Check if DFS Strings Are Palindromes

Problem Description

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.

Key Insights

• The problem involves a tree structure and requires a depth-first search (DFS) traversal.
• The order of child nodes matters, as they must be traversed in increasing order.
• A string is a palindrome if it reads the same forwards and backwards.
• The solution must handle up to 100,000 nodes efficiently, suggesting the need for an optimal approach.

Space and Time Complexity

Time Complexity: O(n) per node, leading to O(n^2) in the worst case if each node's DFS is performed independently. However, optimizations can be made by reusing results. Space Complexity: O(n) for storing the DFS results and the string representation.

Solution

The approach involves using a depth-first search (DFS) to traverse the tree. At each node, we append the character corresponding to that node to a shared string. After traversing all children, we check if the resulting string is a palindrome. We maintain a boolean result for each node indicating whether the DFS string formed starting from that node is a palindrome.

To implement this efficiently:

1. Use a list to keep track of children for each node based on the parent array.
2. Use a recursive DFS function that constructs the string as it traverses.
3. Check for palindromicity at the end of the DFS call for each node.

Code Solutions