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 <= 1051 <= meetings.length <= 105meetings[i].length == 30 <= xi, yi <= n - 1xi != yi1 <= timei <= 1051 <= firstPerson <= n - 1
BFS.
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]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;
}
}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;
}
};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
}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;
}