GuilinDev

Lc1583

05 August 2008

1583. Count Unhappy Friends

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:

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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 {
    public int unhappyFriends(int n, int[][] preferences, int[][] pairs) {
        int count = 0;
        int[][] ranking = new int[n][n];
        int[] pairTab = new int[n];

        for (int person = 0; person < n; person++) {
            int[] friends = preferences[person];
            for (int priority = 0; priority < friends.length; priority++) {
                ranking[person][friends[priority]] = priority;
            }
        }

        for (int[] pair : pairs) {
            pairTab[pair[0]] = pair[1];
            pairTab[pair[1]] = pair[0];
        }

        for (int[] pair : pairs) {
            int first = pair[0];
            int second = pair[1];

            if (ranking[first][second] != 0 && isUnhappy(first, second, preferences, ranking, pairTab)) {
                count++;
            }

            if (ranking[second][first] != 0 && isUnhappy(second, first, preferences, ranking, pairTab)) {
                count++;
            }
        }

        return count;
    }
    
    private boolean isUnhappy(int first, int second, int[][] preferences, int[][] ranking, int[] pairTab){
        int rankOfPrefferredBound = ranking[first][second] - 1;
       
        for (int preferredIdx = 0; preferredIdx <= rankOfPrefferredBound; preferredIdx++) {
            int preferredFriend = preferences[first][preferredIdx];
            int preferredFriendPair = pairTab[preferredFriend];
            if (ranking[preferredFriend][first] < ranking[preferredFriend][preferredFriendPair]) {
                return true;
            }
        } 
        
        return false;
    }
}
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