"""
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.
"""


class Solution:
    def unhappyFriends(
        self, n: int, preferences: List[List[int]], pairs: List[List[int]]
    ) -> int:
        def likeMore(fren, pair):
            prefs = preferences[fren]
            prefs = prefs[: prefs.index(pair)]
            return prefs

        paird = {}
        prefd = {}
        unhappy = set()
        for a, b in pairs:
            paird[a] = b
            paird[b] = a
        for i, pref in enumerate(preferences):
            prefd[i] = pref
        for fren, pair in paird.items():
            if fren in unhappy:
                continue
            prefs = likeMore(fren, pair)
            if len(prefs) > 0:
                for x in prefs:
                    y = paird[x]
                    xprefs = likeMore(x, y)
                    if fren in xprefs:
                        unhappy.add(fren)
                        unhappy.add(x)
                        break
        return len(unhappy)
