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, andu 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 <= 500n is even.preferences.length == npreferences[i].length == n - 10 <= preferences[i][j] <= n - 1preferences[i] does not contain i.- All values in
preferences[i] are unique. pairs.length == n/2pairs[i].length == 2xi != yi0 <= xi, yi <= n - 1- Each person is contained in exactly one pair.
Solutions#
Solution 1#
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| 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
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| 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;
}
}
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| 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;
}
};
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| 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
}
|
