1733. Minimum Number of People to Teach
Description
On a social network consisting of m
users and some friendships between users, two users can communicate with each other if they know a common language.
You are given an integer n
, an array languages
, and an array friendships
where:
- There are
n
languages numbered1
throughn
, languages[i]
is the set of languages theith
user knows, andfriendships[i] = [ui, vi]
denotes a friendship between the usersui
andvi
.
You can choose one language and teach it to some users so that all friends can communicate with each other. Return the minimum number of users you need to teach.
Note that friendships are not transitive, meaning if x
is a friend of y
and y
is a friend of z
, this doesn't guarantee that x
is a friend of z
.
Example 1:
Input: n = 2, languages = [[1],[2],[1,2]], friendships = [[1,2],[1,3],[2,3]] Output: 1 Explanation: You can either teach user 1 the second language or user 2 the first language.
Example 2:
Input: n = 3, languages = [[2],[1,3],[1,2],[3]], friendships = [[1,4],[1,2],[3,4],[2,3]] Output: 2 Explanation: Teach the third language to users 1 and 3, yielding two users to teach.
Constraints:
2 <= n <= 500
languages.length == m
1 <= m <= 500
1 <= languages[i].length <= n
1 <= languages[i][j] <= n
1 <= ui < vi <= languages.length
1 <= friendships.length <= 500
- All tuples
(ui, vi)
are unique languages[i]
contains only unique values
Solutions
Solution 1: Simulation + Statistics
For each friendship, if the sets of languages known by the two people do not intersect, then a language needs to be taught so that the two people can communicate with each other. We put these people into a hash set $s$.
Then in this set $s$, we count the number of people who know each language, and get the maximum number, which we denote as $mx$. So the answer is len(s) - mx
.
The time complexity is $O(m^2 \times k)$. Here, $m$ is the number of languages, and $k$ is the number of friendships.
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