2689. Extract Kth Character From The Rope Tree

Problem Description

The given problem involves a special binary tree known as a "Rope" tree. In a Rope tree, each node can be classified as either a leaf node or an internal node:

  • A leaf node carries a non-empty string in its node.val property and has a node.len value of 0. It also does not have any children.
  • An internal node has an empty string for its node.val property and a node.len value greater than 0. This node type has at least one child, up to a maximum of two children.

For each node in the Rope tree, a string S[node] is defined:

  • If the node is a leaf node, S[node] equals node.val.
  • If the node is an internal node, S[node] is the concatenation of the strings S[node.left] and S[node.right], with the total length of S[node] being equal to node.len.

The problem requires us to find the k-th character in the string S[root], which represents the whole string constructed by the Rope tree starting from the root node.


To solve this problem, we start with a depth-first search (DFS) approach. By using DFS traversal, we can construct the strings for each node starting from the leaves up to the root. This is derived from the fact that an internal node's string S[node] is made by concatenating its left and right children's strings.

Here is a step-by-step explanation of the solution process:

  1. Our goal is to create a recursive function dfs that, given a node, returns the string represented by the subtree rooted at that node.
  2. If the current node is None, we return an empty string since there is no contribution to the parent string.
  3. If the current node is a leaf node (node.len == 0), we simply return the string node.val.
  4. For an internal node, however, we need to concatenate the results of recursively calling dfs on its left and right children. This gives us the entire string represented by the subtree rooted at that internal node.
  5. After calling dfs on the root node to obtain the full string, we can then index into this string to get the character at position k - 1 (since the problem uses 1-based indexing).

It is worth noting that while this approach is intuitive and straightforward, it may not be efficient for large trees since it may involve creating and concatenating very long strings. This solution does not account for cases where the search could be stopped early if k is within the string length of the left subtree, thus avoiding the recursion on the right subtree and saving time and memory. In practice, an optimized solution would take advantage of the information provided in node.len to traverse only the required part of the tree to find the k-th character.

Learn more about Tree and Depth-First Search patterns.

Solution Approach

The implementation of the solution is based on the Depth-First Search (DFS) algorithm. The process traverses the structure of the Rope tree to reconstruct the complete string that the tree represents. The data structure of concern here is the binary tree made up of Rope nodes.

The getKthCharacter method within the Solution class performs this operation:

  1. The method begins by defining a nested function, dfs, that will be used to explore the Rope tree. The dfs function will be called on each node of the tree.

  2. The dfs function checks if the current node is None, which is the base case for the recursion. When encountering a None node, it returns an empty string "".

  3. If the node is a leaf (node.len == 0), it means the node's own val is part of the final string. Therefore, dfs returns node.val directly.

  4. For internal nodes, dfs recursively calls itself on the node.left and node.right children. The function then concatenates the results of these calls to build up the string corresponding to the internal node. This is done using the + operator for concatenation.

  5. The getKthCharacter method initiates the DFS traversal by calling dfs(root), which returns the fully constructed string from the entire tree.

  6. Finally, since the problem requires the k-th character of the string S[root] and the k is 1-based, the method returns the character at the index k - 1.

Here is the key portion of the code, which explains the recursive traversal and concatenation:

1def getKthCharacter(self, root: Optional[object], k: int) -> str:
2    def dfs(root):
3        if root is None:
4            return ""
5        elif root.len == 0:
6            return root.val
7        else:
8            return dfs(root.left) + dfs(root.right)
10    return dfs(root)[k - 1]

The DFS algorithm is ideal for this situation because it allows us to build the string as we dive deeper into the tree and backtrack while maintaining the order of characters as they should appear in the final string. Furthermore, we rely on Python's inherent capabilities in handling strings to concatenate portions of the tree's string values effortlessly.

It's important to reiterate that this solution approach creates the entire string and then indexes into it, which can be highly inefficient if the tree represents a very large string. A more efficient approach would have used the node.len property to directly descend to the node containing the k-th character without constructing the entire string.

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Example Walkthrough

Let's walk through a small example to illustrate the solution approach for the given Rope tree problem. Consider the following Rope tree structure:

1         (internal: len=7)
2              /      \
3     (internal: len=3)  (leaf: val="de")
4           /     \
5(leaf: val="ab") (leaf: val="c")

Our goal is to find the k-th character in the string represented by this Rope tree. For this walkthrough, let’s find the 4th character (k=4).

Following the solution approach:

  1. We begin by invoking the dfs function on the root node (internal with len=7). Since this is an internal node, we need to proceed with its children.

  2. The dfs function is recursively called on the left child (internal with len=3). Again, since this is an internal node, we call dfs on its children.

  3. The dfs function is now called on the left most leaf node (val="ab"). As this is a leaf node, the dfs returns "ab".

  4. The dfs function next proceeds to the sibling of the previous leaf node, which is another leaf node (val="c"). Here, dfs returns "c".

  5. We now have results from the child nodes of the current internal node with len=3. We concatenate the strings "ab" and "c", making up "abc", and return it.

  6. Next, the dfs function is called on the right child of the root node, which is a leaf node (val="de"). Because it's a leaf, dfs returns "de".

  7. Now we are back at the root node, where we concatenate the results from both children, resulting in "abc" + "de" which equals "abcde".

  8. Finally, we look for the 4th character in the constructed string "abcde". The 4th character is d, and since we use 1-based indexing, we return "abcde"[3] which is d.

The complete string from the tree "abcde" is formed as expected, and the fourth character, d, is correctly identified. This example demonstrates the straightforward traversal and concatenation process defined by the solution approach.

Keep in mind that this example is sufficient because the tree is small. However, for larger trees, this approach might not be efficient due to the construction and concatenation of very long strings.

Solution Implementation

1# Definition for a rope tree node.
2class RopeTreeNode:
3    def __init__(self, length=0, value="", left=None, right=None):
4        self.length = length  # The length of the string represented by this node and its descendents
5        self.value = value    # The string value that this node contains
6        self.left = left      # The left child of this node
7        self.right = right    # The right child of this node
9class Solution:
10    def getKthCharacter(self, root: 'RopeTreeNode', k: int) -> str:
11        # Helper function to traverse the rope tree and concatenate the values.
12        def dfs(node):
13            # If the node is None, we are at a leaf, return an empty string
14            if node is None:
15                return ""
16            # If the length of the node is 0, it means it holds a string value (leaf node)
17            if node.length == 0:
18                return node.value
19            # Recursively get the value from left and right child nodes
20            return dfs(node.left) + dfs(node.right)
22        # Conduct a depth-first search to retrieve the full string, then return the k-th character (1-indexed)
23        return dfs(root)[k - 1]
25# Note that the 'Optional' type hint was removed as it was being used without an import from 'typing' and is not necessary here.
2 * Definition for a node in a rope tree data structure.
3 */
4class RopeTreeNode {
5    int length;
6    String value;
7    RopeTreeNode left;
8    RopeTreeNode right;
10    // Constructors for different scenarios
11    RopeTreeNode() {}
12    RopeTreeNode(String val) {
13        this.length = val.length();
14        this.value = val;
15    }
16    RopeTreeNode(int length) {
17        this.length = length;
18        this.value = "";
19    }
20    RopeTreeNode(int length, RopeTreeNode left, RopeTreeNode right) {
21        this.length = length;
22        this.value = "";
23        this.left = left;
24        this.right = right;
25    }
28class Solution {
29    /**
30     * Gets the kth character of the string represented by the rope tree.
31     *
32     * @param root the root node of the rope tree.
33     * @param k the position of the character to find (1-indexed).
34     * @return the kth character in the rope tree.
35     */
36    public char getKthCharacter(RopeTreeNode root, int k) {
37        // Since k is 1-indexed, we subtract 1 to make it 0-indexed.
38        return traverse(root).charAt(k - 1);
39    }
41    /**
42     * DFS traversal to concatenate strings from the rope tree nodes.
43     *
44     * @param node the current node being traversed.
45     * @return the concatenated string from this subtree.
46     */
47    private String traverse(RopeTreeNode node) {
48        // If the node is null, return an empty string.
49        if (node == null) {
50            return "";
51        }
52        // If the node contains a value, return that value.
53        if (!node.value.isEmpty()) {
54            return node.value;
55        }
56        // Recursively traverse the left subtree.
57        String leftString = traverse(node.left);
58        // Recursively traverse the right subtree.
59        String rightString = traverse(node.right);
60        // Concatenate the results from left and right subtrees and return.
61        return leftString + rightString;
62    }
1#include <string>
2#include <functional>
4// Definition for a node of the rope tree data structure.
5struct RopeTreeNode {
6    int len;  // Length of the string represented by subtree rooted at this node
7    std::string val;  // The string value of the node (for leaf nodes)
8    RopeTreeNode *left;  // Pointer to the left child node
9    RopeTreeNode *right;  // Pointer to the right child node
11    // Constructors for the different scenarios when creating a node
12    RopeTreeNode() : len(0), val(""), left(nullptr), right(nullptr) {}
13    explicit RopeTreeNode(std::string s) : len(0), val(std::move(s)), left(nullptr), right(nullptr) {}
14    explicit RopeTreeNode(int x) : len(x), val(""), left(nullptr), right(nullptr) {}
15    RopeTreeNode(int x, RopeTreeNode *left, RopeTreeNode *right) : len(x), val(""), left(left), right(right) {}
18class Solution {
20    // Function to get the kth character in a rope tree
21    char getKthCharacter(RopeTreeNode* root, int k) {
22        // Define a lambda function 'dfs' to perform a depth-first search of the rope tree
23        std::function<std::string(RopeTreeNode*)> dfs = [&](RopeTreeNode* node) -> std::string {
24            if (node == nullptr) {
25                // If node is null, return an empty string
26                return "";
27            }
28            if (node->len == 0) {
29                // If length is zero, we're at a leaf node, return the node's value
30                return node->val;
31            }
33            // Recursively call 'dfs' on the left and right children
34            std::string leftString = dfs(node->left);
35            std::string rightString = dfs(node->right);
37            // Concatenate and return the strings from the left and right subtrees
38            return leftString + rightString;
39        };
41        // Call our 'dfs' function and get the string representing the entire tree
42        std::string ropeString = dfs(root);
44        // Return the kth-1 character as we want to get 0-index based character
45        return ropeString[k - 1];
46    }
1// Structure definition for a node of the Rope tree data structure.
2interface RopeTreeNode {
3    len: number;  // Length of the string represented by the subtree rooted at this node
4    val: string;  // The string value of the node (for leaf nodes)
5    left: RopeTreeNode | null;  // Reference to the left child node
6    right: RopeTreeNode | null;  // Reference to the right child node
9// Function to create a rope tree node given a string
10function createNodeFromString(value: string): RopeTreeNode {
11    return {
12        len: value.length,
13        val: value,
14        left: null,
15        right: null
16    };
19// Function to create a rope tree node given a length and optional children
20function createNodeFromLengthAndChildren(length: number, left?: RopeTreeNode, right?: RopeTreeNode): RopeTreeNode {
21    return {
22        len: length,
23        val: '',
24        left: left || null,
25        right: right || null
26    };
29// Function to get the kth character in a rope tree
30function getKthCharacter(root: RopeTreeNode, k: number): string {
31    // Define a recursive depth-first search function for the rope tree
32    const dfs = (node: RopeTreeNode | null): string => {
33        if (!node) {
34            // If node is null, return an empty string
35            return '';
36        }
37        if (node.len === 0 || !node.left && !node.right) {
38            // If length is zero, we're at a leaf node, return the node's value
39            return node.val;
40        }
42        // Concatenate and return the strings from the left and right subtrees
43        // Only traverse the subtrees if we haven't already found the kth character
44        const leftString = node.left ? dfs(node.left) : '';
45        const rightString = (leftString.length < k && node.right) ? dfs(node.right) : '';
47        // Use the length of the left string and node's length to determine if k is in the left subtree
48        if(k <= leftString.length) {
49            return leftString;
50        } else {
51            // Reduce k by the length of the left string as k is in the right subtree
52            k -= leftString.length;
53            return rightString;
54        }
55    };
57    // Start depth-first search on the root node of the rope tree
58    const ropeString = dfs(root);
60    // Return the kth character (0-index based)
61    return ropeString[k - 1];

Time and Space Complexity

Time Complexity

The provided code uses a Depth-First Search (DFS) approach on a rope tree to find the kth character by concatenating the values of the nodes. The function dfs() is called recursively for each node in the rope tree.

Assuming there are n nodes in the tree, the worst-case scenario is that the tree is imbalanced and takes the form of a linked list. In that case, every call to dfs() would involve a concatenation operation which has a complexity of O(k), where k is the length of the resulting string up to that point. Thus, the overall time complexity for this worst-case scenario would be O(nk).

However, if the tree is balanced and each node has approximately half of the total characters of its parent, the string concatenation work done at each level would amount to O(n) work in total per level since it divides the work by two each time. Considering the height of the tree to be h, and that a balanced binary tree has h = log(n), the time complexity would resemble O(n log(n)).

In the average case, accounting for various possible tree structures, we can expect the time complexity to fall between O(n log(n)) and O(nk).

Space Complexity

Space complexity is primarily affected by the recursive call stack and the space required for the string concatenations. Due to the recursive DFS approach, in the worst case, the height of the call stack would be h, which is n for a degenerate tree (linked list form) and log(n) for a perfectly balanced tree, so the space complexity for the call stack would be O(n) in the worst case, and O(log(n)) in the best case.

Additionally, since each recursive call constructs a string that includes all characters of its subtrees, the function could hold up to O(n) worth of concatenated characters in memory. In the worst case, this would result in a space complexity of O(n + nk) = O(nk).

In the best-case scenario (balanced tree), the space complexity would be O(n + log(n)) = O(n) since string concatenation space would be distributed across the balanced levels of the tree and the highest memory requirement would be in the final string concatenation.

Hence the space complexity varies from O(n) in the best case to O(nk) in the worst case.

Learn more about how to find time and space complexity quickly using problem constraints.

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