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

Master LeetCode problem 2791 (Count Paths That Can Form a Palindrome in a Tree), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2791. Count Paths That Can Form a Palindrome in a Tree

Hard

Dynamic Programming Bit Manipulation Tree Depth-First Search Bitmask

Problem Explanation

Explanation:

To solve this problem, we can use a depth-first search (DFS) algorithm to traverse the tree and keep track of the count of paths that can form a palindrome. We can maintain a frequency count of characters encountered from the root to the current node while traversing the tree. At each node, we can calculate the number of valid pairs of nodes by considering the possible combinations of characters that can form a palindrome.

Algorithm:

Implement a DFS function to traverse the tree recursively.
Keep track of the frequency count of characters encountered on the path from the root to the current node.
Calculate the number of valid pairs at each node by considering the combinations of characters that can form a palindrome.
Update the count of valid pairs accordingly.
Return the total count of valid pairs.

Time Complexity: O(n), where n is the number of nodes in the tree. Space Complexity: O(n) for the recursive stack and auxiliary data structures.

Solution Code

class Solution {
    int count = 0;

    public int countPalindromicPaths(int[] parent, String s) {
        Map<Integer, Map<Character, Integer>> freq = new HashMap<>();
        dfs(0, -1, s, freq, parent);
        return count;
    }

    private Map<Character, Integer> dfs(int node, int parent, String s, Map<Integer, Map<Character, Integer>> freq, int[] parentArr) {
        Map<Character, Integer> nodeFreq = new HashMap<>();
        nodeFreq.put(s.charAt(node), freq.getOrDefault(parent, new HashMap<>()).getOrDefault(s.charAt(node), 0) + 1);
        freq.put(node, nodeFreq);
        
        for (int child = 0; child < parentArr.length; child++) {
            if (parentArr[child] == node && freq.get(child) != null) {
                Map<Character, Integer> childFreq = freq.get(child);
                for (char c : childFreq.keySet()) {
                    int val = childFreq.get(c);
                    count += val * (val - 1) / 2;
                    nodeFreq.put(c, nodeFreq.getOrDefault(c, 0) + val);
                }
            }
        }
        
        for (int child = 0; child < parentArr.length; child++) {
            if (parentArr[child] == node) {
                Map<Character, Integer> childFreq = dfs(child, node, s, freq, parentArr);
                for (char c : childFreq.keySet()) {
                    int val = childFreq.get(c);
                    count += val * (val - 1) / 2;
                    nodeFreq.put(c, nodeFreq.getOrDefault(c, 0) + val);
                }
            }
        }
        
        return nodeFreq;
    }
}

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