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Problem solution · Python

Check if DFS Strings Are Palindromes

Check if DFS Strings Are Palindromes: a Python solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
71 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Check if DFS Strings Are Palindromes, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 71 lines of Python from the credited upstream file 3327.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCheck if DFS Strings Are Palindromes · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def findAnswer(self, parent: list[int], s: str) -> list[bool]:    n = len(parent)    tree = [[] for _ in parent]    start = [0] * n  # start[i] := the start index of `dfsStr` of node i    end = [0] * n  # end[i] := the end index of `dfsStr` of node i    dfsStr = []     for i in range(1, n):      tree[parent[i]].append(i)     self._dfs(tree, 0, 0, s, start, end, dfsStr)    t = '#'.join('@' + ''.join(dfsStr) + '$')    p = self._manacher(t)    return [self._isPalindrome(s, e, p)            for s, e in zip(start, end)]   def _dfs(      self,      tree: list[list[int]],      u: int,      index: int,      s: str,      start: list[int],      end: list[int],      dfsStr: list[str]  ) -> int:    """Returns the start index of the "DFS string" of u's next node."""    start[u] = index    for v in tree[u]:      index = self._dfs(tree, v, index, s, start, end, dfsStr)    end[u] = index    dfsStr.append(s[u])    return index + 1   def _manacher(self, t: str) -> list[int]:    """    Returns an array `p` s.t. `p[i]` is the length of the longest palindrome    centered at `t[i]`, where `t` is a string with delimiters and sentinels.    """    p = [0] * len(t)    center = 0    for i in range(1, len(t) - 1):      rightBoundary = center + p[center]      mirrorIndex = center - (i - center)      if rightBoundary > i:        p[i] = min(rightBoundary - i, p[mirrorIndex])      # Try to expand the palindrome centered at i.      while t[i + 1 + p[i]] == t[i - 1 - p[i]]:        p[i] += 1      # If a palindrome centered at i expands past `rightBoundary`, adjust      # the center based on the expanded palindrome.      if i + p[i] > rightBoundary:        center = i    return p   def _isPalindrome(self, s: int, e: int, p: list[int]) -> bool:    """    Returns true if `dfsStr[s..e]` is a palindrome by using the precomputed    array `p` from the Manacher's algorithm.     The precomputed array `p` is based on the string `t` with delimiters and    sentinels. Let `t = '#'.join('@' + dfsStr + '$')`. Then, the center of    `dfsStr` maps to `t[s + e + 2]` since `dfsStr[s]` maps to `t[2 * s + 2]`    and `dfsStr[e]` maps to `t[2 * e + 2]`. So, the center of `dfsStr` is    `t[(2 * s + 2 + 2 * e + 2) / 2] = t[s + e + 2]`.    """    length = e - s + 1    center = s + e + 2    return p[center] >= length

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