How do you solve Number of Nodes in the Sub-Tree With the Same Label on LeetCode?

Number of Nodes in the Sub-Tree With the Same Label (LeetCode #1519) 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: Using DFS allows us to efficiently count nodes in subtrees without redundant calculations.

What is the time complexity of Number of Nodes in the Sub-Tree With the Same Label?

The optimal solution for Number of Nodes in the Sub-Tree With the Same Label runs in O(n) time and uses O(n) space. The complexity is O(n) because we perform a single DFS traversal of the tree, visiting each node once, and using additional space for the count of labels.

What is the brute force solution for Number of Nodes in the Sub-Tree With the Same Label?

The brute force approach for Number of Nodes in the Sub-Tree With the Same Label is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves traversing each node and counting the nodes in its subtree that share the same label. This is straightforward but inefficient, as it checks every node for every subtree. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Nodes in the Sub-Tree With the Same Label?

Number of Nodes in the Sub-Tree With the Same Label uses the following concepts: Hash Table, Tree, Depth-First Search, Breadth-First Search, Counting. The recommended patterns to study are: Depth-First Search (DFS), Tree Traversal, Counting with Arrays.

What are the interview tips for Number of Nodes in the Sub-Tree With the Same Label?

Always clarify the tree structure and ensure you understand how to traverse it. Discuss your approach with the interviewer before coding to ensure you're on the right track. Consider edge cases, such as trees with only one node or all nodes having the same label.

What are common mistakes when solving Number of Nodes in the Sub-Tree With the Same Label?

Not properly handling the parent node during DFS, leading to incorrect counts. Failing to initialize the count array correctly can result in incorrect results.

#1519

Number of Nodes in the Sub-Tree With the Same Label

Medium

Hash TableTreeDepth-First SearchBreadth-First SearchCountingDepth-First Search (DFS)Tree TraversalCounting with Arrays

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)

The optimal solution uses a single DFS traversal to count labels in the subtree while returning counts up the tree. This avoids redundant calculations and efficiently aggregates results.

⚙️

Algorithm

4 steps

1Step 1: Build the graph from the edges.
2Step 2: Perform a DFS starting from the root node.
3Step 3: Maintain a count array for labels in the subtree.
4Step 4: For each node, update the result based on the counts of its label.

solution.py22 lines

1def countSubtrees(n, edges, labels):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    for a, b in edges:
5        graph[a].append(b)
6        graph[b].append(a)
7
8    result = [0] * n
9    def dfs(node, parent):
10        count = [0] * 26
11        count[ord(labels[node]) - ord('a')] += 1
12        for neighbor in graph[node]:
13            if neighbor != parent:
14                child_count = dfs(neighbor, node)
15                for i in range(26):
16                    count[i] += child_count[i]
17        result[node] = count[ord(labels[node]) - ord('a')]
18        return count
19
20    dfs(0, -1)
21    return result
22

ℹ

Complexity note: The complexity is O(n) because we perform a single DFS traversal of the tree, visiting each node once, and using additional space for the count of labels.

1Using DFS allows us to efficiently count nodes in subtrees without redundant calculations.
2Maintaining a count array for labels helps aggregate results as we backtrack through the tree.

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