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.
Example 4:
Input: n = 6, meetings = [[0,2,1],[1,3,1],[4,5,1]], firstPerson = 1 Output: [0,1,2,3] Explanation: At time 0, person 0 shares the secret with person 1. At time 1, person 0 shares the secret with person 2, and person 1 shares the secret with person 3. Thus, people 0, 1, 2, and 3 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
Similar Questions:
Sort meetings in ascending order of meeting time.
Visit the meetings happening at the same time together.
We can connect the two persons in the same meeting using a UnionFind.
Tricky point here: After traversing this batch of meetings, we reset the persons who don't know the secret by checking if he/she is connected to person 0. With UnionFind, reseting means setting id[x] = x.
In the end, we add all the persons who are connected to person 0 into the answer array.
// OJ: https://leetcode.com/problems/find-all-people-with-secret/
// Author: github.com/lzl124631x
// Time: O(MlogM + (M + N) * logN) where `M` is the length of `meetings`
// Can be reduced to `O(MlogM + (M + N) * alpha(N))`
// Space: O(M + N). Can be reduced to O(N) if we make `ppl` an `unordered_set`.
class UnionFind {
vector<int> id;
public:
UnionFind(int n) : id(n) {
iota(begin(id), end(id), 0);
}
void connect(int a, int b) {
id[find(b)] = find(a);
}
int find(int a) {
return id[a] == a ? a : (id[a] = find(id[a]));
}
bool connected(int a, int b) {
return find(a) == find(b);
}
void reset(int a) { id[a] = a; }
};
class Solution {
public:
vector<int> findAllPeople(int n, vector<vector<int>>& A, int firstPerson) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[2] < b[2]; }); // Sort the meetings in ascending order of meeting time
UnionFind uf(n);
uf.connect(0, firstPerson); // Connect person 0 with the first person
vector<int> ppl;
for (int i = 0, M = A.size(); i < M; ) {
ppl.clear();
int time = A[i][2];
for (; i < M && A[i][2] == time; ++i) { // For all the meetings happening at the same time
uf.connect(A[i][0], A[i][1]); // Connect the two persons
ppl.push_back(A[i][0]); // Add both persons into the pool
ppl.push_back(A[i][1]);
}
for (int n : ppl) { // For each person in the pool, check if he/she's connected with person 0.
if (!uf.connected(0, n)) uf.reset(n); // If not, this person doesn't have secret, reset it.
}
}
vector<int> ans;
for (int i = 0; i < n; ++i) {
if (uf.connected(0, i)) ans.push_back(i); // Push all the persons who are connected with person 0 into answer array
}
return ans;
}
};Complexity Analysis:
Sorting takes O(MlogM) time.
Each meeting is visited and pushed into/popped out of ppl only once. For each visit, the connect/connected takes amortized O(logN) time. So, traversing all the meetings takes O(MlogN) time. Note that if we do union-by-rank, the time complexity can be reduced to O(M * alpha(N)) where alpha(N) is the inverse function of Ackermann function, and is even more efficient than logN. But in LeetCode, the difference is negligible, so I usually just use path compression for simplicity. It's definitly worth mentioning this knowledge during your interview though.
Lastly, traversing all the persons and checking connection with person 0 takes amortized O(NlogN) time.
So, overall the time complexity is amortized O(MlogM + (M + N) * logN), which can be reduced to O(MlogM + (M + N) * alpha(N)) with union-by-rank.
As for space complexity, the Union Find takes O(N) space. The ppl array takes O(M) space in the worst case, but it can be reduced to O(N) if we use unordered_set. I use vector<int> because unordered_set<int> has extra overhead which at times consumes more time/space than vector<int> in LeetCode.
So, overall the (extra) space complexity is O(M + N) which can be reduced to O(N).
https://leetcode.com/problems/find-all-people-with-secret/discuss/1599815/C%2B%2B-Union-Find