Description
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
prefersu
overy
, andu
prefersx
overv
.
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 containi
.- 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.
Solution
Python3
class Solution:
def unhappyFriends(self, n: int, preferences: List[List[int]], pairs: List[List[int]]) -> int:
M = len(pairs)
res = 0
mp = [0] * n
for a, b in pairs:
mp[a] = b
mp[b] = a
rankings = [{ranking : i for i, ranking in enumerate(rankings)} for rankings in preferences]
for i in range(n):
for j in preferences[i]:
if rankings[i][j] < rankings[i][mp[i]] and rankings[j][i] < rankings[j][mp[j]]:
res += 1
break
return res