# 1101. The Earliest Moment When Everyone Become Friends

The Earliest Moment When Everyone Become Friends - LeetCode 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.

• code
``````class Solution:
def earliestAcq(self, logs: List[List[int]], n: int) -> int:
root = [i for i in range(n)]
rank = [0] * n
self.valid_union_count = 0

def find(x):
if x == root[x]:
return x
root[x] = find(root[x])
return root[x]

def union(x, y):
rtx = find(x)
rty = find(y)
if rtx == rty:
return
if rank[x] < rank[y]:
root[rtx] = rty
elif rank[y] < rank[x]:
root[rty] = rtx
else:
root[rty] = rtx
rank[rtx] += 1
self.valid_union_count += 1

logs.sort(key = lambda x: x[0])
for time, i, j in logs:
union(i, j)
if self.valid_union_count == n - 1:
return time

return -1
``````