The Earliest Moment When Everyone Become Friends — LC Medium

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