1101. The Earliest Moment When Everyone Become Friends
There are n people in a social group labeled from 0
to n - 1
. You are given an array logs
where logs[i] = [timestampi, xi, yi]
indicates that xi
and yi
will be friends at the time timestampi
.
Friendship is symmetric. That means if a
is friends with b
, then b
is friends with a
. Also, person a
is acquainted with a person b
if a
is friends with b
, or a
is a friend of someone acquainted with b
.
Return the earliest time for which every person became acquainted with every other person. If there is no such earliest time, return -1
.
Example 1:
Input: logs = [[20190101,0,1],[20190104,3,4],[20190107,2,3],[20190211,1,5],[20190224,2,4],[20190301,0,3],[20190312,1,2],[20190322,4,5]], n = 6
Output: 20190301
Explanation:
The first event occurs at timestamp = 20190101 and after 0 and 1 become friends we have the following friendship groups [0,1], [2], [3], [4], [5].
The second event occurs at timestamp = 20190104 and after 3 and 4 become friends we have the following friendship groups [0,1], [2], [3,4], [5].
The third event occurs at timestamp = 20190107 and after 2 and 3 become friends we have the following friendship groups [0,1], [2,3,4], [5].
The fourth event occurs at timestamp = 20190211 and after 1 and 5 become friends we have the following friendship groups [0,1,5], [2,3,4].
The fifth event occurs at timestamp = 20190224 and as 2 and 4 are already friends anything happens.
The sixth event occurs at timestamp = 20190301 and after 0 and 3 become friends we have that all become friends.
Example 2:
Input: logs = [[0,2,0],[1,0,1],[3,0,3],[4,1,2],[7,3,1]], n = 4
Output: 3
Constraints:
2 <= n <= 100
1 <= logs.length <= 104
logs[i].length == 3
0 <= timestampi <= 109
0 <= xi, yi <= n - 1
xi != yi
- All the values
timestampi
are unique. - All the pairs
(xi, yi)
occur at most one time in the input.
class Solution:
def earliestAcq(self, logs: List[List[int]], n: int) -> int:
# you keep doing the union and find the point where it becomes 1 root. at that point, record the time
logs.sort(key = lambda x: x[0])
self.root = [i for i in range(n)]
self.size = n
for timestamp, vertex1, vertex2 in logs:
self.union(vertex1, vertex2)
if self.size == 1:
return timestamp
return -1
def findRoot(self, target: int) -> int:
while target != self.root[target]:
target = self.root[target]
return target
def union(self, vertex1: int, vertex2: int) -> None:
root1 = self.findRoot(vertex1)
root2 = self.findRoot(vertex2)
if root1 != root2:
self.size -= 1
self.root[root2] = root1
return True
else:
return False