LeetCode #3327 — HARD

Check if DFS Strings Are Palindromes

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

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.

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.
Patterns Used
🌳Tree Traversal→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Tree

Example 1

[-1,0,0,1,1,2]
"aababa"

Example 2

[-1,0,0,0,0]
"aabcb"
Step 02

Core Insight

What unlocks the optimal approach

  • Perform the dfs described from the root of tree, and store the order in which nodes are visited into an array.
  • For any node in the tree, the nodes in its subtree will form a contiguous subarray within the DFS traversal array.
  • Use Manacher’s algorithm to compute the answer for each node in constant time.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3327: Check if DFS Strings Are Palindromes
class Hashing {
    private final long[] p;
    private final long[] h;
    private final long mod;

    public Hashing(String word, long base, int mod) {
        int n = word.length();
        p = new long[n + 1];
        h = new long[n + 1];
        p[0] = 1;
        this.mod = mod;
        for (int i = 1; i <= n; i++) {
            p[i] = p[i - 1] * base % mod;
            h[i] = (h[i - 1] * base + word.charAt(i - 1)) % mod;
        }
    }

    public long query(int l, int r) {
        return (h[r] - h[l - 1] * p[r - l + 1] % mod + mod) % mod;
    }
}

class Solution {
    private char[] s;
    private int[][] pos;
    private List<Integer>[] g;
    private StringBuilder dfsStr = new StringBuilder();

    public boolean[] findAnswer(int[] parent, String s) {
        this.s = s.toCharArray();
        int n = s.length();
        g = new List[n];
        pos = new int[n][0];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int i = 1; i < n; ++i) {
            g[parent[i]].add(i);
        }
        dfs(0);
        final int base = 13331;
        final int mod = 998244353;
        Hashing h1 = new Hashing(dfsStr.toString(), base, mod);
        Hashing h2 = new Hashing(new StringBuilder(dfsStr).reverse().toString(), base, mod);
        boolean[] ans = new boolean[n];
        for (int i = 0; i < n; ++i) {
            int l = pos[i][0], r = pos[i][1];
            int k = r - l + 1;
            long v1 = h1.query(l, l + k / 2 - 1);
            long v2 = h2.query(n + 1 - r, n + 1 - r + k / 2 - 1);
            ans[i] = v1 == v2;
        }
        return ans;
    }

    private void dfs(int i) {
        int l = dfsStr.length() + 1;
        for (int j : g[i]) {
            dfs(j);
        }
        dfsStr.append(s[i]);
        int r = dfsStr.length();
        pos[i] = new int[] {l, r};
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(n)

Approach Breakdown

LEVEL ORDER
O(n) time
O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL
O(n) time
O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.

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