How do you solve Check if DFS Strings Are Palindromes on LeetCode?

Check if DFS Strings Are Palindromes (LeetCode #3327) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The DFS traversal order captures the structure of the tree effectively.

What is the brute force solution for Check if DFS Strings Are Palindromes?

The brute force approach for Check if DFS Strings Are Palindromes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves performing a DFS from each node and constructing the string for each subtree. We then check if the constructed string is a palindrome. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Check if DFS Strings Are Palindromes?

Clarify the tree structure and input format before coding. Discuss the time complexity of your approach during the interview. Practice writing DFS functions, as they are common in tree-related problems.

What are common mistakes when solving Check if DFS Strings Are Palindromes?

Forgetting to sort children in DFS, leading to incorrect traversal order. Not properly checking for palindromes, especially with substrings.

#3327

Check if DFS Strings Are Palindromes

Hard

ArrayHash TableStringTreeDepth-First SearchHash FunctionDepth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the properties of the DFS traversal. By storing the order of nodes visited in a single DFS traversal, we can efficiently check if the substrings corresponding to each node are palindromes without redundant DFS calls.

⚙️

Algorithm

3 steps

1Step 1: Build the tree structure from the parent array.
2Step 2: Perform a single DFS traversal from the root, recording the order of nodes visited in a list.
3Step 3: For each node, extract its corresponding substring from the DFS list and check if it's a palindrome.

solution.py29 lines

1# Full working Python code
2from collections import defaultdict
3
4def checkPalindromeStrings(parent, s):
5    n = len(s)
6    tree = defaultdict(list)
7    for i in range(n):
8        if parent[i] != -1:
9            tree[parent[i]].append(i)
10    
11    dfsOrder = []
12    
13    def dfs(node):
14        dfsOrder.append(s[node])
15        for child in sorted(tree[node]):
16            dfs(child)
17
18    dfs(0)
19    
20    answer = [False] * n
21    
22    for i in range(n):
23        start = dfsOrder.index(s[i])
24        end = start + dfsOrder.count(s[i]) - 1
25        substring = dfsOrder[start:end + 1]
26        answer[i] = substring == substring[::-1]
27    
28    return answer
29

ℹ

Complexity note: The time complexity is O(n) because we perform a single DFS traversal of the tree, visiting each node once. The space complexity is O(n) due to the storage of the DFS order.

1The DFS traversal order captures the structure of the tree effectively.
2Palindromes can be checked efficiently by leveraging string properties.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.