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.

class Solution {
    public int unhappyFriends(int n, int[][] preferences, int[][] pairs) {
            int[] map = new int[n];
            for(int[] pair: pairs){ // Keep record of current matches.
                map[pair[0]] = pair[1];
                map[pair[1]] = pair[0];
            }
            int res = 0;

            Map<Integer, Integer>[] prefer = new Map[n]; // O(1) to fetch the index from the preference array. 

            for(int i = 0; i < n; i++){
                for(int j = 0; j < n-1; j++){
                    if(prefer[i] == null) prefer[i] = new HashMap<>();
                    prefer[i].put(preferences[i][j], j);
                }
            }

            for(int i = 0; i < n; i++){
                for(int j : preferences[i]){
                    if(prefer[j].get(i) < prefer[j].get(map[j]) 
                        && prefer[i].get(map[i]) > prefer[i].get(j)){ // Based on the definition of "unhappy"...
                        res++;
                        break;
                    }
                }
            }
            return res;
    }
}

https://leetcode.com/problems/count-unhappy-friends/discuss/843929/What-a-Bad-Question