"""

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.

"""


from typing import List

class Solution:
    def unhappyFriends(self, n: int, preferences: List[List[int]], pairs: List[List[int]]) -> int:
        prefer = {}
        pair = {}
        for i, p in enumerate(preferences):
            prefer[i] = p
        for a, b in pairs:
            pair[a] = b
            pair[b] = a
        
        counter = 0
        for x, y in pairs:
            x_before_y = []
            for e in prefer[x]:
                if e != y:
                    x_before_y.append(e)
                else:
                    break
            for u in x_before_y:
                v = pair[u]
                unhappy = False
                for e in prefer[u]:
                    if e == v:
                        break
                    if e == x:
                        unhappy = True
                        break
                if unhappy:
                    counter += 1
                    break
            
            y_before_x = []
            for e in prefer[y]:
                if e != x:
                    y_before_x.append(e)
                else:
                    break
            for u in y_before_x:
                v = pair[u]
                unhappy = False
                for e in prefer[u]:
                    if e == v:
                        break
                    if e == y:
                        unhappy = True
                        break
                if unhappy:
                    counter += 1
                    break
        return counter

def test():
    n = 4
    preferences = [[1,2,3],[3,2,0],[3,1,0],[1,2,0]]
    pairs = [[0,1],[2,3]]

    s = Solution()
    print(s.unhappyFriends(n, preferences, pairs))

test()
