How do you solve Count Paths That Can Form a Palindrome in a Tree on LeetCode?

Count Paths That Can Form a Palindrome in a Tree (LeetCode #2791) 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: A string can form a palindrome if at most one character has an odd frequency.

What is the time complexity of Count Paths That Can Form a Palindrome in a Tree?

The optimal solution for Count Paths That Can Form a Palindrome in a Tree runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse each node once during the DFS, and the space complexity is O(n) due to the hashmap storing the mask counts.

What is the brute force solution for Count Paths That Can Form a Palindrome in a Tree?

The brute force approach for Count Paths That Can Form a Palindrome in a Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all pairs of nodes and determining if the characters on the path between them can form a palindrome. This is straightforward but inefficient for larger trees. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Paths That Can Form a Palindrome in a Tree?

Count Paths That Can Form a Palindrome in a Tree uses the following concepts: Hash Table, Bit Manipulation, Tree, Depth-First Search. The recommended patterns to study are: Bit Manipulation, Depth-First Search, Hash Map.

What are the interview tips for Count Paths That Can Form a Palindrome in a Tree?

Clearly explain your thought process and why you choose a particular approach. Be ready to discuss the trade-offs of different methods. Practice DFS and bit manipulation problems to become comfortable with these concepts.

What are common mistakes when solving Count Paths That Can Form a Palindrome in a Tree?

Not considering all pairs correctly, especially when counting paths. Overlooking the need to backtrack in DFS, which can lead to incorrect counts.

#2791

Count Paths That Can Form a Palindrome in a Tree

Hard

Hash TableBit ManipulationTreeDepth-First SearchBit ManipulationDepth-First SearchHash Map

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 uses bit manipulation to track character frequencies efficiently. By maintaining a bitmask for each node that represents the parity of character counts, we can quickly determine if the path can form a palindrome.

⚙️

Algorithm

4 steps

1Step 1: Create a bitmask for each node where each bit represents the parity of the character count from the root to that node.
2Step 2: Use Depth-First Search (DFS) to compute the bitmask for each node.
3Step 3: Count how many times each bitmask appears using a hashmap.
4Step 4: For each bitmask, if it appears k times, it contributes k * (k - 1) / 2 valid pairs. Also, check for each bitmask with one bit flipped (to account for odd counts).

solution.py21 lines

1def countPalindromicPaths(parent, s):
2    from collections import defaultdict
3    n = len(parent)
4    count = 0
5    mask_count = defaultdict(int)
6    mask_count[0] = 1  # Empty path mask
7    def dfs(node, mask):
8        nonlocal count
9        # Update the mask for the current node
10        mask ^= (1 << (ord(s[node]) - ord('a')))
11        count += mask_count[mask]  # Count pairs with the same mask
12        # Check for masks with one bit flipped
13        for i in range(26):
14            count += mask_count[mask ^ (1 << i)]
15        mask_count[mask] += 1
16        for child in range(n):
17            if parent[child] == node:
18                dfs(child, mask)
19        mask_count[mask] -= 1  # Backtrack
20    dfs(0, 0)
21    return count

ℹ

Complexity note: The time complexity is O(n) because we traverse each node once during the DFS, and the space complexity is O(n) due to the hashmap storing the mask counts.

1A string can form a palindrome if at most one character has an odd frequency.
2Using bit manipulation allows us to efficiently track character frequencies and their parities.

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