README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/1700-1799/1733.Minimum%20Number%20of%20People%20to%20Teach/README_EN.md	1983	Biweekly Contest 44 Q2	Greedy Array Hash Table

1733. Minimum Number of People to Teach

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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 i​​​​​​th​​​​ user knows, and
• friendships[i] = [u​​​​​​i​​​, v​​​​​​i] denotes a friendship between the users u​​​​​​​​​​​i​​​​​ 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 <= u​​​​​​i < v​​​​​​i <= languages.length
• 1 <= friendships.length <= 500
• All tuples (u​​​​​i, v​​​​​​i) 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.

Python3

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)

Java

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;
    }
}

C++

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;
    }
};

Go

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)
}