What is the time complexity of Difference Between Maximum and Minimum Price Sum?

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

What data structures are used to solve Difference Between Maximum and Minimum Price Sum?

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

What are the key insights for solving Difference Between Maximum and Minimum Price Sum?

Insight 1: The problem can be solved using dynamic programming and tree traversals (DFS) to efficiently calculate the maximum and minimum path sums for each possible root. Insight 2: We need to consider paths that extend down from a node, as well as paths that go up towards the original root to find the maximum price sum from each node.

Which companies ask LeetCode 2538. Difference Between Maximum and Minimum Price Sum in interviews?

This problem is frequently asked at Directi, Media.net.

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Difference Between Maximum and Minimum Price Sum - LeetCode 2538 Solution

Difference Between Maximum and Minimum Price Sum - Complete Solution Guide

Difference Between Maximum and Minimum Price Sum is LeetCode problem 2538, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There exists an undirected and initially unrooted tree with n nodes indexed from 0 to n - 1 . You are given the integer n and a 2D integer array edges of length n - 1 , where edges[i] = [a i , b i ] indicates that there is an edge between nodes a i and b i in the tree. Each node has an associated price. You are given an integer array price , where price[i] is the price of the i th node. The price sum of a given path is the sum of the prices of all nodes lying on that path. The tree can be rooted

Detailed Explanation

The problem asks us to find the maximum possible 'cost' in an undirected tree. The 'cost' is defined as the difference between the maximum and minimum price sum of all paths starting at a chosen 'root' node. We can select any node as the root and need to find the best root to maximize the cost across all possible paths originating from that root. The input consists of the number of nodes (n), the edges representing the tree structure, and an array of prices associated with each node.

Solution Approach

The provided solution uses a two-pass Depth-First Search (DFS) approach. The first DFS (dfs1) calculates the maximum path sum extending *down* from each node. It stores the top two maximum path sums from each node's children. The second DFS (dfs2) calculates the maximum path sum extending *up* from each node, combining the downward paths calculated earlier. Finally, it iterates through all possible root nodes and updates the global answer by calculating the maximum path starting from the root by combining the upward and downward paths.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the tree from the given edges.
Step 2: Implement `dfs1(u, p)`: This function calculates the maximum path sum extending down from node `u`, excluding its parent `p`. It also stores the top two children with the highest path sums.
Step 3: In `dfs1`, iterate through the neighbors of `u`. If the neighbor is not the parent, recursively call `dfs1` on the neighbor.
Step 4: Update the `path_down[u]` value with the price of the current node plus the maximum path sum from its children. Also update top_children_info to store the maximum two child path values.
Step 5: Implement `dfs2(u, p, up_val)`: This function calculates the maximum path sum extending up from node `u`, given the `up_val` (the maximum path sum from the parent's side).
Step 6: In `dfs2`, iterate through the neighbors of `u`. If the neighbor is not the parent, calculate the new `up_val` for that neighbor based on the top children information of `u` and the current `up_val`.
Step 7: Recursively call `dfs2` on the neighbor with the calculated `new_up_val`.
Step 8: Update the global `ans` variable with the maximum path sum starting at the current node `u`.
Step 9: Call `dfs1` starting from an arbitrary node (e.g., node 0) with no parent (-1).
Step 10: Call `dfs2` starting from node 0 with no parent (-1) and an initial `up_val` of 0.
Step 11: Return the final `ans` value, which represents the maximum possible cost.

Key Insights

Insight 1: The problem can be solved using dynamic programming and tree traversals (DFS) to efficiently calculate the maximum and minimum path sums for each possible root.
Insight 2: We need to consider paths that extend down from a node, as well as paths that go up towards the original root to find the maximum price sum from each node.
Insight 3: To efficiently calculate the path sums going upwards, we can perform a second DFS, utilizing the information gathered from the first DFS about the maximum path sums going downwards.
Insight 4: For each node, we need to store the top two maximum path sums from its children. This allows us to easily calculate the upward path sum when traversing to its children.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: Directi, Media.net.

Difference Between Maximum and Minimum Price Sum | LeetCode 2538

Hard

Array Dynamic Programming Tree Depth-First Search

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

Solution Code

Available in:

import java.util.*;

class Solution {
    private long ans;

    public long maxOutput(int n, int[][] edges, int[] price) {
        if (n == 1) {
            return 0;
        }

        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);
        }

        long[] path_down = new long[n];
        Pair[][] top_children_info = new Pair[n][2];
        for (int i = 0; i < n; i++) {
            top_children_info[i][0] = new Pair(0, -1);
            top_children_info[i][1] = new Pair(0, -1);
        }

        dfs1(0, -1, adj, price, path_down, top_children_info);

        this.ans = 0;

        dfs2(0, -1, 0, adj, price, path_down, top_children_info);

        return this.ans;
    }

    private void dfs1(int u, int p, List<List<Integer>> adj, int[] price, long[] path_down, Pair[][] top_children_info) {
        Pair top1 = new Pair(0, -1);
        Pair top2 = new Pair(0, -1);

        for (int v : adj.get(u)) {
            if (v == p) {
                continue;
            }
            dfs1(v, u, adj, price, path_down, top_children_info);
            long val = path_down[v];
            if (val >= top1.val) {
                top2 = top1;
                top1 = new Pair(val, v);
            } else if (val > top2.val) {
                top2 = new Pair(val, v);
            }
        }

        path_down[u] = price[u] + top1.val;
        top_children_info[u][0] = top1;
        top_children_info[u][1] = top2;
    }

    private void dfs2(int u, int p, long up_val, List<List<Integer>> adj, int[] price, long[] path_down, Pair[][] top_children_info) {
        Pair top1 = top_children_info[u][0];
        Pair top2 = top_children_info[u][1];
        
        this.ans = Math.max(this.ans, Math.max(top1.val, up_val));

        for (int v : adj.get(u)) {
            if (v == p) {
                continue;
            }

            long top1_val = top_children_info[u][0].val;
            int top1_node = top_children_info[u][0].node;
            long top2_val = top_children_info[u][1].val;

            long max_other_child_path;
            if (v == top1_node) {
                max_other_child_path = top2_val;
            } else {
                max_other_child_path = top1_val;
            }

            long new_up_val = price[u] + Math.max(up_val, max_other_child_path);
            dfs2(v, u, new_up_val, adj, price, path_down, top_children_info);
        }
    }

    private static class Pair {
        long val;
        int node;

        public Pair(long val, int node) {
            this.val = val;
            this.node = node;
        }
    }
}

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Problem Info

Description

Problem 2538: Difference Between Maximum and Minimum Price Sum - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Tree Depth-First Search

Companies

DirectiMedia.net

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2538

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

11 steps

1
Step 1: Build the adjacency list representation of the tree from the given edges.
2
Step 2: Implement `dfs1(u, p)`: This function calculates the maximum path sum extending down from node `u`, excluding its parent `p`. It also stores the top two children with the highest path sums.
3
Step 3: In `dfs1`, iterate through the neighbors of `u`. If the neighbor is not the parent, recursively call `dfs1` on the neighbor.
4
Step 4: Update the `path_down[u]` value with the price of the current node plus the maximum path sum from its children. Also update top_children_info to store the maximum two child path values.
5
Step 5: Implement `dfs2(u, p, up_val)`: This function calculates the maximum path sum extending up from node `u`, given the `up_val` (the maximum path sum from the parent's side).
6
Step 6: In `dfs2`, iterate through the neighbors of `u`. If the neighbor is not the parent, calculate the new `up_val` for that neighbor based on the top children information of `u` and the current `up_val`.
7
Step 7: Recursively call `dfs2` on the neighbor with the calculated `new_up_val`.
8
Step 8: Update the global `ans` variable with the maximum path sum starting at the current node `u`.
9
Step 9: Call `dfs1` starting from an arbitrary node (e.g., node 0) with no parent (-1).
10
Step 10: Call `dfs2` starting from node 0 with no parent (-1) and an initial `up_val` of 0.
11
Step 11: Return the final `ans` value, which represents the maximum possible cost.

💡

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 algorithm performs two DFS traversals of the tree, each visiting each node once. Therefore, the time complexity is O(n). The adjacency list, `path_down`, and `top_children_info` arrays all require O(n) space, resulting in a space complexity of 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

Brute Force

For each node, treat it as the root and calculate all possible paths starting from it. Find the maximum and minimum price sums for each root, and then the maximum cost across all possible roots.

Time: O(n^2) or higher depending on path enumerationSpace: O(n) for storing paths temporarily

Trade-offs: This approach is simple to understand but highly inefficient for larger trees. It is only suitable for very small trees where the n is very low (e.g. n <= 10).

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming and tree traversals (DFS) to efficiently calculate the maximum and minimum path sums for each possible root.
2Insight 2: We need to consider paths that extend down from a node, as well as paths that go up towards the original root to find the maximum price sum from each node.
3Insight 3: To efficiently calculate the path sums going upwards, we can perform a second DFS, utilizing the information gathered from the first DFS about the maximum path sums going downwards.
4Insight 4: For each node, we need to store the top two maximum path sums from its children. This allows us to easily calculate the upward path sum when traversing to its children.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: n = 1. If there's only one node, the cost is 0 because there are no paths to compare. The code explicitly handles this base case.
Edge case 2: A tree with skewed structure. The DFS approach correctly handles skewed trees by considering both downward and upward paths from each node.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering paths that go *up* the tree. Many naive approaches only consider paths going down from the root, leading to incorrect results.

✗Common mistake 2: Incorrectly calculating the maximum path sum when a node has multiple children. It's important to keep track of the top two path sums to correctly calculate the upward path sum during the second DFS.

✗Common mistake 3: Forgetting the case where n=1, which results in 0 cost since there are no two different paths to calculate the cost difference.

✅

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: Network routing. Finding the most efficient path through a network can be analogous to finding the maximum price sum path in a tree. Different paths may have different costs (e.g., latency, bandwidth), and we want to minimize/maximize the cost difference between the best and worst routes.

🏦Real-world scenario 2: Supply chain optimization. Imagine a supply chain where each node represents a warehouse and the edges represent transportation routes. The price at each node can be the cost of storing goods. Finding the 'maximum cost' difference can help identify bottlenecks or areas where significant cost variations exist, leading to optimization opportunities.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: June 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Difference Between Maximum and Minimum Price Sum"?

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

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