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:

• 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.

思路：这题就是一个建模的过程，没什么算法；不是什么好的面试题；

class Solution {
    public int unhappyFriends(int n, int[][] preferences, int[][] pairs) {
        HashMap> preHashMap = new HashMap<>();
        for(int i = 0; i < preferences.length; i++) {
            preHashMap.putIfAbsent(i, new ArrayList());
            for(int j = 0; j < preferences[i].length; j++) {
                preHashMap.get(i).add(preferences[i][j]);
            }
        }
        
        HashMap pairMap = new HashMap<>();
        for(int i = 0; i < pairs.length; i++) {
            int a = pairs[i][0];
            int b = pairs[i][1];
            pairMap.put(a, b);
            pairMap.put(b, a);
        }
        
        int count = 0;
        for(int i = 0; i < n; i++) {
            int pair = pairMap.get(i);
            List prelist = getPrelist(preHashMap.get(i), pair);
            if(unhappy(prelist, i, pair, pairMap, preHashMap)) {
                count++;
            }
        }
        return count;
    }
    
    private List getPrelist(List list, int pair) {
        List res = new ArrayList();
        for(int i = 0; i < list.size(); i++) {
            if(list.get(i) != pair) {
                res.add(list.get(i));
            } else {
                break;
            }
        }
        return res;
    }
    /*
    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.
    */
    private boolean unhappy(List prelist, int x, int y, HashMap pairMap, 
                            HashMap> preHashMap) {
        for(Integer u: prelist) {
            Integer v = pairMap.get(u);
            if(prefer(x, u, y, preHashMap) && prefer(u, x, v, preHashMap)) {
                return true;
            }
        }
        return false;
    }
    
    //  x prefers u over y,
    private boolean prefer(int x, int u, int y, HashMap> preHashMap) {
        List xprelist = preHashMap.get(x);
        int uindex = -1;
        int yindex = -1;
        for(int i = 0; i < xprelist.size(); i++) {
            if(xprelist.get(i) == u) {
                uindex = i;
            }
            if(xprelist.get(i) == y) {
                yindex = i;
            }
        }
        return uindex != -1 && yindex != -1 && uindex < yindex;
    }
}