Class Solution
Medium
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:
xprefersuovery, anduprefersxoverv.
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:
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1 is paired with 0 but prefers 3 over 0, and
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3 prefers 1 over 2.
Friend 3 is unhappy because:
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3 is paired with 2 but prefers 1 over 2, and
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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 <= 500nis even.preferences.length == npreferences[i].length == n - 10 <= preferences[i][j] <= n - 1preferences[i]does not containi.- 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.
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Constructor Summary
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Method Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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unhappyFriends
public int unhappyFriends(int n, int[][] preferences, int[][] pairs)
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