1583.count-unhappy-friends.kt

/*
 * @lc app=leetcode id=1583 lang=kotlin
 *
 * [1583] Count Unhappy Friends
 *
 * https://leetcode.com/problems/count-unhappy-friends/description/
 *
 * algorithms
 * Medium (51.45%)
 * Likes:    48
 * Dislikes: 219
 * Total Accepted:    5.7K
 * Total Submissions: 11K
 * Testcase Example:  '4\n[[1,2,3],[3,2,0],[3,1,0],[1,2,0]]\n[[0,1],[2,3]]'
 *
 * 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.
 * 
 * 
 */

// @lc code=start
class Solution {
    fun unhappyFriends(n: Int, preferences: Array<IntArray>, pairs: Array<IntArray>): Int {
        var matrix = Array(n){ IntArray(n) }
        for (i in 0 until n) {
            for (j in 0 until n-1) {
                matrix[i][preferences[i][j]] = j
            }
        }

        var unhappy = BooleanArray(n)
        var i = 0
        var j = 0
        while (i+i < n) {
            val x = pairs[i][0]
            val y = pairs[i][1]
            j = 0
            while (j+j < n) {

                if (i == j) {
                    j++
                    continue
                }

                val u = pairs[j][0]
                val v = pairs[j][1]

                if ((matrix[x][u] < matrix[x][y] && matrix[u][x] < matrix[u][v])
                 || (matrix[x][v] < matrix[x][y] && matrix[v][x] < matrix[v][u])) unhappy[x] = true;
                if ((matrix[y][u] < matrix[y][x] && matrix[u][y] < matrix[u][v])
                 || (matrix[y][v] < matrix[y][x] && matrix[v][y] < matrix[v][u])) unhappy[y] = true;

                j++
            }
            i++
        }
        var ans = 0
        for (i in 0 until n)
            if (unhappy[i]) ans++
        return ans
    }
}
// @lc code=end