1583 Count Unhappy Friends

🔀 Pick One

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.
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class Solution:
    def unhappyFriends(self, n: int, preferences: List[List[int]], pairs: List[List[int]]) -> int:
        d = {}
        for p in pairs:
            d[p[0]] = p[1]
            d[p[1]] = p[0]

        prefer = {}
        for i in range(n):
            prefer[i] = {}
            for j in range(n-1): 
                prefer[i][preferences[i][j]] = j
        
        res = 0
        for x in range(n):
            for u in preferences[x]:
                '''
                x is unhappy if 
                  x is paired with y and there is a
                  u paired with v 
                x prefers u over y, and
                u prefers x over v.
                '''
                y = d[x]
                v = d[u]
                if (
                    prefer[x].get(u) < prefer[x].get(y)
                    and
                    prefer[u].get(x) < prefer[u].get(v)
                ):
                    res += 1
                    break
        return res