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1583. Count Unhappy Friends

Difficulty Topics

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 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.

Solution

count-unhappy-friends.py
class Solution:
    def unhappyFriends(self, n: int, pref: List[List[int]], pairs: List[List[int]]) -> int:
        res = set()

        # helper function to check for happiness
        def checkHappiness(x,y,u,v):
             if rank[x][u] < rank[x][y] and rank[u][x] < rank[u][v]:
                res.add(x)
                res.add(u)

        # store the pref according in map with index
        rank = [{v:i for i,v in enumerate(row)} for row in pref] 

        for i in range(n//2):
            x, y = pairs[i]
            for j in range(i+1, n//2):
                    u,v = pairs[j]
                    # loop through permutations of pairs
                    checkHappiness(x, y, u, v)
                    checkHappiness(y, x, u, v)
                    checkHappiness(x, y, v, u)
                    checkHappiness(y, x, v, u)

        return len(res)