1583. Count Unhappy Friends

Problem

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&nbsp;<= n - 1

• Each person is contained in exactly one pair.

Solution (Java)

class Solution {
    public int unhappyFriends(int n, int[][] preferences, int[][] pairs) {
        int unhappyFriends = 0;
        Map<Integer, Integer> assignedPair = new HashMap<>();
        for (int[] pair : pairs) {
            assignedPair.put(pair[0], pair[1]);
            assignedPair.put(pair[1], pair[0]);
        }
        for (int[] pair : pairs) {
            if (isUnHappy(pair[1], pair[0], preferences, assignedPair)) {
                unhappyFriends++;
            }
            if (isUnHappy(pair[0], pair[1], preferences, assignedPair)) {
                unhappyFriends++;
            }
        }
        return unhappyFriends;
    }

    private boolean isUnHappy(
            int self,
            int assignedFriend,
            int[][] preferences,
            Map<Integer, Integer> assignedPairs) {
        int[] preference = preferences[self];
        int assignedFriendPreferenceIndex = findIndex(preference, assignedFriend);
        for (int i = 0; i <= assignedFriendPreferenceIndex; i++) {
            int preferredFriend = preference[i];
            int preferredFriendAssignedFriend = assignedPairs.get(preferredFriend);
            if (preferredFriendAssignedFriend == self) {
                return false;
            }
            int candidateAssignedFriendIndex =
                    findIndex(preferences[preferredFriend], preferredFriendAssignedFriend);
            if (isPreferred(self, preferences[preferredFriend], candidateAssignedFriendIndex)) {
                return true;
            }
        }
        return false;
    }

    private boolean isPreferred(int self, int[] preference, int boundary) {
        for (int i = 0; i <= boundary; i++) {
            if (self == preference[i]) {
                return true;
            }
        }
        return false;
    }

    private int findIndex(int[] preference, int assignedFriend) {
        for (int i = 0; i < preference.length; i++) {
            if (preference[i] == assignedFriend) {
                return i;
            }
        }
        return 0;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).