What is the time complexity of Maximum Path Quality of a Graph?

The optimal solution has a time complexity of O(V * (2 * E)^(V-1)).

What is the space complexity of Maximum Path Quality of a Graph?

The space complexity is O(V + E).

What data structures are used to solve Maximum Path Quality of a Graph?

This problem uses Array, Backtracking, Graph. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Path Quality of a Graph?

Insight 1: Since the graph isn't guaranteed to be a DAG and revisits are allowed, backtracking is necessary to explore different paths. Without backtracking, we cannot explore all possibilities of paths that may yield the maximal quality. Insight 2: We need to keep track of the number of times each node has been visited in a given path to ensure that the node's value is added to the path's quality only once. Use of a `visited_counts` array is crucial to achieve this. The `current_quality` in dfs represents the sum of unique values visited during the current path.

Which companies ask LeetCode 2065. Maximum Path Quality of a Graph in interviews?

This problem is frequently asked at Atlassian, DoorDash, SAP.

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Maximum Path Quality of a Graph - LeetCode 2065 Solution

Maximum Path Quality of a Graph - Complete Solution Guide

Maximum Path Quality of a Graph is LeetCode problem 2065, 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 is an undirected graph with n nodes numbered from 0 to n - 1 ( inclusive ). You are given a 0-indexed integer array values where values[i] is the value of the i th node. You are also given a 0-indexed 2D integer array edges , where each edges[j] = [u j , v j , time j ] indicates that there is an undirected edge between the nodes u j and v j , and it takes time j seconds to travel between the two nodes. Finally, you are given an integer maxTime . A valid path in the graph is any path that s

Detailed Explanation

The problem describes an undirected graph where each node has a value and each edge has a time associated with it. We need to find a valid path starting and ending at node 0, with a total travel time not exceeding `maxTime`. The goal is to maximize the *quality* of the path, which is the sum of the values of the *unique* nodes visited. A node's value is only added to the quality once, even if visited multiple times. The graph is not guaranteed to be connected and each node has at most 4 edges.

Solution Approach

The solution uses a Depth-First Search (DFS) algorithm with backtracking to explore all possible paths from node 0 back to node 0 within the given time constraint. The DFS function recursively explores the graph, tracking the current time, the current quality of the path, and the number of times each node has been visited. It maintains a `visited_counts` array to avoid adding a node's value to the path quality more than once in a single path. The global variable `max_quality` is updated whenever a valid path (starting and ending at 0 within the time limit) yields a better quality than the current maximum.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the graph from the given `edges` array. Each entry in the adjacency list will store a list of neighboring nodes and the time it takes to travel to them.
Step 2: Initialize a `visited_counts` array to keep track of the number of times each node has been visited in the current path.
Step 3: Define a recursive DFS function that takes the current node `u`, the `current_time` taken to reach `u`, and the `current_quality` of the path so far as input.
Step 4: Inside the DFS function, if this is the first visit to the current node `u`, add its value to the `current_quality` and increment the count in `visited_counts`.
Step 5: If the current node is 0 (the starting node), update the `max_quality` with the maximum of the current `max_quality` and `current_quality`.
Step 6: Iterate through the neighbors of the current node `u`. For each neighbor `v` and the time `time_to_v` to reach `v`, check if `current_time + time_to_v <= maxTime`. If so, recursively call the DFS function with `v`, `current_time + time_to_v`, and `current_quality`.
Step 7: After exploring all neighbors of `u`, backtrack by decrementing the count of node `u` in the `visited_counts` array. This allows the algorithm to explore other possible paths.
Step 8: Start the DFS traversal from node 0 with initial `current_time` and `current_quality` as 0.
Step 9: Return the final `max_quality`.

Key Insights

Insight 1: Since the graph isn't guaranteed to be a DAG and revisits are allowed, backtracking is necessary to explore different paths. Without backtracking, we cannot explore all possibilities of paths that may yield the maximal quality.
Insight 2: We need to keep track of the number of times each node has been visited in a given path to ensure that the node's value is added to the path's quality only once. Use of a `visited_counts` array is crucial to achieve this. The `current_quality` in dfs represents the sum of unique values visited during the current path.
Insight 3: The maximum allowed time `maxTime` acts as a constraint, pruning the search space and preventing infinite loops in cyclic graphs. The path has to start and end at node 0, implying that the path is a 'round trip'.
Insight 4: Understanding that the problem is essentially a constrained optimization problem on a graph. The constraints are the maximum time allowed, and the optimization objective is to maximize the quality of the path. This helps frame the problem as a search for the best path.

Complexity Analysis

Time Complexity: O(V * (2 * E)^(V-1))

Space Complexity: O(V + E)

Topics

This problem involves: Array, Backtracking, Graph.

Companies

Asked at: Atlassian, DoorDash, SAP.

Maximum Path Quality of a Graph | LeetCode 2065

Hard

Array Backtracking Graph

Time: O(V * (2 * E)^(V-1))Space: O(V + E)

Solution Code

Available in:

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

class Solution {
    public int maximalPathQuality(int[] values, int[][] edges, int maxTime) {
        int n = values.length;
        List<List<int[]>> 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];
            int time = edge[2];
            adj.get(u).add(new int[]{v, time});
            adj.get(v).add(new int[]{u, time});
        }

        int max_quality = 0;
        
        int[] visited_counts = new int[n];

        class Helper {
            void dfs(int u, int current_time, int current_quality) {
                // If this is the first visit to node u on the current path
                if (visited_counts[u] == 0) {
                    current_quality += values[u];
                }
                
                visited_counts[u]++;

                // If we are at node 0, this forms a valid path
                if (u == 0) {
                    max_quality = Math.max(max_quality, current_quality);
                }

                // Explore neighbors
                for (int[] neighbor : adj.get(u)) {
                    int v = neighbor[0];
                    int time_to_v = neighbor[1];
                    if (current_time + time_to_v <= maxTime) {
                        dfs(v, current_time + time_to_v, current_quality);
                    }
                }
                
                // Backtrack: undo the visit to node u for the current path
                visited_counts[u]--;
            }
        }

        Helper helper = new Helper();

        // Start the traversal from node 0
        helper.dfs(0, 0, 0);
        
        return max_quality;
    }
}

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Problem Info

Description

Problem 2065: Maximum Path Quality of a Graph - Detailed explanation and solution approaches.

Topics

Array Backtracking Graph

Companies

AtlassianDoorDashSAP

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:2065

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

9 steps

1
Step 1: Build the adjacency list representation of the graph from the given `edges` array. Each entry in the adjacency list will store a list of neighboring nodes and the time it takes to travel to them.
2
Step 2: Initialize a `visited_counts` array to keep track of the number of times each node has been visited in the current path.
3
Step 3: Define a recursive DFS function that takes the current node `u`, the `current_time` taken to reach `u`, and the `current_quality` of the path so far as input.
4
Step 4: Inside the DFS function, if this is the first visit to the current node `u`, add its value to the `current_quality` and increment the count in `visited_counts`.
5
Step 5: If the current node is 0 (the starting node), update the `max_quality` with the maximum of the current `max_quality` and `current_quality`.
6
Step 6: Iterate through the neighbors of the current node `u`. For each neighbor `v` and the time `time_to_v` to reach `v`, check if `current_time + time_to_v <= maxTime`. If so, recursively call the DFS function with `v`, `current_time + time_to_v`, and `current_quality`.
7
Step 7: After exploring all neighbors of `u`, backtrack by decrementing the count of node `u` in the `visited_counts` array. This allows the algorithm to explore other possible paths.
8
Step 8: Start the DFS traversal from node 0 with initial `current_time` and `current_quality` as 0.
9
Step 9: Return the final `max_quality`.

💡

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(V * (2 * E)^(V-1))

💾

Space Complexity

O(V + E)

📝

Why?

The time complexity arises from the exponential nature of the DFS traversal. The space complexity is determined by the size of the adjacency list (to store the graph) and the `visited_counts` array, along with the maximum recursion depth.

📖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

Dijkstra's Algorithm with Pruning

Use Dijkstra's algorithm to find the shortest path from node 0 to all other nodes. Then, perform a modified DFS, pruning paths that exceed `maxTime` based on the shortest path distances calculated by Dijkstra's. This may offer slight improvements if Dijkstra pre-computation drastically reduces DFS branching.

Time: O(E log V + (V * E^(V-1))), where E log V is due to Dijkstra and (V * E^(V-1)) the DFS.Space: O(V + E)

Trade-offs: Dijkstra pre-computation introduces overhead but can efficiently prune paths. Suitable if significant pruning is expected. Potentially more efficient than pure DFS for certain graph structures, but the overall worst-case complexity can still be exponential.

✨

Key Insights

1Insight 1: Since the graph isn't guaranteed to be a DAG and revisits are allowed, backtracking is necessary to explore different paths. Without backtracking, we cannot explore all possibilities of paths that may yield the maximal quality.
2Insight 2: We need to keep track of the number of times each node has been visited in a given path to ensure that the node's value is added to the path's quality only once. Use of a `visited_counts` array is crucial to achieve this. The `current_quality` in dfs represents the sum of unique values visited during the current path.
3Insight 3: The maximum allowed time `maxTime` acts as a constraint, pruning the search space and preventing infinite loops in cyclic graphs. The path has to start and end at node 0, implying that the path is a 'round trip'.
4Insight 4: Understanding that the problem is essentially a constrained optimization problem on a graph. The constraints are the maximum time allowed, and the optimization objective is to maximize the quality of the path. This helps frame the problem as a search for the best path.

📚

Key Terms Explained

For Students

Array

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

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: The graph is disconnected. The solution handles this correctly because the DFS only explores paths reachable from node 0. If node 0 is isolated, the result will be the value of node 0 if maxTime allows return to it directly (time 0) otherwise 0 if there are no other nodes and the value of node 0 is 0, or the quality remains 0.
Edge case 2: `maxTime` is very small or zero. The solution correctly handles this by only exploring neighbors that can be reached within the remaining time. If no such neighbors exist, the DFS backtracks. If `maxTime` is zero, only the node 0 is visited.
Edge case 3: A node has a very large value. The solution handles large values correctly by using the `current_quality` variable, which can accumulate large values.
Edge case 4: The graph contains cycles. The solution leverages the `visited_counts` and `maxTime` to control the depth and duration of the path to avoid infinite loops and exceeding the time constraint.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using backtracking. Without backtracking, the algorithm will not explore all possible paths and may miss the optimal solution.

✗Common mistake 2: Adding a node's value multiple times to the path quality if the node is visited multiple times. The `visited_counts` array is crucial to avoid this.

✗Common mistake 3: Not checking the time constraint before exploring a neighbor. This can lead to exploring invalid paths and exceeding the `maxTime` limit.

✗Common mistake 4: Integer overflow issues for large values or long paths. Use appropriate data types (e.g., `long` in Java/C++) to prevent this, especially when accumulating the path quality.

✅

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: Pathfinding with resource constraints. For example, finding the most valuable route for a delivery truck, considering time limits and the value of each location (e.g., priority customers).

🏦Real-world scenario 2: Network optimization. Determining the best path for data transmission across a network to maximize the amount of data transferred while staying within latency constraints.

🎮Real-world scenario 3: Game AI. Developing an AI agent that needs to navigate a game map to collect resources while minimizing travel time and maximizing the value of resources collected.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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 25, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

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.

How do I avoid duplicate solutions in backtracking?

Sort the input first, then skip consecutive duplicates at the same recursion level: if (i > start && nums[i] == nums[i-1]) continue. For permutations, use a visited array or swap-based approach. Always consider if order matters (permutations vs combinations).

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Maximum Path Quality of a Graph"?

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 backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?