Count Unhappy Friends

Problem

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.

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

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.

Solution

Method 1 – Preference Ranking and Pair Simulation

Intuition

A friend is unhappy if they prefer someone else over their current pair, and that someone also prefers them over their own pair. We can use a ranking matrix to quickly check preferences and simulate all possible pairs to find unhappy friends.

Approach

1. Build a ranking matrix rank[i][j] representing how much person i prefers person j (lower is more preferred).
2. Build a pair array where pair[x] = y means x is paired with y.
3. For each person x, for every person u that x prefers over their pair y:
   • If u prefers x over their own pair v, mark x as unhappy.
4. Count and return the number of unhappy friends.

Code

C++

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

Go

func unhappyFriends(n int, preferences [][]int, pairs [][]int) int {
    rank := make([][]int, n)
    for i := range rank {
        rank[i] = make([]int, n)
        for j, v := range preferences[i] {
            rank[i][v] = j
        }
    }
    pair := make([]int, n)
    for _, p := range pairs {
        pair[p[0]] = p[1]
        pair[p[1]] = p[0]
    }
    ans := 0
    for x := 0; x < n; x++ {
        y := pair[x]
        for i := 0; preferences[x][i] != y; i++ {
            u := preferences[x][i]
            v := pair[u]
            if rank[u][x] < rank[u][v] {
                ans++
                break
            }
        }
    }
    return ans
}

Java

class Solution {
    public int unhappyFriends(int n, int[][] preferences, int[][] pairs) {
        int[][] rank = new int[n][n];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                rank[i][preferences[i][j]] = j;
        int[] pair = new int[n];
        for (int[] p : pairs) {
            pair[p[0]] = p[1];
            pair[p[1]] = p[0];
        }
        int ans = 0;
        for (int x = 0; x < n; x++) {
            int y = pair[x];
            for (int i = 0; preferences[x][i] != y; i++) {
                int u = preferences[x][i];
                int v = pair[u];
                if (rank[u][x] < rank[u][v]) {
                    ans++;
                    break;
                }
            }
        }
        return ans;
    }
}

Kotlin

class Solution {
    fun unhappyFriends(n: Int, preferences: Array<IntArray>, pairs: Array<IntArray>): Int {
        val rank = Array(n) { IntArray(n) }
        for (i in 0 until n)
            for (j in 0 until n)
                rank[i][preferences[i][j]] = j
        val pair = IntArray(n)
        for (p in pairs) {
            pair[p[0]] = p[1]
            pair[p[1]] = p[0]
        }
        var ans = 0
        for (x in 0 until n) {
            val y = pair[x]
            for (i in 0 until n) {
                if (preferences[x][i] == y) break
                val u = preferences[x][i]
                val v = pair[u]
                if (rank[u][x] < rank[u][v]) {
                    ans++
                    break
                }
            }
        }
        return ans
    }
}

Python

class Solution:
    def unhappyFriends(self, n: int, preferences: list[list[int]], pairs: list[list[int]]) -> int:
        rank = [[0]*n for _ in range(n)]
        for i in range(n):
            for j, v in enumerate(preferences[i]):
                rank[i][v] = j
        pair = [0]*n
        for x, y in pairs:
            pair[x] = y
            pair[y] = x
        ans = 0
        for x in range(n):
            y = pair[x]
            for u in preferences[x]:
                if u == y:
                    break
                v = pair[u]
                if rank[u][x] < rank[u][v]:
                    ans += 1
                    break
        return ans

Rust

impl Solution {
    pub fn unhappy_friends(n: i32, preferences: Vec<Vec<i32>>, pairs: Vec<Vec<i32>>) -> i32 {
        let n = n as usize;
        let mut rank = vec![vec![0; n]; n];
        for i in 0..n {
            for (j, &v) in preferences[i].iter().enumerate() {
                rank[i][v as usize] = j;
            }
        }
        let mut pair = vec![0; n];
        for p in pairs.iter() {
            pair[p[0] as usize] = p[1] as usize;
            pair[p[1] as usize] = p[0] as usize;
        }
        let mut ans = 0;
        for x in 0..n {
            let y = pair[x];
            for &u in preferences[x].iter() {
                if u as usize == y { break; }
                let v = pair[u as usize];
                if rank[u as usize][x] < rank[u as usize][v] {
                    ans += 1;
                    break;
                }
            }
        }
        ans
    }
}

TypeScript

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

Complexity

• ⏰ Time complexity: O(n^2) for building the rank matrix and checking all pairs.
• 🧺 Space complexity: O(n^2) for the rank matrix and O(n) for pairs.