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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 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
}

  • 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;
}

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