# Solving Leetcode Interviews in Seconds with AI: Check if DFS Strings Are Palindromes


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3327" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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.  

	# Explanation
	Here's the breakdown of the solution:

*   **Build the Tree:** Create an adjacency list representation of the tree from the `parent` array. This allows efficient traversal of the tree's children.
*   **DFS and Palindrome Check:** For each node `i`, perform a Depth-First Search (DFS) starting from that node. During the DFS, construct the `dfsStr` as specified in the problem statement. After the DFS completes, check if the `dfsStr` is a palindrome.
*   **Store Results:** Store the boolean result (palindrome or not) for each node in the `answer` array.

*   **Runtime Complexity:** O(n^2), where n is the number of nodes in the tree. This is because, in the worst case, each DFS traversal could visit all n nodes, and palindrome check takes O(n).
    **Storage Complexity:** O(n).  This accounts for the adjacency list representation of the tree, the `answer` array, and the maximum depth of the recursion stack, which can be n in a skewed tree.

	
	# Code
	```python
	def solve():
    def is_palindrome(s):
        return s == s[::-1]

    def dfs(node, adj, s):
        nonlocal dfs_str
        for child in sorted(adj[node]):
            dfs(child, adj, s)
        dfs_str += s[node]

    n = len(parent)
    adj = [[] for _ in range(n)]
    for i in range(1, n):
        adj[parent[i]].append(i)

    answer = [False] * n
    for i in range(n):
        global dfs_str
        dfs_str = ""
        dfs(i, adj, s)
        answer[i] = is_palindrome(dfs_str)

    return answer

parent = [-1,0,0,1,1,2]
s = "aababa"
print(solve())

parent = [-1,0,0,0,0]
s = "aabcb"
print(solve())
	```