LeetCode 1583. Count Unhappy Friends Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Enumeration

We use an array d to record the closeness between each pair of friends, where 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 p to record the paired friend for each friend.

We enumerate each friend x. For x's paired friend y, we find the closeness d[x][y] of x to y. Then, we enumerate other friends u who are closer than d[x][y]. If there exists a friend u such that the closeness d[u][x] of u to x is higher than 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(n²), and the space complexity is O(n²). Here, n is the number of friends.

PythonJavaC++GoTypeScriptRust

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

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