1583. Count Unhappy Friends

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.

 

Solutions

Solution: Simulation

• Time complexity: O(n²)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number} n
 * @param {number[][]} preferences
 * @param {number[][]} pairs
 * @return {number}
 */
const unhappyFriends = function (n, preferences, pairs) {
  const pairPreferenceMap = pairs.reduce((map, [x, y]) => {
    map.set(x, preferences[x].indexOf(y));
    return map.set(y, preferences[y].indexOf(x));
  }, new Map());
  let result = 0;

  for (let friend = 0; friend < n; friend++) {
    const pairPreference = pairPreferenceMap.get(friend);

    for (let index = 0; index < pairPreference; index++) {
      const partner = preferences[friend][index];

      if (preferences[partner].indexOf(friend) < pairPreferenceMap.get(partner)) {
        result += 1;
        break;
      }
    }
  }
  return result;
};