How do you solve Find Subtree Sizes After Changes on LeetCode?

Find Subtree Sizes After Changes (LeetCode #3331) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding tree traversal is crucial for solving problems involving parent-child relationships.

What is the brute force solution for Find Subtree Sizes After Changes?

The brute force approach for Find Subtree Sizes After Changes is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check for each node if there exists an ancestor with the same character. This involves traversing the tree for each node, leading to a lot of redundant checks. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Subtree Sizes After Changes?

Find Subtree Sizes After Changes uses the following concepts: Array, Hash Table, String, Tree, Depth-First Search. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Subtree Sizes After Changes?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as trees with only one node or all nodes having the same character. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving Find Subtree Sizes After Changes?

Not properly handling the case where a character has not been seen before. Failing to reset the last seen character after traversing back up the tree.

#3331

Find Subtree Sizes After Changes

Medium

ArrayHash TableStringTreeDepth-First SearchHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

This approach uses a depth-first search (DFS) to traverse the tree while keeping track of the most recent ancestor for each character. This allows us to efficiently find the closest ancestor with the same character without redundant checks.

⚙️

Algorithm

3 steps

1Step 1: Initialize a map to store the most recent node for each character as we perform DFS.
2Step 2: During DFS, for each node, check if the character has been seen before. If so, update its parent accordingly.
3Step 3: After making all parent changes, perform another DFS to calculate the size of the subtree for each node.

solution.py38 lines

1def findSubtreeSizes(parent, s):
2    n = len(parent)
3    new_parent = parent[:]
4    last_seen = {}
5    def dfs(node):
6        nonlocal last_seen
7
8        # Save the current character and its node
9        char = s[node]
10        prev = last_seen.get(char, -1)
11
12        # Update the last seen character
13        last_seen[char] = node
14
15        for child in range(n):
16            if parent[child] == node:
17                # If the child has the same character, update its parent
18                if s[child] == char:
19                    new_parent[child] = prev if prev != -1 else node
20                dfs(child)
21
22        # Restore the last seen character
23        if prev != -1:
24            last_seen[char] = prev
25        else:
26            del last_seen[char]
27
28    dfs(0)
29    subtree_size = [0] * n
30    def calculate_size(node):
31        size = 1
32        for child in range(n):
33            if new_parent[child] == node:
34                size += calculate_size(child)
35        subtree_size[node] = size
36        return size
37    calculate_size(0)
38    return subtree_size

ℹ

Complexity note: The time complexity is O(n) because we traverse each node once during the DFS, and we use a map to store the last seen nodes for each character, which takes linear space.

1Understanding tree traversal is crucial for solving problems involving parent-child relationships.
2Using a map to track the last seen nodes for characters can significantly reduce the complexity of finding ancestors.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.