# 1583. Count Unhappy Friends

## Description

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

For each person ipreferences[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

• class Solution {
public int unhappyFriends(int n, int[][] preferences, int[][] pairs) {
int[][] d = new int[n][n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n - 1; ++j) {
d[i][preferences[i][j]] = j;
}
}
int[] p = new int[n];
for (var e : pairs) {
int x = e[0], y = e[1];
p[x] = y;
p[y] = x;
}
int ans = 0;
for (int x = 0; x < n; ++x) {
int y = p[x];
int find = 0;
for (int i = 0; i < d[x][y]; ++i) {
int u = preferences[x][i];
if (d[u][x] < d[u][p[u]]) {
find = 1;
break;
}
}
ans += find;
}
return ans;
}
}

• class Solution {
public:
int unhappyFriends(int n, vector<vector<int>>& preferences, vector<vector<int>>& pairs) {
int d[n][n];
int p[n];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n - 1; ++j) {
d[i][preferences[i][j]] = j;
}
}
for (auto& e : pairs) {
int x = e[0], y = e[1];
p[x] = y;
p[y] = x;
}
int ans = 0;
for (int x = 0; x < n; ++x) {
int y = p[x];
int find = 0;
for (int i = 0; i < d[x][y]; ++i) {
int u = preferences[x][i];
if (d[u][x] < d[u][p[u]]) {
find = 1;
break;
}
}
ans += find;
}
return ans;
}
};

• class Solution:
def unhappyFriends(
self, n: int, preferences: List[List[int]], pairs: List[List[int]]
) -> int:
d = [{p: i for i, p in enumerate(v)} for v in preferences]
p = {}
for x, y in pairs:
p[x] = y
p[y] = x
ans = 0
for x in range(n):
y = p[x]
ans += any(d[u][x] < d[u][p[u]] for u in preferences[x][: d[x][y]])
return ans


• func unhappyFriends(n int, preferences [][]int, pairs [][]int) (ans int) {
d := make([][]int, n)
p := make([]int, n)
for i := range d {
d[i] = make([]int, n)
for j := 0; j < n-1; j++ {
d[i][preferences[i][j]] = j
}
}
for _, e := range pairs {
x, y := e[0], e[1]
p[x] = y
p[y] = x
}
for x := 0; x < n; x++ {
y := p[x]
find := 0
for i := 0; i < d[x][y]; i++ {
u := preferences[x][i]
if d[u][x] < d[u][p[u]] {
find = 1
break
}
}
ans += find
}
return
}