LeetCode in Kotlin

2246. Longest Path With Different Adjacent Characters

Hard

You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1.

You are also given a string s of length n, where s[i] is the character assigned to node i.

Return the length of the longest path in the tree such that no pair of adjacent nodes on the path have the same character assigned to them.

Example 1:

Input: parent = [-1,0,0,1,1,2], s = “abacbe”

Output: 3

Explanation: The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -> 1 -> 3. The length of this path is 3, so 3 is returned.

It can be proven that there is no longer path that satisfies the conditions.

Example 2:

Input: parent = [-1,0,0,0], s = “aabc”

Output: 3

Explanation: The longest path where each two adjacent nodes have different characters is the path: 2 -> 0 -> 3. The length of this path is 3, so 3 is returned.

Constraints:

• n == parent.length == s.length
• 1 <= n <= 105
• 0 <= parent[i] <= n - 1 for all i >= 1
• parent[0] == -1
• parent represents a valid tree.
• s consists of only lowercase English letters.

Solution

import java.util.LinkedList

class Solution {
    fun longestPath(parent: IntArray, s: String): Int {
        // for first max length
        val first = IntArray(s.length)
        first.fill(0)
        // for second max length
        val second = IntArray(s.length)
        second.fill(0)
        // for number of children for this node
        val children = IntArray(s.length)
        children.fill(0)
        for (i in 1 until s.length) {
            // calculate all children for each node
            children[parent[i]]++
        }
        // for traversal from leafs to root
        val st = LinkedList<Int>()
        // put all leafs
        for (i in 1 until s.length) {
            if (children[i] == 0) {
                st.add(i)
            }
        }
        // traversal from leafs to root
        while (st.isNotEmpty()) {
            // fetch current node
            val i = st.pollLast()
            // if we in root - ignore it
            if (i == 0) {
                continue
            }
            if (--children[parent[i]] == 0) {
                // decrease number of children by parent node and if number of children
                st.add(parent[i])
            }
            // is equal 0 - our parent became a leaf
            // if letters isn't equal
            if (s[parent[i]] != s[i]) {
                // fetch maximal path from node
                val maxi = 1 + Math.max(first[i], second[i])
                // and update maximal first and second path from parent
                if (maxi >= first[parent[i]]) {
                    second[parent[i]] = first[parent[i]]
                    first[parent[i]] = maxi
                } else if (second[parent[i]] < maxi) {
                    second[parent[i]] = maxi
                }
            }
        }
        // fetch answer
        var ans = 0
        for (i in 0 until s.length) {
            ans = Math.max(ans, first[i] + second[i])
        }
        return ans + 1
    }
}

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