2791. Count Paths That Can Form a Palindrome in a Tree
Problem Description
In this problem, we are given a tree with n
nodes, where each node is numbered from 0
to n - 1
. The tree is rooted at node 0
, and we are provided a parent array named parent
that defines the tree's structure. The parent[i]
value represents the parent node of node i
, and for the root node 0
, we have parent[0] == -1
because it does not have any parent.
Along with the tree, we have a string s
of length n
that assigns characters to the edges in the tree. The character s[i]
corresponds to the edge between node i
and its parent parent[i]
. Note that s[0]
can be ignored because it is associated with the root, which has no parent.
The task is to find the number of distinct pairs of nodes (u, v)
where u < v
and the series of characters assigned to the edges in the path from u
to v
can be rearranged to form a palindrome. A palindrome is a sequence that reads the same backward and forward.
Flowchart Walkthrough
To analyze Leetcode 2791, "Count Paths That Can Form a Palindrome in a Tree," using the provided algorithm flowchart, let's go through the decision nodes in the flowchart to determine the appropriate algorithm:
-
Is it a graph?
- Yes: The problem involves a tree, which is a specialized type of graph with no cycles and is connected.
-
Is it a tree?
- Yes: The structure described in the problem is explicitly mentioned as a tree.
-
DFS:
- Since it is a tree, Depth-First Search (DFS) is typically used for tree-related problems to explore all paths or perform operations on paths, such as checking for palindrome conditions from any node to any other node.
The flowchart confirms that DFS is the suitable choice for solving this problem as the tree structure lends itself well to recursive traversal techniques like DFS, which can efficiently explore all possible paths for checking palindrome conditions.
For a visual representation and further understanding of the decision-making process, please refer to the Flowchart.
Intuition
To solve this problem, we need to find all paths in the tree that can create palindromes when the edge characters are rearranged. However, since directly checking every path would be too slow, we approach the problem using a bitwise XOR technique.
We begin with the observation that, for a string of characters to be rearranged into a palindrome, each character must appear an even number of times except for at most one character which can be in the middle. In terms of edges in the path, it means that every character should connect an even number of edges except for at most one.
Translating this idea into an algorithm, we use a bitwise representation to keep track of even and odd counts of edge characters along any path. We create a 26-bit integer to represent the count of each letter in the alphabet, where each bit corresponds to a specific character. If a bit is set, it means an odd count of that character; if it is not set, it means an even count.
Now, through depth-first search (DFS), we traverse the tree starting at the root node and calculate the XOR value representing the character count at each node. For each node, we update the global answer (ans
) by counting how many times we've seen this XOR pattern (since the path from the root to this node has the same palindrome check as some previous path), and how many times we've seen this XOR pattern with exactly one bit flipped (since that represents just a single character being odd, which is allowable for palindromes).
The overall process includes accumulating the XOR value while descending through the tree, checking palindromic conditions at each step, and updating the global count of valid paths.
The provided solution keeps track of the counts using a Counter dictionary (cnt
), which maps each 26-bit integer to the number of times that bit pattern has previously occurred during the DFS. The dfs()
function is called recursively to traverse the tree, and the ans
variable, defined as nonlocal, is used to accumulate the total number of palindromic paths.
Learn more about Tree, Depth-First Search, Dynamic Programming and Bitmask patterns.
Solution Approach
The solution uses a depth-first search (DFS) algorithm to traverse the tree, taking advantage of a bitwise representation to encode the frequency of edge characters along the path from the root to any given node.
Data Structures Used:
- A defaultdict (imported from
collections
module) namedg
to represent the adjacency list of the tree, where each node has a list of tuples containing its child node and the corresponding edge character's bit representation. - A Counter (also from
collections
module) namedcnt
to keep track of the number of times we've seen a particular 26-bit representation of character frequencies.
The DFS Algorithm and Bitwise Patterns:
Within the provided code, the dfs()
function is responsible for exploring the tree:
- It accepts two arguments:
i
, the current node index, andxor
, the sum of edge character bit representations up to this node. ans
is a nonlocal variable that accumulates the number of valid paths that can form a palindrome. It is incremented both when an existingxor
value is found incnt
and when a single-bit difference version ofxor
is found incnt
.- As the function traverses the tree, it calculates the new
xor
value for each child by using the bitwise XOR operation (^
) between the currentxor
and the bit representation of the child edge's character. - A loop through 26 different bit flips (
for k in range(26):
) is used to check if there's a valid path that has exactly one character with an odd count, which is still acceptable for palindrome formation. If such a condition is found,ans
is also incremented for these occurrences. - The
cnt
dictionary keeps track of how many times we've seen this particularxor
pattern by incrementingcnt[x]
. - The recursion continues by calling
dfs(j, x)
for each childj
.
Building the Bit Representation:
- For each character in the edge, a 26-bit integer is built by shifting
1
to the leftord(s[i]) - ord('a')
times, which indicates the position of the character in the alphabet (e.g., 'a' corresponds to shifting 0 times as it's the first letter, 'b' corresponds to shifting 1 time, and so on). - This way, each bit of the integer represents the count's parity of a corresponding character on the path from the root to the current node.
Initialization and Invocation:
- The function
countPalindromePaths
initializes the adjacency listg
and the Countercnt
with{0: 1}
, which corresponds to the bit representation of an empty path (all bits set to zero, meaning even count for every character). - It then calls
dfs(0, 0)
, starting the traversal from the root node with an initialxor
of0
.
By using these techniques, the code efficiently finds all palindromic paths within the tree.
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Start EvaluatorExample Walkthrough
Let's use a small example to illustrate the solution approach. Consider the following tree with n = 4
nodes and its associated parent array and string:
parent = [-1, 0, 1, 1] s = "abca"
This defines a tree where:
- Node
0
is the root asparent[0] == -1
. - Node
1
is the child of node0
. - Nodes
2
and3
are children of node1
.
The string s
colors the edges as follows:
- Edge from node
1
to node0
(the parent) with character'a'
. - Edge from node
2
to node1
with character'b'
. - Edge from node
3
to node1
with character'c'
.
Thus, the tree looks like this:
0 / a / 1 / \ b c / \ 2 3
In the bitwise representation:
- For edge 'a' (from 0 to 1), the bit is
1 << (ord('a') - ord('a')) = 1 << 0 = 1
. - For edge 'b' (from 1 to 2), the bit is
1 << (ord('b') - ord('a')) = 1 << 1 = 2
. - For edge 'c' (from 1 to 3), the bit is
1 << (ord('c') - ord('a')) = 1 << 2 = 4
.
Now, we'll walk through the DFS algorithm with this example:
-
We start at the root node
0
with an initialxor
value of0
. -
We visit the child node
1
. Edge0-1
is labeled'a'
, so thexor
value becomes0 ^ 1 = 1
. -
At node
1
, there are two children. We explore each of them in turn:- Visiting node
2
, the edge1-2
is'b'
, so thexor
becomes1 ^ 2 = 3
. We now have a path0 - 1 - 2
with edge characters'a', 'b'
, which gives a bitwise pattern of011
telling us the count of character 'a' is odd, 'b' is odd, and 'c' is even (hence not set). - Visiting node
3
, the edge1-3
is'c'
, so thexor
becomes1 ^ 4 = 5
. The path0 - 1 - 3
has edge characters'a', 'c'
, resulting in a bitwise pattern of101
.
- Visiting node
-
For each of these nodes
2
and3
, we check the presence of thexor
incnt
and check for each possible single-bit flip.- When visiting node
2
, thexor
value before visiting children is3
, which has not been seen before, socnt[3]
is set to1
. - For node
3
, similarly, thexor
value is5
, andcnt[5]
is set to1
. - There are no single-bit difference patterns for
3
or5
incnt
yet, soans
stays the same for now.
- When visiting node
-
We do not increment the global
ans
during this initial pass as the DFS is still in progress. -
Finally, after traversing the entire tree, we find that we have the following pair that can form a palindrome:
- Nodes
(2, 3)
: The path characters are'b'
and'c'
. By reordering the characters, we can't form a palindrome as there must be an even count of each character or an even count of all characters except one.
- Nodes
At the end of our DFS, we find that there are no valid palindromic paths in this particular example, so the answer (ans
) is 0
.
Solution Implementation
1from collections import defaultdict, Counter
2
3class Solution:
4 def countPalindromePaths(self, parents: List[int], characters: str) -> int:
5 # Helper function to perform Depth-First Search (DFS)
6 def dfs(node_index: int, xor_value: int):
7 # Accessing the variable 'answer' from the outer scope
8 nonlocal answer
9 for child_index, char_value in adjacency_list[node_index]:
10 # Computing the XOR of the current path's characters
11 new_xor = xor_value ^ char_value
12 # If the XOR matches any of our previous XOR values, it's a palindrome path
13 answer += xor_count[new_xor]
14 # Check if toggling any single bit (corresponding to changing one character)
15 # results in a palindrome path
16 for k in range(26):
17 answer += xor_count[new_xor ^ (1 << k)]
18 # Count the occurrences of this XOR value along the path
19 xor_count[new_xor] += 1
20 # Recursive DFS call to continue the search
21 dfs(child_index, new_xor)
22 # When backtracking, remove the count of the current node's XOR from the count map
23 xor_count[xor_value] -= 1
24
25 # Size of the tree (number of nodes)
26 n = len(parents)
27 # Adjacency list to represent the tree and store (child_index, binary representation of the character)
28 adjacency_list = defaultdict(list)
29 for i in range(1, n):
30 parent_index = parents[i]
31 adjacency_list[parent_index].append((i, 1 << (ord(characters[i]) - ord('a'))))
32
33 # The answer to keep track of the number of palindrome paths
34 answer = 0
35 # Counter to store the number of times an XOR value has been seen on a path
36 xor_count = Counter({0: 1})
37
38 # Start DFS from the root node with XOR value 0 (since no characters have been visited yet)
39 dfs(0, 0)
40
41 # Return the final count of palindrome paths
42 return answer
43
44# Note: List[int] and the function type signature should ideally be imported:
45# from typing import List
46# if you are testing this code outside of a preset environment like LeetCode,
47# otherwise it would throw a NameError for the undefined name 'List'.
48
1class Solution {
2 // Graph representation: an array of lists, where each list contains pairs (child index, character bitmask)
3 private List<int[]>[] graph;
4 // Counter for each state of the XOR of the path (nodeMask)
5 private Map<Integer, Integer> countMap = new HashMap<>();
6 // Answer to accumulate the number of possible palindrome paths
7 private long answer;
8
9 public long countPalindromePaths(List<Integer> parents, String s) {
10 int n = parents.size();
11 graph = new List[n];
12 countMap.put(0, 1);
13 // Initialize each element of the graph with an empty ArrayList
14 Arrays.setAll(graph, k -> new ArrayList<>());
15 // Create the graph from the parent list
16 for (int i = 1; i < n; ++i) {
17 int parentIndex = parents.get(i);
18 graph[parentIndex].add(new int[] {i, 1 << (s.charAt(i) - 'a')});
19 }
20 // Start DFS traversal from root node=0 with initial XOR value 0
21 dfs(0, 0);
22 return answer;
23 }
24
25 // DFS function to traverse the tree and calculate palindrome paths
26 private void dfs(int node, int nodeMask) {
27 // Iterate for every edge connected to the current node
28 for (int[] edge : graph[node]) {
29 int child = edge[0], charBitmask = edge[1];
30 int xor = nodeMask ^ charBitmask;
31
32 // Update the answer by counting valid palindrome paths ending at the current node
33 answer += countMap.getOrDefault(xor, 0);
34 // Try flipping each bit to spot palindrome pairs
35 for (int k = 0; k < 26; ++k) {
36 answer += countMap.getOrDefault(xor ^ (1 << k), 0);
37 }
38 // Add the new XOR to the map with a count, if it is already present add to its count
39 countMap.merge(xor, 1, Integer::sum);
40 // Continue to DFS for the child node
41 dfs(child, xor);
42 }
43 }
44}
45
1class Solution {
2public:
3 long long countPalindromePaths(vector<int>& parent, string s) {
4 int nodeCount = parent.size(); // Get the number of nodes in the tree
5 vector<vector<pair<int, int>>> graph(nodeCount); // Create an adjacency list for the graph
6 unordered_map<int, int> countMap; // Store the count of each bitmask
7
8 countMap[0] = 1; // Initialize the count for an empty path to 1
9
10 // Build the graph from the parent relationships
11 for (int i = 1; i < nodeCount; ++i) {
12 int parentNode = parent[i];
13 // Each edge has a bitmask corresponding to the character of the child node
14 graph[parentNode].emplace_back(i, 1 << (s[i] - 'a'));
15 }
16
17 long long answer = 0; // Store the final count of palindromic paths
18 // Define a lambda function for depth-first search on the graph
19 function<void(int, int)> dfs = [&](int node, int bitmask) {
20 // Traverse all child nodes of the current node
21 for (auto [childNode, value] : graph[node]) {
22 int newXor = bitmask ^ value; // Compute new bitmask after including current node
23 answer += countMap[newXor]; // Increase count if the bitmask represents a palindrome so far
24
25 // Increase count for each palindromic path ending with one character difference
26 for (int k = 0; k < 26; ++k) {
27 answer += countMap[newXor ^ (1 << k)];
28 }
29
30 ++countMap[newXor]; // Mark the current path
31 dfs(childNode, newXor); // Recurse for child
32 }
33 };
34
35 dfs(0, 0); // Start DFS from the root node with an empty bitmask
36 return answer; // Return the final count of palindromic paths
37 }
38};
39
1function countPalindromePaths(parent: number[], charSequence: string): number {
2 const nodeCount = parent.length; // Total number of nodes
3 // Graph representation: An array of tuples for each node's children
4 const graph: [number, number][][] = Array.from({ length: nodeCount }, () => []);
5
6 // Construct the graph's adjacency list
7 for (let i = 1; i < nodeCount; ++i) {
8 const charBitMask = 1 << (charSequence.charCodeAt(i) - 97); // Bit-representation for character
9 graph[parent[i]].push([i, charBitMask]);
10 }
11
12 // Map to count occurrences of XOR values
13 const xorCount: Map<number, number> = new Map();
14 xorCount.set(0, 1); // Initial XOR count for root
15
16 let palindromePathsCount = 0; // Variable to track total palindrome paths found
17
18 // Recursive DFS to explore all paths
19 const dfs = (nodeIndex: number, xorValue: number): void => {
20 for (const [childIndex, charMask] of graph[nodeIndex]) {
21 const newXor = xorValue ^ charMask; // Calculate new XOR
22 palindromePathsCount += xorCount.get(newXor) || 0; // Increment paths count if a palindrome is found
23
24 // Check for possible single character difference palindromes
25 for (let k = 0; k < 26; ++k) {
26 palindromePathsCount += xorCount.get(newXor ^ (1 << k)) || 0;
27 }
28
29 // Update the count for the current XOR
30 xorCount.set(newXor, (xorCount.get(newXor) || 0) + 1);
31
32 // Continue DFS with the child node and new XOR
33 dfs(childIndex, newXor);
34 }
35 };
36
37 // Start DFS at the root with initial XOR of 0
38 dfs(0, 0);
39
40 // Return the total number of palindrome paths found
41 return palindromePathsCount;
42}
43
Time and Space Complexity
The given Python code defines a method countPalindromePaths
that returns the number of palindromic paths in a tree-like structure, where each node is labeled with a character from string s
. The tree structure is defined by the parent
list, where parent[i]
is the parent of the i-th
node.
Time Complexity
The time complexity of the code is O(N * 26)
, where N
is the number of nodes in the tree. Here's the breakdown:
- The
dfs
function is called recursively for each node exactly once, resulting inN
calls. - Inside the
dfs
function, for each nodei
, it iterates over all its children followed by a loop running 26 times (for each lowercase alphabet letter). - Every step inside the inner
for
loop has constant time operations except for the recursivedfs
call that happens once for each child node.
Therefore, we have a single dfs
call per node and a fixed number of additional iterations per node, resulting in a time complexity of O(N * 26)
.
Space Complexity
The space complexity of the code is O(N)
, where N
is the number of nodes in the tree. Here's why:
- The
g
dictionary andcnt
counter have at mostN
entries, as they store vertices and the number of times certain XOR values occur. - The
dfs
call stack can grow up toO(N)
in the case of a tree that degenerates into a linked list. - The other variables like
x
andans
use constant space.
As a result, the total space complexity is determined by the size of the tree and bookkeeping variables, leading to O(N)
.
In conclusion, the given Python code has a time complexity of O(N * 26)
and a space complexity of O(N)
.
Learn more about how to find time and space complexity quickly using problem constraints.
What's the output of running the following function using input [30, 20, 10, 100, 33, 12]
?
1def fun(arr: List[int]) -> List[int]:
2 import heapq
3 heapq.heapify(arr)
4 res = []
5 for i in range(3):
6 res.append(heapq.heappop(arr))
7 return res
8
1public static int[] fun(int[] arr) {
2 int[] res = new int[3];
3 PriorityQueue<Integer> heap = new PriorityQueue<>();
4 for (int i = 0; i < arr.length; i++) {
5 heap.add(arr[i]);
6 }
7 for (int i = 0; i < 3; i++) {
8 res[i] = heap.poll();
9 }
10 return res;
11}
12
1class HeapItem {
2 constructor(item, priority = item) {
3 this.item = item;
4 this.priority = priority;
5 }
6}
7
8class MinHeap {
9 constructor() {
10 this.heap = [];
11 }
12
13 push(node) {
14 // insert the new node at the end of the heap array
15 this.heap.push(node);
16 // find the correct position for the new node
17 this.bubble_up();
18 }
19
20 bubble_up() {
21 let index = this.heap.length - 1;
22
23 while (index > 0) {
24 const element = this.heap[index];
25 const parentIndex = Math.floor((index - 1) / 2);
26 const parent = this.heap[parentIndex];
27
28 if (parent.priority <= element.priority) break;
29 // if the parent is bigger than the child then swap the parent and child
30 this.heap[index] = parent;
31 this.heap[parentIndex] = element;
32 index = parentIndex;
33 }
34 }
35
36 pop() {
37 const min = this.heap[0];
38 this.heap[0] = this.heap[this.size() - 1];
39 this.heap.pop();
40 this.bubble_down();
41 return min;
42 }
43
44 bubble_down() {
45 let index = 0;
46 let min = index;
47 const n = this.heap.length;
48
49 while (index < n) {
50 const left = 2 * index + 1;
51 const right = left + 1;
52
53 if (left < n && this.heap[left].priority < this.heap[min].priority) {
54 min = left;
55 }
56 if (right < n && this.heap[right].priority < this.heap[min].priority) {
57 min = right;
58 }
59 if (min === index) break;
60 [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61 index = min;
62 }
63 }
64
65 peek() {
66 return this.heap[0];
67 }
68
69 size() {
70 return this.heap.length;
71 }
72}
73
74function fun(arr) {
75 const heap = new MinHeap();
76 for (const x of arr) {
77 heap.push(new HeapItem(x));
78 }
79 const res = [];
80 for (let i = 0; i < 3; i++) {
81 res.push(heap.pop().item);
82 }
83 return res;
84}
85
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