Given a tree encoded as a string, for every node, compute the sum of its subtree values and return the sums tree encoded as a string.
For example, the following tree is given as an encoded string "1{2{3,4},5{6}}" :
1 / \ 2 5 / \ \ 3 4 6After computing the sums, the tree will look like the following:
21 / \ 9 11 / \ \ 3 4 6Therefore the output will be "21{9{3,4},11{6}}"
Example 1:
Input: "1{2{3,4},5{6}}" Output: "21{9{3,4},11{6}}"
Example 2:
Input: "1{1,1{1,1,1},1}" Output: "7{1,4{1,1,1},1}"
The core challenge of this problem is to parse the encoded string representation of a tree, compute the sum of values for each subtree, and then re-encode the tree with these sums. This problem is significant in scenarios where hierarchical data needs to be processed and summarized, such as in organizational structures or file systems.
Potential pitfalls include correctly parsing the string representation and ensuring that the sums are computed accurately for each subtree.
To solve this problem, we need to:
A naive solution might involve manually parsing the string and using nested loops to compute sums, but this would be inefficient and error-prone.
An optimized approach involves using a recursive function to parse the string and compute sums simultaneously. This ensures that each node and its subtree are processed in a single pass.
Here is a step-by-step breakdown of the algorithm:
import java.util.*;
public class EncodedTreeSums {
// Helper class to represent a tree node
static class TreeNode {
int value;
List children;
TreeNode(int value) {
this.value = value;
this.children = new ArrayList<>();
}
}
// Function to parse the encoded string and build the tree
private static TreeNode parseTree(String s, int[] index) {
int value = 0;
while (index[0] < s.length() && Character.isDigit(s.charAt(index[0]))) {
value = value * 10 + (s.charAt(index[0]) - '0');
index[0]++;
}
TreeNode node = new TreeNode(value);
if (index[0] < s.length() && s.charAt(index[0]) == '{') {
index[0]++; // skip '{'
while (s.charAt(index[0]) != '}') {
node.children.add(parseTree(s, index));
if (s.charAt(index[0]) == ',') {
index[0]++; // skip ','
}
}
index[0]++; // skip '}'
}
return node;
}
// Function to compute subtree sums and re-encode the tree
private static int computeSums(TreeNode node, StringBuilder sb) {
int sum = node.value;
sb.append(node.value);
if (!node.children.isEmpty()) {
sb.append('{');
for (int i = 0; i < node.children.size(); i++) {
if (i > 0) {
sb.append(',');
}
sum += computeSums(node.children.get(i), sb);
}
sb.append('}');
}
node.value = sum;
return sum;
}
// Main function to process the encoded tree string
public static String encodedTreeSums(String s) {
int[] index = {0};
TreeNode root = parseTree(s, index);
StringBuilder sb = new StringBuilder();
computeSums(root, sb);
return sb.toString();
}
public static void main(String[] args) {
String input1 = "1{2{3,4},5{6}}";
String input2 = "1{1,1{1,1,1},1}";
System.out.println(encodedTreeSums(input1)); // Output: "21{9{3,4},11{6}}"
System.out.println(encodedTreeSums(input2)); // Output: "7{1,4{1,1,1},1}"
}
}
The time complexity of this approach is O(n), where n is the number of nodes in the tree. This is because each node is processed exactly once. The space complexity is also O(n) due to the recursion stack and the storage of the tree structure.
Potential edge cases include:
To test the solution comprehensively, consider the following test cases:
When approaching such problems, it is helpful to:
Understanding and solving problems involving tree structures is crucial for many applications in computer science. By practicing such problems, you can improve your problem-solving skills and gain a deeper understanding of data structures and algorithms.
For further reading and practice, consider the following resources: