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

The optimal solution has a time complexity of O(n).

What data structures are used to solve Number of Nodes in the Sub-Tree With the Same Label?

This problem uses Hash Table, Tree, Depth-First Search, Breadth-First Search, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Nodes in the Sub-Tree With the Same Label?

Insight 1: The core idea is to use Depth-First Search (DFS) to traverse the tree and compute the counts recursively. Each node's subtree count depends on the counts from its children. Insight 2: A key data structure for storing label counts within the subtree is an array of size 26 (one for each lowercase English letter). This allows efficient aggregation of counts from child subtrees.

Which companies ask LeetCode 1519. Number of Nodes in the Sub-Tree With the Same Label in interviews?

This problem is frequently asked at Samsung.

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Number of Nodes in the Sub-Tree With the Same Label - LeetCode 1519 Solution

Number of Nodes in the Sub-Tree With the Same Label - Complete Solution Guide

Number of Nodes in the Sub-Tree With the Same Label is LeetCode problem 1519, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a tree (i.e. a connected, undirected graph that has no cycles) consisting of n nodes numbered from 0 to n - 1 and exactly n - 1 edges . The root of the tree is the node 0 , and each node of the tree has a label which is a lower-case character given in the string labels (i.e. The node with the number i has the label labels[i] ). The edges array is given on the form edges[i] = [a i , b i ] , which means there is an edge between nodes a i and b i in the tree. Return an array of size n

Detailed Explanation

The problem requires us to traverse a tree and, for each node, count the number of nodes in its subtree (including itself) that have the same label as that node. The tree is given by the number of nodes `n`, a list of edges connecting the nodes, and a string `labels` where `labels[i]` represents the label of node `i`. The task is to return an array `ans` where `ans[i]` is the count of nodes in the subtree rooted at node `i` that have the same label as node `i`.

Solution Approach

The provided solution uses a Depth-First Search (DFS) algorithm to traverse the tree. For each node, it recursively visits its children, collecting the label counts from their subtrees. It then adds the node's own label to the count and stores the count of the matching label in the result array. The DFS function returns an array representing the label counts for the entire subtree rooted at that node.

Step-by-Step Algorithm

Step 1: Build an adjacency list representation of the tree from the given `edges`. The adjacency list stores the neighbors of each node.
Step 2: Initialize an array `ans` of size `n` to store the final results. All elements are initially 0.
Step 3: Define a recursive DFS function that takes the current `node` and its `parent` as input.
Step 4: Inside the DFS function, create an array `node_counts` of size 26 to store the counts of each label in the subtree rooted at the current `node`. Initialize all counts to 0.
Step 5: Iterate through the children of the current `node`. If a child is not the parent (to avoid cycles), recursively call the DFS function on the child. Add the label counts returned by the child DFS to `node_counts`.
Step 6: Increment the count of the current `node`'s label in the `node_counts` array.
Step 7: Store the count of the current `node`'s label in the `ans` array at the index corresponding to the current `node`.
Step 8: Return the `node_counts` array.
Step 9: Call the DFS function starting from the root node (node 0) with a parent of -1.
Step 10: Return the `ans` array.

Key Insights

Insight 1: The core idea is to use Depth-First Search (DFS) to traverse the tree and compute the counts recursively. Each node's subtree count depends on the counts from its children.
Insight 2: A key data structure for storing label counts within the subtree is an array of size 26 (one for each lowercase English letter). This allows efficient aggregation of counts from child subtrees.
Insight 3: Efficiently build an adjacency list representation of the tree using the given edges. This allows us to easily iterate through the neighbors of any given node.
Insight 4: Handle edge cases such as a single node, disconnected subgraphs, or invalid input, although the constraints guarantee a connected graph.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Tree, Depth-First Search, Breadth-First Search, Counting.

Companies

Asked at: Samsung.

Number of Nodes in the Sub-Tree With the Same Label | LeetCode 1519

Medium

Hash Table Tree Depth-First Search Breadth-First Search Counting

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int[] countSubTrees(int n, int[][] edges, String labels) {
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            adj.get(v).add(u);
        }

        int[] ans = new int[n];

        class Helper {
            int[] dfs(int node, int parent) {
                int[] node_counts = new int[26];

                for (int child : adj.get(node)) {
                    if (child != parent) {
                        int[] child_counts = dfs(child, node);
                        for (int i = 0; i < 26; i++) {
                            node_counts[i] += child_counts[i];
                        }
                    }
                }

                int label_index = labels.charAt(node) - 'a';
                node_counts[label_index] += 1;
                ans[node] = node_counts[label_index];

                return node_counts;
            }
        }

        Helper helper = new Helper();
        helper.dfs(0, -1);
        return ans;
    }
}

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Problem Info

Description

Problem 1519: Number of Nodes in the Sub-Tree With the Same Label - Detailed explanation and solution approaches.

Topics

Hash Table Tree Depth-First Search Breadth-First Search Counting

Companies

Samsung

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1519

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Build an adjacency list representation of the tree from the given `edges`. The adjacency list stores the neighbors of each node.
2
Step 2: Initialize an array `ans` of size `n` to store the final results. All elements are initially 0.
3
Step 3: Define a recursive DFS function that takes the current `node` and its `parent` as input.
4
Step 4: Inside the DFS function, create an array `node_counts` of size 26 to store the counts of each label in the subtree rooted at the current `node`. Initialize all counts to 0.
5
Step 5: Iterate through the children of the current `node`. If a child is not the parent (to avoid cycles), recursively call the DFS function on the child. Add the label counts returned by the child DFS to `node_counts`.
6
Step 6: Increment the count of the current `node`'s label in the `node_counts` array.
7
Step 7: Store the count of the current `node`'s label in the `ans` array at the index corresponding to the current `node`.
8
Step 8: Return the `node_counts` array.
9
Step 9: Call the DFS function starting from the root node (node 0) with a parent of -1.
10
Step 10: Return the `ans` array.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The DFS algorithm visits each node once, and the operations within each node (building the counts array and summing the child arrays) take O(1) or O(26) which is constant time. Therefore, the overall time complexity is O(n). The adjacency list takes O(n) space, and the recursion stack can go up to O(n) in the worst case (skewed tree). The counts array takes O(1) space because it is a fixed size of 26. The answer array takes O(n). Thus the total space is O(n).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative DFS

Implement DFS iteratively using a stack to avoid recursion overhead. This can sometimes be more space-efficient for very deep trees.

Time: O(n)Space: O(n)

Trade-offs: Can reduce recursion stack space, but might be slightly more complex to implement compared to recursive DFS.

✨

Key Insights

1Insight 1: The core idea is to use Depth-First Search (DFS) to traverse the tree and compute the counts recursively. Each node's subtree count depends on the counts from its children.
2Insight 2: A key data structure for storing label counts within the subtree is an array of size 26 (one for each lowercase English letter). This allows efficient aggregation of counts from child subtrees.
3Insight 3: Efficiently build an adjacency list representation of the tree using the given edges. This allows us to easily iterate through the neighbors of any given node.
4Insight 4: Handle edge cases such as a single node, disconnected subgraphs, or invalid input, although the constraints guarantee a connected graph.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Single node tree (n=1). The algorithm will correctly return [1] because the DFS will be called once, and the label count for that node will be 1.
Edge case 2: Skewed tree. The DFS will still work correctly, although the recursion stack can become deep, contributing to the space complexity, it is still O(n).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the parent node correctly, leading to cycles in the DFS and incorrect counts.

✗Common mistake 2: Forgetting to add the current node's label count to the `node_counts` array before returning.

✗Common mistake 3: Using inefficient data structures for the adjacency list or label counts, leading to higher time complexity.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing social networks. Nodes could represent users, edges represent connections, and labels represent interests. The algorithm can be used to find the number of users with similar interests within a certain user's network.

🏦Real-world scenario 2: File system analysis. Nodes represent directories/files, edges represent the directory structure, and labels represent file types. The algorithm could determine how many files of the same type exist within a directory's subdirectories.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 19, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Number of Nodes in the Sub-Tree With the Same Label"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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