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] = [x_i, y_i] denotes x_i is paired with y_i and y_i is paired with x_i.

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
x_i != y_i
0 <= x_i, y_i <= 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.

Python3

class Solution:
    def unhappyFriends(
        self, n: int, preferences: List[List[int]], pairs: List[List[int]]
    ) -> int:
        d = [{x: j for j, x in enumerate(p)} for p in preferences]
        p = {}
        for x, y in pairs:
            p[x] = y
            p[y] = x
        ans = 0
        for x in range(n):
            y = p[x]
            for i in range(d[x][y]):
                u = preferences[x][i]
                v = p[u]
                if d[u][x] < d[u][v]:
                    ans += 1
                    break
        return ans

Java

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];
            for (int i = 0; i < d[x][y]; ++i) {
                int u = preferences[x][i];
                int v = p[u];
                if (d[u][x] < d[u][v]) {
                    ++ans;
                    break;
                }
            }
        }
        return ans;
    }
}

C++

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

Go

func unhappyFriends(n int, preferences [][]int, pairs [][]int) (ans int) {
	d := make([][]int, n)
	for i := range d {
		d[i] = make([]int, n)
	}

	for i := 0; i < n; i++ {
		for j := 0; j < n-1; j++ {
			d[i][preferences[i][j]] = j
		}
	}

	p := make([]int, n)
	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]
		for i := 0; i < d[x][y]; i++ {
			u := preferences[x][i]
			v := p[u]
			if d[u][x] < d[u][v] {
				ans++
				break
			}
		}
	}

	return
}

TypeScript

function unhappyFriends(n: number, preferences: number[][], pairs: number[][]): number {
    const d: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n - 1; ++j) {
            d[i][preferences[i][j]] = j;
        }
    }
    const p: number[] = Array(n).fill(0);
    for (const [x, y] of pairs) {
        p[x] = y;
        p[y] = x;
    }
    let ans = 0;
    for (let x = 0; x < n; ++x) {
        const y = p[x];
        for (let i = 0; i < d[x][y]; ++i) {
            const u = preferences[x][i];
            const v = p[u];
            if (d[u][x] < d[u][v]) {
                ++ans;
                break;
            }
        }
    }
    return ans;
}