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g1501_1600.s1583_count_unhappy_friends.readme.md Maven / Gradle / Ivy

1583\. Count Unhappy Friends

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:

*   `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.