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 numbered 1
through n
, languages[i]
is the set of languages the ith
user knows, andfriendships[i] = [ui, vi]
denotes a friendship between the users ui
and vi
.
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#
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| class Solution:
def minimumTeachings(
self, n: int, languages: List[List[int]], friendships: List[List[int]]
) -> int:
def check(u, v):
for x in languages[u - 1]:
for y in languages[v - 1]:
if x == y:
return True
return False
s = set()
for u, v in friendships:
if not check(u, v):
s.add(u)
s.add(v)
cnt = Counter()
for u in s:
for l in languages[u - 1]:
cnt[l] += 1
return len(s) - max(cnt.values(), default=0)
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| class Solution {
public int minimumTeachings(int n, int[][] languages, int[][] friendships) {
Set<Integer> s = new HashSet<>();
for (var e : friendships) {
int u = e[0], v = e[1];
if (!check(u, v, languages)) {
s.add(u);
s.add(v);
}
}
if (s.isEmpty()) {
return 0;
}
int[] cnt = new int[n + 1];
for (int u : s) {
for (int l : languages[u - 1]) {
++cnt[l];
}
}
int mx = 0;
for (int v : cnt) {
mx = Math.max(mx, v);
}
return s.size() - mx;
}
private boolean check(int u, int v, int[][] languages) {
for (int x : languages[u - 1]) {
for (int y : languages[v - 1]) {
if (x == y) {
return true;
}
}
}
return false;
}
}
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| class Solution {
public:
int minimumTeachings(int n, vector<vector<int>>& languages, vector<vector<int>>& friendships) {
unordered_set<int> s;
for (auto& e : friendships) {
int u = e[0], v = e[1];
if (!check(u, v, languages)) {
s.insert(u);
s.insert(v);
}
}
if (s.empty()) {
return 0;
}
vector<int> cnt(n + 1);
for (int u : s) {
for (int& l : languages[u - 1]) {
++cnt[l];
}
}
return s.size() - *max_element(cnt.begin(), cnt.end());
}
bool check(int u, int v, vector<vector<int>>& languages) {
for (int x : languages[u - 1]) {
for (int y : languages[v - 1]) {
if (x == y) {
return true;
}
}
}
return false;
}
};
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| func minimumTeachings(n int, languages [][]int, friendships [][]int) int {
check := func(u, v int) bool {
for _, x := range languages[u-1] {
for _, y := range languages[v-1] {
if x == y {
return true
}
}
}
return false
}
s := map[int]bool{}
for _, e := range friendships {
u, v := e[0], e[1]
if !check(u, v) {
s[u], s[v] = true, true
}
}
if len(s) == 0 {
return 0
}
cnt := make([]int, n+1)
for u := range s {
for _, l := range languages[u-1] {
cnt[l]++
}
}
return len(s) - slices.Max(cnt)
}
|