LeetCode 1583 Count Unhappy Friends - Medium

1583. Count Unhappy Friends — Medium

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Problem

1583. Count Unhappy Friends -- Medium

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

class Solution:
    def unhappyFriends(self, n: int, preferences: List[List[int]], pairs: List[List[int]]) -> int:
        rank = [[0] * n for _ in range(n)]
        for i, preference in enumerate(preferences):
            for j, friend in enumerate(preference):
                rank[i][friend] = j
 
        pairedWith = [0] * n
        for i, j in pairs:
            pairedWith[i] = j
            pairedWith[j] = i
 
        count = 0
        for x in range(n):
            y = pairedWith[x]
            for u in preferences[x]:
                if u == y:
                    break
                v = pairedWith[u]
                if rank[u][x] < rank[u][v]:
                    count += 1
                    break
        return count