1583

1583. Count Unhappy Friends (Medium)

You are given a list of preferences for n friends, where n is always even.

For each person i, preferences[i] contains a list of friends sorted in the order of preference. In other words, a friend earlier in the list is more preferred than a friend later in the list. Friends in each list are denoted by integers from 0 to n-1.

All the friends are divided into pairs. The pairings are given in a list pairs, where pairs[i] = [xi, yi] denotes xi is paired with yi and yi is paired with xi.

However, this pairing may cause some of the friends to be unhappy. A friend x is unhappy if x is paired with y and there exists a friend u who is paired with v but:

• x prefers u over y, and
• u prefers x over v.

Return the number of unhappy friends.

 

Example 1:

Input: n = 4, preferences = [[1, 2, 3], [3, 2, 0], [3, 1, 0], [1, 2, 0]], pairs = [[0, 1], [2, 3]]
Output: 2
Explanation:
Friend 1 is unhappy because:
- 1 is paired with 0 but prefers 3 over 0, and
- 3 prefers 1 over 2.
Friend 3 is unhappy because:
- 3 is paired with 2 but prefers 1 over 2, and
- 1 prefers 3 over 0.
Friends 0 and 2 are happy.

Example 2:

Input: n = 2, preferences = [[1], [0]], pairs = [[1, 0]]
Output: 0
Explanation: Both friends 0 and 1 are happy.

Example 3:

Input: n = 4, preferences = [[1, 3, 2], [2, 3, 0], [1, 3, 0], [0, 2, 1]], pairs = [[1, 3], [0, 2]]
Output: 4

 

Constraints:

• 2 <= n <= 500
• n is even.
• preferences.length == n
• preferences[i].length == n - 1
• 0 <= preferences[i][j] <= n - 1
• preferences[i] does not contain i.
• All values in preferences[i] are unique.
• pairs.length == n/2
• pairs[i].length == 2
• xi != yi
• 0 <= xi, yi <= n - 1
• Each person is contained in exactly one pair.

Related Topics:
Array

Solution 1.

Note that given the definition, if x is unhappy due to (x, y) and (u, v), then u is unhappy too.

// OJ: https://leetcode.com/problems/count-unhappy-friends/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N^2)
class Solution {
    unordered_map<int, unordered_map<int, int>> m;
    unordered_set<int> s;
    void unhappy(int x, int y, int u, int v) {
        if (m[x][y] < m[x][u] || m[u][v] < m[u][x]) return;
        s.insert(x);
        s.insert(u);
    }
public:
    int unhappyFriends(int n, vector<vector<int>>& pref, vector<vector<int>>& pairs) {
        for (int i = 0; i < pref.size(); ++i) {
            for (int j = 0; j < pref[i].size(); ++j) m[i][pref[i][j]] = j;
        }
        for (int i = 0; i < pairs.size(); ++i) {
            int x = pairs[i][0], y = pairs[i][1];
            for (int j = i + 1; j < pairs.size(); ++j) {
                int u = pairs[j][0], v = pairs[j][1];
                unhappy(x, y, u, v);
                unhappy(x, y, v, u);
                unhappy(y, x, u, v);
                unhappy(y, x, v, u);
            }
        }
        return s.size();
    }
};

It's more performant if we use vector.

// OJ: https://leetcode.com/problems/count-unhappy-friends/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N^2)
class Solution {
    vector<vector<int>> pref;
    vector<bool> unhappy;
    void check(int x, int y, int u, int v) {
        if (pref[x][y] < pref[x][u] || pref[u][v] < pref[u][x]) return;
        unhappy[x] = true;
        unhappy[u] = true;
    }
public:
    int unhappyFriends(int n, vector<vector<int>>& preferences, vector<vector<int>>& pairs) {
        pref.assign(n, vector<int>(n));
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < preferences[i].size(); ++j) pref[i][preferences[i][j]] = j;
        }
        unhappy.assign(n, false);
        for (int i = 0; i < pairs.size(); ++i) {
            int x = pairs[i][0], y = pairs[i][1];
            for (int j = i + 1; j < pairs.size(); ++j) {
                int u = pairs[j][0], v = pairs[j][1];
                check(x, y, u, v);
                check(x, y, v, u);
                check(y, x, u, v);
                check(y, x, v, u);
            }
        }
        return accumulate(begin(unhappy), end(unhappy), 0);
    }
};