1101. The Earliest Moment When Everyone Become Friends.py

# There are n people in a social group labeled from 0 to n - 1. You are given an array logs where logs[i] = [timestampi, x_i, y_i] indicates that x_i and y_i 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 <= 10^4
# logs[i].length == 3
# 0 <= timestampi <= 10^9
# 0 <= x_i, y_i <= n - 1
# x_i != y_i
# All the values timestampi are unique.
# All the pairs (x_i, y_i) occur at most one time in the input.


class Solution:
    def earliestAcq(self, logs: List[List[int]], n: int) -> int:
        # First, we need to sort the events in chronological order.
        logs.sort(key=lambda i: i[0])

        uf = UnionFind(n)
        for ts, x, y in logs:
            uf.union(x, y)
            if uf.get_count() == 1:
                return ts

        # More than one groups left, i.e. not everyone is connected.
        return -1


class UnionFind:
    def __init__(self, size):
        self.root = [i for i in range(size)]
        self.rank = [1] * size
        self.count = size

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

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        if rootX != rootY:
            if self.rank[rootX] > self.rank[rootY]:
                self.root[rootY] = rootX
            elif self.rank[rootX] < self.rank[rootY]:
                self.root[rootX] = rootY
            else:
                self.root[rootY] = rootX
                self.rank[rootX] += 1
            self.count -= 1

    def get_count(self):
        return self.count