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[学习笔记] Leetcode 1583. Count Unhappy Friends

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发表于 2020-9-16 21:22:40 | 显示全部楼层 |阅读模式

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

For each person i, preferences 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 = [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.length == n - 1
0 <= preferences[j] <= n - 1
preferences does not contain i.
All values in preferences are unique.
pairs.length == n/2
pairs.length == 2
xi != yi
0 <= xi, yi <= n - 1
Each person is contained in exactly one pair.

  1. class Solution:
  2.     def unhappyFriends(self, n: int, preferences: List[List[int]], pairs: List[List[int]]) -> int:
  3.         hashmap = collections.defaultdict(lambda: collections.defaultdict(int))
  4.         for i in range(len(preferences)):
  5.             for j in range(len(preferences[i])):
  6.                 first = i
  7.                 second = preferences[i][j]
  8.                 idx = j
  9.                
  10.                 hashmap[first][second] = idx
  11.         
  12.         unhappy = [0] * n
  13.         m = len(pairs)
  14.         
  15.         for i in range(m):
  16.             x = pairs[i][0]
  17.             y = pairs[i][1]
  18.             
  19.             x_y_prefer, y_x_prefer = hashmap[x][y], hashmap[y][x]
  20.             
  21.             for j in range(i + 1, m):
  22.                 u = pairs[j][0]
  23.                 v = pairs[j][1]

  24.                 x_u_prefer, x_v_prefer = hashmap[x][u], hashmap[x][v]
  25.                 y_u_prefer, y_v_prefer = hashmap[y][u], hashmap[y][v]
  26.                
  27.                 u_v_prefer, u_x_prefer, u_y_prefer = hashmap[u][v], hashmap[u][x], hashmap[u][y]
  28.                 v_u_prefer, v_x_prefer, v_y_prefer = hashmap[v][u], hashmap[v][x], hashmap[v][y]
  29.                
  30.                 if x_u_prefer < x_y_prefer and u_x_prefer < u_v_prefer:
  31.                     unhappy[x] = unhappy[u] = 1
  32.                 if y_v_prefer < y_x_prefer and v_y_prefer < v_u_prefer:
  33.                     unhappy[y] = unhappy[v] = 1
  34.                 if x_v_prefer < x_y_prefer and v_x_prefer < v_u_prefer:
  35.                     unhappy[x] = unhappy[v] = 1
  36.                 if y_u_prefer < y_x_prefer and u_y_prefer < u_v_prefer:
  37.                     unhappy[y] = unhappy[u] = 1
  38.         return sum(unhappy)
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