LeetCode 3327: Check if DFS Strings Are Palindromes Solution

Master LeetCode problem 3327 (Check if DFS Strings Are Palindromes), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3327. Check if DFS Strings Are Palindromes

Hard

Array Hash Table String Tree Depth-First Search Hash Function

Problem Explanation

Explanation

To solve this problem, we will perform a Depth-First Search (DFS) traversal on the tree rooted at node 0. At each node, we will recursively traverse its children in increasing order and construct a string dfsStr by appending the characters of the nodes. After visiting all children, we check if the constructed string dfsStr is a palindrome or not. Based on this check, we update the boolean array answer accordingly. We repeat this process for all nodes in the tree.

Time Complexity

The time complexity of this solution is O(n) where n is the number of nodes in the tree.

Space Complexity

The space complexity of this solution is O(n) for the boolean array answer, the string dfsStr, and the recursive stack used in the DFS traversal.

Solution Code

class Solution {
    public boolean[] checkIfDfsStringsArePalindromes(int[] parent, String s) {
        int n = parent.length;
        boolean[] answer = new boolean[n];
        
        ArrayList<ArrayList<Integer>> adjList = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adjList.add(new ArrayList<>());
        }
        
        for (int i = 1; i < n; i++) {
            adjList.get(parent[i]).add(i);
        }
        
        for (int i = 0; i < n; i++) {
            StringBuilder dfsStr = new StringBuilder();
            dfs(i, adjList, s, dfsStr);
            answer[i] = isPalindrome(dfsStr.toString());
        }
        
        return answer;
    }
    
    private void dfs(int x, ArrayList<ArrayList<Integer>> adjList, String s, StringBuilder dfsStr) {
        for (int child : adjList.get(x)) {
            dfs(child, adjList, s, dfsStr);
        }
        dfsStr.append(s.charAt(x));
    }
    
    private boolean isPalindrome(String str) {
        int left = 0, right = str.length() - 1;
        while (left < right) {
            if (str.charAt(left) != str.charAt(right)) {
                return false;
            }
            left++;
            right--;
        }
        return true;
    }
}

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