstPerson: number): number[] {
    let parent: Array<number> = Array.from({ length: n + 1 }, (v, i) => i);
    parent[firstPerson] = 0;

    function findParent(index: number): number {
        if (parent[index] != index) parent[index] = findParent(parent[index]);
        return parent[index];
    }

    let map = new Map<number, Array<Array<number>>>();
    for (let meeting of meetings) {
        const time = meeting[2];
        let members: Array<Array<number>> = map.get(time) || new Array();
        members.push(meeting);
        map.set(time, members);
    }
    const times = [...map.keys()].sort((a, b) => a - b);
    for (let time of times) {
        // round 1
        for (let meeting of map.get(time)) {
            let [a, b] = meeting;
            if (!parent[findParent(a)] || !parent[findParent(b)]) {
                parent[findParent(a)] = 0;
                parent[findParent(b)] = 0;
            }
            parent[findParent(a)] = parent[findParent(b)];
        }
        // round 2
        for (let meeting of map.get(time)) {
            let [a, b] = meeting;
            if (!parent[findParent(a)] || !parent[findParent(b)]) {
                parent[findParent(a)] = 0;
                parent[findParent(b)] = 0;
            } else {
                parent[a] = a;
                parent[b] = b;
            }
        }
    }

    let ans = new Array<number>();
    for (let i = 0; i <= n; i++) {
        if (!parent[findParent(i)]) {
            ans.push(i);
        }
    }
    return ans;
}

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