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:
        unhappy = [0] * n
        for i in range(n//2):
            a, b = pairs[i]
            b_a_idx, a_b_idx = preferences[b].index(a), preferences[a].index(b)
            for j in range(i+1, n//2):
                c, d = pairs[j]
                c_a_idx = preferences[c].index(a)
                c_b_idx = preferences[c].index(b)
                c_d_idx = preferences[c].index(d)
                
                d_a_idx = preferences[d].index(a) 
                d_b_idx = preferences[d].index(b) 
                d_c_idx = preferences[d].index(c) 
                
                a_c_idx = preferences[a].index(c) 
                a_d_idx = preferences[a].index(d) 
                
                b_c_idx = preferences[b].index(c) 
                b_d_idx = preferences[b].index(d) 
                
                # a <-> c
                if c_a_idx < c_d_idx and a_c_idx < a_b_idx: unhappy[a] = unhappy[c] = 1
                # a <-> d
                if d_a_idx < d_c_idx and a_d_idx < a_b_idx: unhappy[a] = unhappy[d] = 1
                # b <-> c
                if c_b_idx < c_d_idx and b_c_idx < b_a_idx: unhappy[b] = unhappy[c] = 1
                # b <-> d
                if d_b_idx < d_c_idx and b_d_idx < b_a_idx: unhappy[b] = unhappy[d] = 1
        return sum(unhappy)

class Solution:
    def unhappyFriends(self, n: int, preferences: List[List[int]], pairs: List[List[int]]) -> int:
        unhappy = [0] * n
        for i in range(n//2):
            a, b = pairs[i]
            b_a_idx, a_b_idx = preferences[b].index(a), preferences[a].index(b)
            for j in range(i+1, n//2):
                c, d = pairs[j]
                c_a_idx = preferences[c].index(a)
                c_b_idx = preferences[c].index(b)
                c_d_idx = preferences[c].index(d)
                
                d_a_idx = preferences[d].index(a) 
                d_b_idx = preferences[d].index(b) 
                d_c_idx = preferences[d].index(c) 
                
                a_c_idx = preferences[a].index(c) 
                a_d_idx = preferences[a].index(d) 
                
                b_c_idx = preferences[b].index(c) 
                b_d_idx = preferences[b].index(d) 
                
                # a <-> c
                if c_a_idx < c_d_idx and a_c_idx < a_b_idx: unhappy[a] = unhappy[c] = 1
                # a <-> d
                if d_a_idx < d_c_idx and a_d_idx < a_b_idx: unhappy[a] = unhappy[d] = 1
                # b <-> c
                if c_b_idx < c_d_idx and b_c_idx < b_a_idx: unhappy[b] = unhappy[c] = 1
                # b <-> d
                if d_b_idx < d_c_idx and b_d_idx < b_a_idx: unhappy[b] = unhappy[d] = 1
        return sum(unhappy)
