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
prefersu
overy
, andu
prefersx
overv
.
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 containi
.- 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 1: Enumeration
We use an array $\textit{d}$ to record the closeness between each pair of friends, where $\textit{d}[i][j]$ represents the closeness of friend $i$ to friend $j$ (the smaller the value, the closer they are). Additionally, we use an array $\textit{p}$ to record the paired friend for each friend.
We enumerate each friend $x$. For $x$'s paired friend $y$, we find the closeness $\textit{d}[x][y]$ of $x$ to $y$. Then, we enumerate other friends $u$ who are closer than $\textit{d}[x][y]$. If there exists a friend $u$ such that the closeness $\textit{d}[u][x]$ of $u$ to $x$ is higher than $\textit{d}[u][y]$, then $x$ is an unhappy friend, and we increment the result by one.
After the enumeration, we obtain the number of unhappy friends.
The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the number of friends.
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