What is the time complexity of Count Nodes With the Highest Score?

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

What data structures are used to solve Count Nodes With the Highest Score?

This problem uses Array, Tree, Depth-First Search, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Nodes With the Highest Score?

Insight 1: The key observation is that removing a node splits the tree into multiple subtrees. These subtrees are the left subtree, the right subtree (if they exist), and the subtree containing the parent of the removed node. If a subtree does not exist, its size is implicitly 1 in the score calculation. Insight 2: Depth-First Search (DFS) is the most suitable algorithm to traverse the tree and calculate the size of each subtree efficiently.

Which companies ask LeetCode 2049. Count Nodes With the Highest Score in interviews?

This problem is frequently asked at DoorDash, Visa.

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Count Nodes With the Highest Score - LeetCode 2049 Solution

Count Nodes With the Highest Score - Complete Solution Guide

Count Nodes With the Highest Score is LeetCode problem 2049, 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

There is a binary tree rooted at 0 consisting of n nodes. The nodes are labeled from 0 to n - 1 . You are given a 0-indexed integer array parents representing the tree, where parents[i] is the parent of node i . Since node 0 is the root, parents[0] == -1 . Each node has a score . To find the score of a node, consider if the node and the edges connected to it were removed . The tree would become one or more non-empty subtrees. The size of a subtree is the number of the nodes in it. The score of t

Detailed Explanation

The problem asks us to find the number of nodes in a binary tree that have the highest score. The tree is represented by a `parents` array, where `parents[i]` is the parent of node `i`. The root node has `parents[0] == -1`. The score of a node is calculated as the product of the sizes of the subtrees formed when that node and its connecting edges are removed from the tree. The goal is to return the count of nodes with the maximum calculated score.

Solution Approach

The provided solution uses Depth-First Search (DFS) to traverse the binary tree and calculate the score for each node. For each node, the DFS function calculates the sizes of its left and right subtrees. It then calculates the score by multiplying the sizes of the left subtree, the right subtree, and the 'parent subtree' (the remaining part of the tree when the node is removed). During the DFS traversal, the solution keeps track of the maximum score encountered so far and the count of nodes with that maximum score. After traversing all nodes, the solution returns the count of nodes with the maximum score.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the binary tree from the `parents` array. The adjacency list `adj[i]` stores the children of node `i`.
Step 2: Initialize `max_score` to 0 and `count` to 0. These variables will store the highest score encountered and the number of nodes with that score, respectively.
Step 3: Define a recursive DFS function `dfs(u)` that calculates the size of the subtree rooted at node `u`. The `dfs` function does the following:
Step 4: Initialize `size` to 1 (the size of the current node `u`) and `score_prod` to 1.
Step 5: Iterate through the children `v` of node `u` in the adjacency list `adj[u]`.
Step 6: Recursively call `dfs(v)` to get the size `child_size` of the subtree rooted at child `v`.
Step 7: Update `size` by adding `child_size` to it, and update `score_prod` by multiplying it with `child_size`.
Step 8: Calculate the size of the 'parent subtree' as `parent_part_size = n - size`.
Step 9: Calculate the score of the current node `u` as `current_score = score_prod * max(1, parent_part_size)`. We use `max(1, parent_part_size)` to ensure that if `parent_part_size` is 0 (meaning `u` is the root), we multiply by 1 instead of 0.
Step 10: Update `max_score` and `count`: if `current_score` is greater than `max_score`, update `max_score` to `current_score` and reset `count` to 1. If `current_score` is equal to `max_score`, increment `count`.
Step 11: Return the size of the subtree rooted at `u`.
Step 12: Call the `dfs` function starting from the root node (node 0).
Step 13: Return the final value of `count`.

Key Insights

Insight 1: The key observation is that removing a node splits the tree into multiple subtrees. These subtrees are the left subtree, the right subtree (if they exist), and the subtree containing the parent of the removed node. If a subtree does not exist, its size is implicitly 1 in the score calculation.
Insight 2: Depth-First Search (DFS) is the most suitable algorithm to traverse the tree and calculate the size of each subtree efficiently.
Insight 3: Since the number of nodes can be up to 10^5, we need to use `long` data type to store the scores, as the product of subtree sizes can exceed the range of `int`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Tree, Depth-First Search, Binary Tree.

Companies

Asked at: DoorDash, Visa.

Count Nodes With the Highest Score | LeetCode 2049

Medium

Array Tree Depth-First Search Binary Tree

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

Solution Code

Available in:

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

class Solution {
    public int countHighestScoreNodes(int[] parents) {
        int n = parents.length;
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int i = 0; i < n; i++) {
            int p = parents[i];
            if (p != -1) {
                adj.get(p).add(i);
            }
        }

        long max_score = 0;
        int count = 0;

        class DFSResult {
            long max_score;
            int count;

            public DFSResult(long max_score, int count) {
                this.max_score = max_score;
                this.count = count;
            }
        }


        class DFSHelper {
            long max_score;
            int count;
            DFSHelper(long max_score, int count){
                this.max_score = max_score;
                this.count = count;
            }

            public int dfs(int u, List<List<Integer>> adj, int n) {

                int size = 1;
                long score_prod = 1;

                for (int v : adj.get(u)) {
                    int child_size = dfs(v, adj, n);
                    size += child_size;
                    score_prod *= child_size;
                }

                int parent_part_size = n - size;

                long current_score = score_prod * Math.max(1, parent_part_size);

                if (current_score > max_score) {
                    max_score = current_score;
                    count = 1;
                } else if (current_score == max_score) {
                    count += 1;
                }

                return size;
            }
        }
        DFSHelper helper = new DFSHelper(0,0);
        helper.dfs(0, adj, n);


        return helper.count;
    }
}

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Problem Info

Description

Problem 2049: Count Nodes With the Highest Score - Detailed explanation and solution approaches.

Topics

Array Tree Depth-First Search Binary Tree

Companies

DoorDashVisa

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2049

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

13 steps

1
Step 1: Build the adjacency list representation of the binary tree from the `parents` array. The adjacency list `adj[i]` stores the children of node `i`.
2
Step 2: Initialize `max_score` to 0 and `count` to 0. These variables will store the highest score encountered and the number of nodes with that score, respectively.
3
Step 3: Define a recursive DFS function `dfs(u)` that calculates the size of the subtree rooted at node `u`. The `dfs` function does the following:
4
Step 4: Initialize `size` to 1 (the size of the current node `u`) and `score_prod` to 1.
5
Step 5: Iterate through the children `v` of node `u` in the adjacency list `adj[u]`.
6
Step 6: Recursively call `dfs(v)` to get the size `child_size` of the subtree rooted at child `v`.
7
Step 7: Update `size` by adding `child_size` to it, and update `score_prod` by multiplying it with `child_size`.
8
Step 8: Calculate the size of the 'parent subtree' as `parent_part_size = n - size`.
9
Step 9: Calculate the score of the current node `u` as `current_score = score_prod * max(1, parent_part_size)`. We use `max(1, parent_part_size)` to ensure that if `parent_part_size` is 0 (meaning `u` is the root), we multiply by 1 instead of 0.
10
Step 10: Update `max_score` and `count`: if `current_score` is greater than `max_score`, update `max_score` to `current_score` and reset `count` to 1. If `current_score` is equal to `max_score`, increment `count`.
11
Step 11: Return the size of the subtree rooted at `u`.
12
Step 12: Call the `dfs` function starting from the root node (node 0).
13
Step 13: Return the final value of `count`.

💡

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 adjacency list uses O(n) space because each node can have up to n-1 children in the worst case. The DFS visits each node once, so the time complexity is O(n). The recursion depth can be O(n) in the worst case, using O(n) space for the call stack.

📖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 Depth-First Search

Instead of using recursion, we can implement DFS iteratively using a stack. This can help avoid potential stack overflow issues for very large trees.

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

Trade-offs: Iterative DFS uses slightly more memory due to the explicit stack, but it avoids the recursion limit. It might be slightly harder to implement and read than the recursive version.

✨

Key Insights

1Insight 1: The key observation is that removing a node splits the tree into multiple subtrees. These subtrees are the left subtree, the right subtree (if they exist), and the subtree containing the parent of the removed node. If a subtree does not exist, its size is implicitly 1 in the score calculation.
2Insight 2: Depth-First Search (DFS) is the most suitable algorithm to traverse the tree and calculate the size of each subtree efficiently.
3Insight 3: Since the number of nodes can be up to 10^5, we need to use `long` data type to store the scores, as the product of subtree sizes can exceed the range of `int`.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: A tree with only two nodes. The code correctly handles this by computing the score of each node based on removing it and multiplying the sizes of the remaining subtrees.
Edge case 2: A skewed tree where almost all nodes are children of a single parent. The DFS algorithm still correctly calculates the subtree sizes and scores for each node in this scenario.
Edge case 3: The root node. The algorithm correctly handles the root node by calculating parent_part_size as n - size, where size is the size of the entire subtree rooted at 0. Therefore, the product with the parent size becomes 1 if it's the root being removed.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using `int` instead of `long` to store the scores. The product of subtree sizes can easily exceed the maximum value of an `int`, leading to incorrect results. This will cause integer overflow issues.

✗Common mistake 2: Not handling the case where the parent subtree size is 0 (when the current node is the root). In this case, you need to multiply by 1 instead of 0 to get the correct score.

✗Common mistake 3: Not creating a proper adjacency list which reflects the parent-child relationship. The code must properly create a child to parent relationship.

✅

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 network structures. The algorithm can be adapted to find critical nodes in a network, where removing a critical node significantly impacts the connectivity and size of the remaining network components.

🏦Real-world scenario 2: File system analysis. The tree structure of a file system can be analyzed to identify directories that, if removed, would result in the largest number of files being affected.

🎮Real-world scenario 3: Social Network analysis: Analyzing the influence of a user in a social network based on the number of connections that will be severed if a specific user is removed.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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 9, 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

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

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 "Count Nodes With the Highest Score"?

For this medium-level Array 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.

🤖 Need help with this problem?