Check if DFS Strings Are Palindromes

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TreesStringsRecursion

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the structure representing the graph? Is it an adjacency list or an adjacency matrix, and what data types do the nodes and edges use?
What constitutes a "DFS string" exactly? Is it the sequence of node values visited during a standard depth-first search traversal, and is the starting node pre-defined?
Can the graph be disconnected or contain cycles? If so, how should the DFS handle these cases, and how does that affect the creation of the DFS string?
If multiple DFS traversals result in palindromes, should I return `true` if any DFS string is a palindrome, or do I need to check all possible DFS strings?
What is the maximum possible number of nodes in the graph, and what is the maximum length of a single node's value when converted to a string?

Brute Force Solution

Approach

We're given a tree-like structure and we want to see if paths from the top to the bottom create palindrome strings. The brute force way is to explore every single possible path and then individually check if the string formed by that path is a palindrome.

Here's how the algorithm would work step-by-step:

Start at the very top of the tree.
Explore one path all the way down to the bottom.
As you go down that path, remember all the letters you encounter.
Once you reach the bottom, you have a string of letters.
Check if that string is a palindrome. Does it read the same forwards and backward?
Now, go back to where you had a choice of paths and try a different path.
Again, remember the letters, form a string, and check if it's a palindrome.
Keep doing this until you've explored every single possible path from the top to a bottom.
If at least one of the strings formed by these paths is a palindrome, then we have our answer.

Code Implementation

def check_dfs_strings_are_palindromes(tree):    def is_palindrome(string):        return string == string[::-1]
    def depth_first_search(node, current_path):        current_path += node['value']        # If we've reached a leaf node, check if the path is a palindrome
        if not node['children']:            if is_palindrome(current_path):                return True            else:                return False        # Recursively explore each child of the current node
        for child in node['children']:            if depth_first_search(child, current_path):                return True        return False
    # Start DFS from the root node with an empty path
    if not tree:        return False    return depth_first_search(tree, '')

Big(O) Analysis

Time Complexity

O(N * M) – The algorithm explores all possible paths from the root to the leaves in the tree. In the worst-case scenario, where the tree is balanced and each node has a constant number of children, the number of paths can be proportional to N, where N represents the number of leaf nodes. For each of these paths, which has a length proportional to M, where M is the maximum depth of the tree, the algorithm constructs a string and checks if it is a palindrome which takes O(M) time. Therefore, the overall time complexity is O(N * M), where N is the number of paths (leaf nodes) and M is the depth of the tree (length of the path).

Space Complexity

O(H) – The dominant space complexity comes from the recursion stack. In the worst-case scenario, the algorithm explores a path from the root to a leaf, resulting in recursive calls that add function call stack frames. The maximum depth of the recursion is determined by the height (H) of the tree. Therefore, the auxiliary space used by the recursion stack is proportional to the height of the tree, leading to a space complexity of O(H).

Optimal Solution

Approach

The core idea is to explore all possible strings created during a Depth First Search (DFS) traversal and efficiently check if each string is a palindrome. We use recursion to explore the possible strings and a simple palindrome check.

Here's how the algorithm would work step-by-step:

Start a recursive function that takes the current node being visited and the current string built so far as input.
Add the current node's value to the current string.
Check if the updated string is a palindrome.
If the current node has any children, call the recursive function on each of them, passing the updated string along. This explores each path in the implied tree.
The recursion stops when you reach a node without children. The final string will be tested if it is a palindrome. If you determine that the string is a palindrome return true. Otherwise, return false.
Combine the results: if at least one of the strings made by following a path is a palindrome, the answer is true; otherwise, it is false.

Code Implementation

def check_dfs_strings_are_palindromes(root):
    def is_palindrome(input_string):
        return input_string == input_string[::-1]

    def dfs(node, current_string):
        current_string += node.value

        # Check for palindrome after adding node value
        if is_palindrome(current_string):
            is_current_string_palindrome = True
        else:
            is_current_string_palindrome = False

        # If the node has no children.
        if not node.children:
            return is_current_string_palindrome

        # This boolean will store the combined result.
        at_least_one_palindrome = False

        # Explore all children.
        for child in node.children:
            at_least_one_palindrome = at_least_one_palindrome or dfs(child, current_string)

        return at_least_one_palindrome

    # Handle the case where the tree is empty.
    if not root:
        return False

    # Initiate DFS with an empty string.
    return dfs(root, "")

class Node:
    def __init__(self, value):
        self.value = value
        self.children = []

Big(O) Analysis

Time Complexity

O(n*m) – The time complexity is dominated by the Depth First Search (DFS) traversal and the palindrome checks. The DFS visits each node in the implicit tree which could have a depth of n (the number of nodes, in the worst case where the tree is a linked list) and for each path of length n, a string of length n is formed. The palindrome check performed on each such string takes O(m) time where m is the length of the string. Therefore, the overall time complexity is O(n*m), where n is the number of nodes in the implied tree and m is the average length of a string generated by a path from root to a leaf which, in the worst case, could be O(n) making the complexity O(n^2).

Space Complexity

O(N) – The space complexity is dominated by the depth of the recursion and the string concatenation within each recursive call. The recursion depth can go as deep as the number of nodes in the tree, which we define as N, leading to a maximum call stack size of O(N). Additionally, in each recursive call, a new string is created by adding the current node's value to the existing string; in the worst case, the length of the string grows up to O(N). Thus, the space used by the recursive call stack and the string concatenation scales linearly with N, resulting in O(N) auxiliary space.

Edge Cases

Case	How to Handle
Null or empty input string	Return true (or an empty list) as an empty string is considered a palindrome.
String with a single character	Return true as a single character string is a palindrome.
Maximum string length exceeding memory limitations	Check and handle potential stack overflow issues during recursion by limiting string length or using iterative DFS.
DFS yields no strings	If DFS produces no strings, return an empty list or a boolean indicating no palindromes were found.
All DFS strings are identical	The palindrome check should still correctly identify if this identical string is a palindrome or not.
DFS leads to cyclical paths creating infinite recursion	Implement cycle detection in the DFS to avoid infinite loops, potentially using a 'visited' set.
String contains non-alphanumeric characters	Filter out non-alphanumeric characters or specify the palindrome check to only consider alphanumeric characters.
Stack Overflow with deep recursion	Consider iterative DFS using a stack data structure if recursive DFS risks stack overflow for large trees.

Null or empty input string

How to Handle:

Return true (or an empty list) as an empty string is considered a palindrome.

String with a single character

How to Handle:

Return true as a single character string is a palindrome.

Maximum string length exceeding memory limitations

How to Handle:

Check and handle potential stack overflow issues during recursion by limiting string length or using iterative DFS.

DFS yields no strings

How to Handle:

If DFS produces no strings, return an empty list or a boolean indicating no palindromes were found.

All DFS strings are identical

How to Handle:

The palindrome check should still correctly identify if this identical string is a palindrome or not.

DFS leads to cyclical paths creating infinite recursion

How to Handle:

Implement cycle detection in the DFS to avoid infinite loops, potentially using a 'visited' set.

String contains non-alphanumeric characters

How to Handle:

Filter out non-alphanumeric characters or specify the palindrome check to only consider alphanumeric characters.

Stack Overflow with deep recursion

How to Handle:

Consider iterative DFS using a stack data structure if recursive DFS risks stack overflow for large trees.