The Earliest Moment When Everyone Become Friends [LC#1101]

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 timestamp i.

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.

Intuition

Since the friendship is symetric, wheever two people become friends, we can merge their acquaintance sets togethere. When there is only 1 such sets, everyone is aquainted. For finding earliest time, we simply go through the logs in chronological order.

Code

def earliestAcq(self, logs: List[List[int]], n: int) -> int:
    dsu = DSU()
    for i in range(n):
        dsu.find(i)
    for timestamp, x, y in sorted(logs):
        if dsu.union(x, y):
            if dsu.components == 1:
                return timestamp
    return -1

Time Complexity

• $T(n) = O(n)$ since union is $O(\alpha(n))$. $S(n) = O(n)$