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

/**
 * @param {number} n
 * @param {number[][]} preferences
 * @param {number[][]} pairs
 * @return {number}
 */
var unhappyFriends = function(n, preferences, pairs) {
    var preferenceMap = Array(n).fill(0).map(() => Array(n));
    for (var i = 0; i < preferences.length; i++) {
        for (var j = 0; j < preferences[i].length; j++) {
            preferenceMap[i][preferences[i][j]] = j;
        }
    }
    var pairMap = Array(n);
    for (var m = 0; m < pairs.length; m++) {
        pairMap[pairs[m][0]] = pairs[m][1];
        pairMap[pairs[m][1]] = pairs[m][0];
    }
    var res = 0;
    for (var k = 0; k < n; k++) {
        judge(preferenceMap, pairMap, k, n) && res++;
    }
    return res;
};

var judge = function(preferenceMap, pairMap, i, n) {
    for (var k = 0; k < n; k++) {
        if (k === i || k === pairMap[i]) continue;
        if (preferenceMap[i][pairMap[i]] > preferenceMap[i][k]
            && preferenceMap[k][pairMap[k]] > preferenceMap[k][i]) {
            return true;
        }
    }
    return false;
};

Explain:

nope.

Complexity:

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