# 1583. Count Unhappy Friends

## 1. Question

You are given a list of preferences for n friends, where n is always even.

For each personi, 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.

## 2. Examples

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


## 3. 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 inpreferences[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.

## 5. Solutions

class Solution {
public int unhappyFriends(int n, int[][] preferences, int[][] pairs) {
int[][] order = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n - 1; j++) {
order[i][preferences[i][j]] = j;
}
}
int[] match = new int[n];
for (int[] pair : pairs) {
int person0 = pair[0], person1 = pair[1];
match[person0] = person1;
match[person1] = person0;
}
int unhappyCount = 0;
for (int x = 0; x < n; x++) {
int y = match[x];
int index = order[x][y];
for (int i = 0; i < index; i++) {
int u = preferences[x][i];
int v = match[u];
if (order[u][x] < order[u][v]) {
unhappyCount++;
break;
}
}
}
return unhappyCount;
}
}
// 题解自官方。
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