LeetCode in Kotlin
2092. Find All People With Secret
Hard
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
Solution
@Suppress("NAME_SHADOWING")
class Solution {
fun findAllPeople(n: Int, meetings: Array<IntArray>, firstPerson: Int): List<Int> {
meetings.sortWith { a: IntArray, b: IntArray -> a[2] - b[2] }
val uf = UF(n)
// base
uf.union(0, firstPerson)
// for every time we have a pool of people that talk to each other
// if someone knows a secret proir to this meeting - all pool will too
// if not - reset unions from this pool
var i = 0
while (i < meetings.size) {
val curTime = meetings[i][2]
val pool: MutableSet<Int> = HashSet()
while (i < meetings.size && curTime == meetings[i][2]) {
val currentMeeting = meetings[i]
uf.union(currentMeeting[0], currentMeeting[1])
pool.add(currentMeeting[0])
pool.add(currentMeeting[1])
i++
}
// meeting that took place now should't affect future
// meetings if people don't know the secret
for (j in pool) {
if (!uf.connected(0, j)) {
uf.reset(j)
}
}
}
// if the person is conneted to 0 - they know a secret
val ans: MutableList<Int> = ArrayList()
for (j in 0 until n) {
if (uf.connected(j, 0)) {
ans.add(j)
}
}
return ans
}
// regular union find
private class UF(n: Int) {
private val parent: IntArray
private val rank: IntArray
init {
parent = IntArray(n)
rank = IntArray(n)
for (i in 0 until n) {
parent[i] = i
}
}
fun union(p: Int, q: Int) {
val rootP = find(p)
val rootQ = find(q)
if (rootP == rootQ) {
return
}
if (rank[rootP] < rank[rootQ]) {
parent[rootP] = rootQ
} else {
parent[rootQ] = rootP
rank[rootP]++
}
}
fun find(p: Int): Int {
var p = p
while (parent[p] != p) {
p = parent[parent[p]]
}
return p
}
fun connected(p: Int, q: Int): Boolean {
return find(p) == find(q)
}
fun reset(p: Int) {
parent[p] = p
rank[p] = 0
}
}
}