Extract Kth Character From The Rope Tree - Practice Coding Problems

Extract Kth Character From The Rope Tree - Problem

Imagine you have a special binary tree called a Rope Tree that efficiently stores and manages large strings. This data structure is commonly used in text editors and programming environments to handle massive documents without loading everything into memory at once.

You're given the root of a rope binary tree and an integer k. Each node has additional properties beyond the typical left/right children:

node.val: A string containing only lowercase English letters (possibly empty)
node.len: A non-negative integer representing the total length

There are two types of nodes:

Leaf nodes: No children, node.len = 0, and node.val contains the actual string data
Internal nodes: Have at least one child, node.len > 0, and node.val is empty

The string represented by any node S[node] is defined as:

If it's a leaf: S[node] = node.val
If it's internal: S[node] = concat(S[node.left], S[node.right])

Your task: Return the k-th character (1-indexed) from the string S[root] without actually constructing the entire string.

Input & Output

example_1.py — Basic Rope Tree

$ Input: root = [Node("", Node("hello"), Node("world"))], k = 6

› Output: "w"

💡 Note: The rope tree represents the string "helloworld". The 6th character (1-indexed) is 'w' from the beginning of "world".

example_2.py — Complex Tree Structure

$ Input: root = [Node("", Node("", Node("ab"), Node("cd")), Node("ef"))], k = 4

› Output: "d"

💡 Note: The tree represents "abcdef". Left subtree gives "abcd", right subtree gives "ef". The 4th character is 'd'.

example_3.py — Single Character

$ Input: root = [Node("x")], k = 1

› Output: "x"

💡 Note: Simple case with just one leaf node containing "x". The 1st character is 'x'.

Constraints

The number of nodes in the tree is in the range [1, 10³]
0 ≤ node.val.length ≤ 50
node.val contains only lowercase English letters
1 ≤ k ≤ total length of the rope tree string
For leaf nodes: node.len == 0 and node.val is non-empty
For internal nodes: node.len > 0 and node.val is empty

Visualization

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Understanding the Visualization

Start at Root

Begin navigation at root node, note total length information

Compare Lengths

Check if target position k falls in left subtree by comparing with left subtree length

Choose Direction

Go left if k ≤ left_length, otherwise go right with adjusted k value

Reach Destination

Continue until reaching leaf node, then return character at final position

Key Takeaway

🎯 Key Insight: By utilizing the length information stored in internal nodes, we can perform binary search navigation through the rope tree, achieving optimal O(h) time complexity instead of O(n) string construction.

Asked in

G Google 25 a Amazon 18 ⊞ Microsoft 15 f Meta 12

The optimal solution uses binary search on the rope tree structure, leveraging the length information in internal nodes to navigate directly to the target character. Instead of building the complete string, we compare k with left subtree lengths to determine which branch contains our target, achieving O(h) time complexity where h is tree height.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Build Full String)	O(n)	O(m + h)	Traverse entire tree to build complete string, then return k-th character
Binary Search on Tree (Optimal)	O(h)	O(h)	Use node.len values to binary search through the rope tree structure

Brute Force (Build Full String) — Algorithm Steps

Perform DFS traversal of the entire tree
For each leaf node encountered, append node.val to result string
For internal nodes, recursively process left subtree then right subtree
Once complete string is built, return character at index k-1
Handle edge cases where k is out of bounds

Visualization

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Step-by-Step Walkthrough

Start DFS

Begin depth-first traversal from root node

Visit Leaves

Collect string values from all leaf nodes in order

Concatenate

Build complete string by concatenating all leaf values

Index Access

Return character at position k-1 from complete string

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

// Definition for a rope tree node.
struct RopeTreeNode {
    char* val;
    struct RopeTreeNode* left;
    struct RopeTreeNode* right;
    int len;
};

char* buildString(struct RopeTreeNode* node) {
    if (!node) {
        char* empty = malloc(1);
        empty[0] = '\0';
        return empty;
    }
    
    // Leaf node - return copy of the actual string
    if (node->len == 0) {
        char* result = malloc(strlen(node->val) + 1);
        strcpy(result, node->val);
        return result;
    }
    
    // Internal node - concatenate left and right subtrees
    char* leftStr = buildString(node->left);
    char* rightStr = buildString(node->right);
    
    char* result = malloc(strlen(leftStr) + strlen(rightStr) + 1);
    strcpy(result, leftStr);
    strcat(result, rightStr);
    
    free(leftStr);
    free(rightStr);
    
    return result;
}

char getKthCharacter(struct RopeTreeNode* root, int k) {
    char* completeString = buildString(root);
    char result = (k <= strlen(completeString)) ? completeString[k-1] : ' ';
    free(completeString);
    return result;
}

int main() {
    int k;
    scanf("%d", &k);
    
    // Create demo rope tree
    struct RopeTreeNode* leaf1 = malloc(sizeof(struct RopeTreeNode));
    leaf1->val = "hello";
    leaf1->left = leaf1->right = NULL;
    leaf1->len = 0;
    
    struct RopeTreeNode* leaf2 = malloc(sizeof(struct RopeTreeNode));
    leaf2->val = "world";
    leaf2->left = leaf2->right = NULL;
    leaf2->len = 0;
    
    struct RopeTreeNode* root = malloc(sizeof(struct RopeTreeNode));
    root->val = "";
    root->left = leaf1;
    root->right = leaf2;
    root->len = 10;
    
    printf("%c\n", getKthCharacter(root, k));
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Must visit every node in the tree to build complete string, where n is total number of nodes

✓ Linear Growth

Space Complexity

O(m + h)

O(m) for storing the complete string where m is total length, plus O(h) for recursion stack depth h

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Build Full String) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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