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:
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 |
Example 2:
1 | Input: logs = [[0,2,0],[1,0,1],[3,0,3],[4,1,2],[7,3,1]], n = 4 |
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.
1 | class Solution { |
After optimization:
1 |
|
If rooty is root, its rank will be increased.
Then it will be more likely to be parent. The graph looks like a star and the root is in the center.