LeetCode in Kotlin

2581. Count Number of Possible Root Nodes

Hard

Alice has an undirected tree with n nodes labeled from 0 to n - 1. The tree is represented as a 2D integer array edges of length n - 1 where edges[i] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the tree.

Alice wants Bob to find the root of the tree. She allows Bob to make several guesses about her tree. In one guess, he does the following:

Chooses two distinct integers u and v such that there exists an edge [u, v] in the tree.
He tells Alice that u is the parent of v in the tree.

Bob’s guesses are represented by a 2D integer array guesses where guesses[j] = [u_j, v_j] indicates Bob guessed u_j to be the parent of v_j.

Alice being lazy, does not reply to each of Bob’s guesses, but just says that at least k of his guesses are true.

Given the 2D integer arrays edges, guesses and the integer k, return the number of possible nodes that can be the root of Alice’s tree. If there is no such tree, return 0.

Example 1:

Input: edges = [[0,1],[1,2],[1,3],[4,2]], guesses = [[1,3],[0,1],[1,0],[2,4]], k = 3

Output: 3

Explanation:

Root = 0, correct guesses = [1,3], [0,1], [2,4]

Root = 1, correct guesses = [1,3], [1,0], [2,4]

Root = 2, correct guesses = [1,3], [1,0], [2,4]

Root = 3, correct guesses = [1,0], [2,4]

Root = 4, correct guesses = [1,3], [1,0]

Considering 0, 1, or 2 as root node leads to 3 correct guesses.

Example 2:

Input: edges = [[0,1],[1,2],[2,3],[3,4]], guesses = [[1,0],[3,4],[2,1],[3,2]], k = 1

Output: 5

Explanation:

Root = 0, correct guesses = [3,4]

Root = 1, correct guesses = [1,0], [3,4]

Root = 2, correct guesses = [1,0], [2,1], [3,4]

Root = 3, correct guesses = [1,0], [2,1], [3,2], [3,4]

Root = 4, correct guesses = [1,0], [2,1], [3,2]

Considering any node as root will give at least 1 correct guess.

Constraints:

edges.length == n - 1
2 <= n <= 10⁵
1 <= guesses.length <= 10⁵
0 <= a_i, b_i, u_j, v_j <= n - 1
a_i != b_i
u_j != v_j
edges represents a valid tree.
guesses[j] is an edge of the tree.
guesses is unique.
0 <= k <= guesses.length

Solution

import java.util.ArrayList
import java.util.HashSet

@Suppress("NAME_SHADOWING")
class Solution {
    private lateinit var parents: IntArray
    private lateinit var graph: Array<MutableList<Int>?>
    private lateinit var guess: Array<HashSet<Int>?>
    private var ans = 0

    fun rootCount(edges: Array<IntArray>, guesses: Array<IntArray>, k: Int): Int {
        val n = edges.size + 1
        graph = arrayOfNulls(n)
        guess = arrayOfNulls(n)
        for (i in 0 until n) {
            graph[i] = ArrayList()
            guess[i] = HashSet<Int>()
        }
        // Create tree
        for (i in edges.indices) {
            graph[edges[i][0]]!!.add(edges[i][1])
            graph[edges[i][1]]!!.add(edges[i][0])
        }
        // Create guess array
        for (i in guesses.indices) {
            guess[guesses[i][0]]!!.add(guesses[i][1])
        }
        parents = IntArray(n)
        fill(0, -1)
        var c = 0
        for (i in 1 until n) {
            if (guess[parents[i]]!!.contains(i)) c++
        }
        if (c >= k) ans++
        for (i in graph[0]!!) dfs(i, 0, c, k)
        return ans
    }

    // Fill the parent array
    private fun fill(v: Int, p: Int) {
        parents[v] = p
        for (child in graph[v]!!) {
            if (child == p) continue
            fill(child, v)
        }
    }

    // Use DFS to make all nodes as root one by one
    private fun dfs(v: Int, p: Int, c: Int, k: Int) {
        var c = c
        if (guess[p]!!.contains(v)) c--
        if (guess[v]!!.contains(p)) c++
        if (c >= k) ans++
        for (child in graph[v]!!) if (child != p) dfs(child, v, c, k)
    }
}

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