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

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count Paths That Can Form a Palindrome in a Tree?

This problem uses Dynamic Programming, Bit Manipulation, Tree, Depth-First Search, Bitmask. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count Paths That Can Form a Palindrome in a Tree?

Insight 1: A string can be rearranged to form a palindrome if and only if the number of characters appearing an odd number of times is at most 1. Insight 2: We can use bit manipulation to efficiently track the parity (even or odd) of character counts along the path from the root to each node. Each bit in a bitmask represents a character, and a set bit indicates an odd count.

Which companies ask LeetCode 2791. Count Paths That Can Form a Palindrome in a Tree in interviews?

This problem is frequently asked at thoughtspot.

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Count Paths That Can Form a Palindrome in a Tree - LeetCode 2791 Solution

Q: How to solve Count Paths That Can Form a Palindrome in a Tree?

The solution uses a depth-first search (DFS) or breadth-first search (BFS) to traverse the tree. For each node, a bitmask is maintained, representing the XOR of the characters encountered along the path from the root to that node. After computing the bitmasks for all nodes, the solution iterates through each node and counts the number of other nodes whose path XOR yields either 0 (already a palindrome) or a power of 2 (can become a palindrome by changing a single character). A hash map is used to store counts of each mask.

Count Paths That Can Form a Palindrome in a Tree - Complete Solution Guide

Count Paths That Can Form a Palindrome in a Tree is LeetCode problem 2791, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1 . The tree is represented by a 0-indexed 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 the edge between i and parent[i] . s[0] can be ignored. Return the number of pairs of nodes (u, v) such that u < v and the

Detailed Explanation

The problem asks us to count the number of pairs of nodes (u, v) in a tree, where u < v, such that the path between u and v can form a palindrome after rearranging the edge labels (characters) along the path. The tree is represented by a `parent` array, and edge labels are given by a string `s`. A palindrome is a string that reads the same forwards and backward. The core challenge is efficiently determining if the characters along a path can be rearranged to form a palindrome.

Solution Approach

The solution uses a depth-first search (DFS) or breadth-first search (BFS) to traverse the tree. For each node, a bitmask is maintained, representing the XOR of the characters encountered along the path from the root to that node. After computing the bitmasks for all nodes, the solution iterates through each node and counts the number of other nodes whose path XOR yields either 0 (already a palindrome) or a power of 2 (can become a palindrome by changing a single character). A hash map is used to store counts of each mask.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the tree from the `parent` array.
Step 2: Perform a BFS or DFS traversal of the tree, starting from the root (node 0).
Step 3: During traversal, compute the `path_masks` array for each node `i`, where `path_masks[i]` represents the XOR of all characters encountered on the path from the root to node `i`.
Step 4: Iterate through all nodes, and for each node 'i' check with previously visited nodes 'j' if the path from i to j can be rearranged to a palindrome.
Step 5: Use a hash map `counts` to store the frequency of each mask. When visiting node `i`, first find how many nodes `j` have the same `path_mask` (path from `i` to `j` is a palindrome), and then check for nodes whose `path_mask` differs by only one bit (path can be rearranged into a palindrome by changing one character).
Step 6: Update `counts` to reflect visiting the new node `i`.

Key Insights

Insight 1: A string can be rearranged to form a palindrome if and only if the number of characters appearing an odd number of times is at most 1.
Insight 2: We can use bit manipulation to efficiently track the parity (even or odd) of character counts along the path from the root to each node. Each bit in a bitmask represents a character, and a set bit indicates an odd count.
Insight 3: Using the XOR operation to update the bitmask as we traverse the tree allows us to quickly determine the bitmask for any node based on its parent's bitmask and the edge label connecting them.
Insight 4: For path (u, v), the path from u to v is the path from root to u XOR path from root to v. The common path from root to LCA will cancel out and result in path from u to v.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Dynamic Programming, Bit Manipulation, Tree, Depth-First Search, Bitmask.

Companies

Asked at: thoughtspot.

Count Paths That Can Form a Palindrome in a Tree | LeetCode 2791

Hard

Dynamic Programming Bit Manipulation Tree Depth-First Search Bitmask

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public long countPalindromePaths(List<Integer> parent, String s) {
        int n = parent.size();
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int i = 1; i < n; i++) {
            adj.get(parent.get(i)).add(i);
        }

        int[] pathMasks = new int[n];

        Queue<Integer> q = new LinkedList<>();
        q.offer(0);

        while (!q.isEmpty()) {
            int u = q.poll();
            for (int v : adj.get(u)) {
                int charCode = s.charAt(v) - 'a';
                pathMasks[v] = pathMasks[u] ^ (1 << charCode);
                q.offer(v);
            }
        }

        Map<Integer, Integer> counts = new HashMap<>();
        long ans = 0;
        for (int mask : pathMasks) {
            ans += counts.getOrDefault(mask, 0);

            for (int i = 0; i < 26; i++) {
                int targetMask = mask ^ (1 << i);
                ans += counts.getOrDefault(targetMask, 0);
            }

            counts.put(mask, counts.getOrDefault(mask, 0) + 1);
        }

        return ans;
    }
}

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Problem Info

Description

Problem 2791: Count Paths That Can Form a Palindrome in a Tree - Detailed explanation and solution approaches.

Topics

Dynamic Programming Bit Manipulation Tree Depth-First Search Bitmask

Companies

thoughtspot

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2791

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Build the adjacency list representation of the tree from the `parent` array.
2
Step 2: Perform a BFS or DFS traversal of the tree, starting from the root (node 0).
3
Step 3: During traversal, compute the `path_masks` array for each node `i`, where `path_masks[i]` represents the XOR of all characters encountered on the path from the root to node `i`.
4
Step 4: Iterate through all nodes, and for each node 'i' check with previously visited nodes 'j' if the path from i to j can be rearranged to a palindrome.
5
Step 5: Use a hash map `counts` to store the frequency of each mask. When visiting node `i`, first find how many nodes `j` have the same `path_mask` (path from `i` to `j` is a palindrome), and then check for nodes whose `path_mask` differs by only one bit (path can be rearranged into a palindrome by changing one character).
6
Step 6: Update `counts` to reflect visiting the new node `i`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the tree traversal and the mask counting process. The space complexity comes from storing the tree structure and the masks. The hash map stores unique mask values.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Depth-First Search (DFS) with Recursion

Instead of BFS, DFS can be used to traverse the tree and calculate path masks. The core logic remains the same.

Time: O(n)Space: O(n) due to the recursion stack in the worst case (skewed tree).

Trade-offs: DFS might be simpler to implement recursively, but BFS can be slightly more memory-efficient in some tree structures due to the iterative nature and potentially shallower recursion depth compared to DFS in certain scenarios.

✨

Key Insights

1Insight 1: A string can be rearranged to form a palindrome if and only if the number of characters appearing an odd number of times is at most 1.
2Insight 2: We can use bit manipulation to efficiently track the parity (even or odd) of character counts along the path from the root to each node. Each bit in a bitmask represents a character, and a set bit indicates an odd count.
3Insight 3: Using the XOR operation to update the bitmask as we traverse the tree allows us to quickly determine the bitmask for any node based on its parent's bitmask and the edge label connecting them.
4Insight 4: For path (u, v), the path from u to v is the path from root to u XOR path from root to v. The common path from root to LCA will cancel out and result in path from u to v.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: A tree with a single node. In this case, the number of pairs is 0, and the algorithm correctly returns 0.
Edge case 2: A tree where all edge labels are the same character. In this case, any path can form a palindrome if the path length is even. The algorithm correctly handles this by tracking the counts of masks.
Edge case 3: A tree with a long path where the path is already a palindrome. The XOR will be 0 and will be correctly counted.
Edge case 4: The constraint u < v is important and needs to be handled implicitly by counting only pairs encountered previously by the current node.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the path is already a palindrome (mask == 0).

✗Common mistake 2: Not accounting for the constraint u < v, which could lead to double-counting pairs.

✗Common mistake 3: Using the wrong data structure for counting masks, which could lead to a higher time complexity. A hash map is efficient for this task.

✗Common mistake 4: Incorrectly calculating the XOR mask, potentially due to off-by-one errors or not converting characters to integers correctly.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network routing: Imagine a network where the edges represent connections and the labels represent the types of connections. We might want to find paths between nodes that have a certain property (e.g., can be reconfigured into a specific sequence of connections).

🏦Real-world scenario 2: Bioinformatics: In phylogenetic trees, the path between two species represents their evolutionary relationship. The edge labels could represent certain genetic mutations. Finding palindromic paths could indicate interesting evolutionary events.

🎮Real-world scenario 3: Data compression: Finding palindromic sequences can be used to identify repetitive patterns, which can be used for data compression.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 31, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Paths That Can Form a Palindrome in a Tree"?

For this hard-level Dynamic Programming problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether bit manipulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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