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Problem solution · C++

Check if DFS Strings Are Palindromes

Check if DFS Strings Are Palindromes: a C++ 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
81 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

  • 81 lines of C++ from the credited upstream file 3327.cpp.
  • The implementation visibly relies on sequence storage.
  • 6 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 · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  vector<bool> findAnswer(vector<int>& parent, string s) {    const int n = parent.size();    vector<bool> ans(n);    vector<vector<int>> tree(n);    vector<int> start(n);  // start[i] := the start index of `dfsStr` of node i    vector<int> end(n);    // end[i] := the end index of `dfsStr` of node i    string dfsStr;     for (int i = 1; i < n; ++i)      tree[parent[i]].push_back(i);     dfs(tree, 0, /*index=*/0, s, start, end, dfsStr);    const string t = join('@' + dfsStr + '$', /*delimiter=*/'#');    const vector<int> p = manacher(t);     for (int i = 0; i < n; ++i)      ans[i] = isPalindrome(start[i], end[i], p);     return ans;  }  private:  // Returns the start index of the "DFS string" of u's next node.  int dfs(const vector<vector<int>>& tree, int u, int index, const string& s,          vector<int>& start, vector<int>& end, string& dfsStr) {    start[u] = index;    for (const int v : tree[u])      index = dfs(tree, v, index, s, start, end, dfsStr);    end[u] = index;    dfsStr += s[u];    return index + 1;  }   // Returns an array `p` s.t. `p[i]` is the length of the longest palindrome  // centered at `t[i]`, where `t` is a string with delimiters and sentinels.  vector<int> manacher(const string& t) {    vector<int> p(t.length());    int center = 0;    for (int i = 1; i < t.length() - 1; ++i) {      const int rightBoundary = center + p[center];      const int mirrorIndex = center - (i - center);      if (rightBoundary > i)        p[i] = min(rightBoundary - i, p[mirrorIndex]);      // Try to expand the palindrome centered at i.      while (t[i + 1 + p[i]] == t[i - 1 - p[i]])        ++p[i];      // If a palindrome centered at i expands past `rightBoundary`, adjust      // the center based on the expanded palindrome.      if (i + p[i] > rightBoundary)        center = i;    }    return p;  }   // Returns true if `dfsStr[s..e]` is a palindrome by using the precomputed  // array `p` from the Manacher's algorithm.  //  // The precomputed array `p` is based on the string `t` with delimiters and  // sentinels. Let `t = '#'.join('@' + dfsStr + '$')`. 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]`.  bool isPalindrome(int s, int e, const vector<int>& p) {    const int length = e - s + 1;    const int center = s + e + 2;    return p[center] >= length;  }   string join(const string& s, char delimiter) {    string joined;    for (int i = 0; i < s.length() - 1; ++i) {      joined += s[i];      joined += delimiter;    }    joined += s.back();    return joined;  }};

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