LeetCode in Kotlin

3327. Check if DFS Strings Are Palindromes

Hard

You are given a tree rooted at node 0, consisting of n nodes numbered from 0 to n - 1. The tree is represented by an 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.

Consider an empty string dfsStr, and define a recursive function dfs(int x) that takes a node x as a parameter and performs the following steps in order:

• Iterate over each child y of x in increasing order of their numbers, and call dfs(y).
• Add the character s[x] to the end of the string dfsStr.

Note that dfsStr is shared across all recursive calls of dfs.

You need to find a boolean array answer of size n, where for each index i from 0 to n - 1, you do the following:

• Empty the string dfsStr and call dfs(i).
• If the resulting string dfsStr is a palindrome, then set answer[i] to true. Otherwise, set answer[i] to false.

Return the array answer.

A palindrome is a string that reads the same forward and backward.

Example 1:

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

Output: [true,true,false,true,true,true]

Explanation:

• Calling dfs(0) results in the string dfsStr = "abaaba", which is a palindrome.
• Calling dfs(1) results in the string dfsStr = "aba", which is a palindrome.
• Calling dfs(2) results in the string dfsStr = "ab", which is not a palindrome.
• Calling dfs(3) results in the string dfsStr = "a", which is a palindrome.
• Calling dfs(4) results in the string dfsStr = "b", which is a palindrome.
• Calling dfs(5) results in the string dfsStr = "a", which is a palindrome.

Example 2:

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

Output: [true,true,true,true,true]

Explanation:

Every call on dfs(x) results in a palindrome string.

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 only of lowercase English letters.

Solution

import kotlin.math.min

class Solution {
    private val e: MutableList<MutableList<Int?>?> = ArrayList<MutableList<Int?>?>()
    private val stringBuilder = StringBuilder()
    private var s: String? = null
    private var now = 0
    private var n = 0
    private lateinit var l: IntArray
    private lateinit var r: IntArray
    private lateinit var p: IntArray
    private lateinit var c: CharArray

    private fun dfs(x: Int) {
        l[x] = now + 1
        for (v in e[x]!!) {
            dfs(v!!)
        }
        stringBuilder.append(s!![x])
        r[x] = ++now
    }

    private fun matcher() {
        c[0] = '~'
        c[1] = '#'
        for (i in 1..n) {
            c[2 * i + 1] = '#'
            c[2 * i] = stringBuilder[i - 1]
        }
        var j = 1
        var mid = 0
        var localR = 0
        while (j <= 2 * n + 1) {
            if (j <= localR) {
                p[j] = min(p[(mid shl 1) - j], (localR - j + 1))
            }
            while (c[j - p[j]] == c[j + p[j]]) {
                ++p[j]
            }
            if (p[j] + j > localR) {
                localR = p[j] + j - 1
                mid = j
            }
            ++j
        }
    }

    fun findAnswer(parent: IntArray, s: String): BooleanArray {
        n = parent.size
        this.s = s
        for (i in 0 until n) {
            e.add(ArrayList<Int?>())
        }
        for (i in 1 until n) {
            e[parent[i]]!!.add(i)
        }
        l = IntArray(n)
        r = IntArray(n)
        dfs(0)
        c = CharArray(2 * n + 10)
        p = IntArray(2 * n + 10)
        matcher()
        val ans = BooleanArray(n)
        for (i in 0 until n) {
            val mid = (2 * r[i] - 2 * l[i] + 1) / 2 + 2 * l[i]
            ans[i] = p[mid] - 1 >= r[i] - l[i] + 1
        }
        return ans
    }
}

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