leetcode.jp 博客
本网站所有文字及图片内容均来自网络,每周定时更新,仅供个人学习与研究,请勿用于商业用途。谢谢合作。

1583. Count Unhappy Friends

You are given a list of preferences for n friends, where n is always even.

For each person ipreferences[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.

Difficulty:

Medium

Lock:

Normal

Company:

Bloomberg

Solution(Chinese):

LEETCODE 1583. Count Unhappy Friends 解题思路分析