Count Unhappy Friends

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:

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:


Solution:

class Solution {
    public int unhappyFriends(int n, int[][] preferences, int[][] pairs) {
        Map<Integer, int[]> map = new HashMap();
        for (int[] pair : pairs) {
            for (int i : pair) {
                map.put(i, pair);
            }
        }
        int count = 0;
        for (int i = 0; i < n; i ++) {
            if (unhappy(i, preferences, map)) {
                count ++;
            }
        }
        return count;
    }
    
    private boolean unhappy(int i, int[][] prefs, Map<Integer, int[]> map) {
        int[] iPref = prefs[i];
        int[] iPair = map.get(i);
        int j = -1;
        for (int val : iPair) {
            if (val != i) {
                j = val;
            }
        }
        if (iPref[0] == j) return false;
        for (int k : iPref) {
            if (k == j) break;
            int[] kPair = map.get(k);
            int q = -1;
            for (int val : kPair) {
                if (val != k) {
                    q = val;
                }
            }
            // System.out.println(i +", " + j + ", " + k + "," + q);
            for (int val : prefs[k]) {
                if (val == i) {
                    return true;
                }
                if (val == q) {
                    break;
                }
            }
        }
        
        return false;
    }
}