Best Unique-Node Value on a Timed Return Walk

Quick Overview

Best Unique-Node Value on a Timed Return Walk. An undirected graph has node values and edges [u,v,travelTime]. Start at 0, return to 0 within maxTime, and maximize the sum of values of distinct visited nodes. Repeated visits are allowed but score a node only once. A zero-edge walk is allowed. Each node has at most four incident edges.

Best Unique-Node Value on a Timed Return Walk

Company: Nutanix

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: hard

Interview Round: Other

### Problem An undirected graph has vertices numbered from `0` through `n - 1`, with value `values[i]` at vertex `i`. Each edge `[u, v, travelTime]` takes `travelTime` seconds to traverse in either direction. Start at vertex `0` and return to it within `maxTime` seconds. The value of a walk is the sum of the values of the distinct vertices it visits: revisiting a vertex does not score it again. Return the greatest value of a valid walk. You may stay at vertex `0` without traversing an edge. Each vertex has at most four incident edges. ```cpp class Solution { public: int maximalPathQuality(vector<int>& values, vector<vector<int>>& edges, int maxTime); }; ``` ### Examples **Example 1:** ```text values = [3,8], edges = [[0,1,10]], maxTime = 20 Output: 11 Explanation: The round trip 0->1->0 visits both nodes within the budget. ``` **Example 2:** ```text values = [6,100], edges = [[0,1,10]], maxTime = 10 Output: 6 Explanation: Visiting node 1 leaves no time to return, so staying at 0 is best. ``` ### Constraints - `1 <= values.length <= 1000`; `0 <= values[i] <= 10^8`. - `0 <= edges.length <= 2000`; edges are unique pairs with `0 <= u < v < n`. - Edge times and maxTime lie in `[10,100]`. - Maximum degree is four; the graph may be disconnected. ### Clarifying Questions Only walks ending back at zero contribute an answer. Visiting a node repeatedly does not add its value again. ```hint Inspect the structure How does the minimum edge duration bound the depth of exhaustive walk exploration? ``` ### Reasoning checkpoints - Enforce the time limit on a walk that finishes back at its starting vertex. - Score each distinct visited vertex once while allowing repeated visits and traversals. - Justify that the search or optimization covers every feasible return walk under the supplied bounds. ### Optional extensions What additional information could rule out a partial walk that cannot return to the start in time?

Overview: Best Unique-Node Value on a Timed Return Walk. An undirected graph has node values and edges [u,v,travelTime]. Start at 0, return to 0 within maxTime, and maximize the sum of values of distinct visited nodes. Repeated visits are allowed but score a node only once. A zero-edge walk is allowed. Each node has at most four incident edges.

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Nutanix
Aug 8, 2026
hardSoftware EngineerOtherCoding & Algorithms
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Problem

An undirected graph has vertices numbered from 0 through n - 1, with value values[i] at vertex i. Each edge [u, v, travelTime] takes travelTime seconds to traverse in either direction. Start at vertex 0 and return to it within maxTime seconds. The value of a walk is the sum of the values of the distinct vertices it visits: revisiting a vertex does not score it again. Return the greatest value of a valid walk. You may stay at vertex 0 without traversing an edge. Each vertex has at most four incident edges.

class Solution {
public:
    int maximalPathQuality(vector<int>& values, vector<vector<int>>& edges, int maxTime);
};

Examples

Example 1:

values = [3,8], edges = [[0,1,10]], maxTime = 20
Output: 11
Explanation: The round trip 0->1->0 visits both nodes within the budget.

Example 2:

values = [6,100], edges = [[0,1,10]], maxTime = 10
Output: 6
Explanation: Visiting node 1 leaves no time to return, so staying at 0 is best.

Constraints

  • 1 <= values.length <= 1000 ; 0 <= values[i] <= 10^8 .
  • 0 <= edges.length <= 2000 ; edges are unique pairs with 0 <= u < v < n .
  • Edge times and maxTime lie in [10,100] .
  • Maximum degree is four; the graph may be disconnected.

Clarifying Questions Guidance

Only walks ending back at zero contribute an answer. Visiting a node repeatedly does not add its value again.

Reasoning checkpoints

  • Enforce the time limit on a walk that finishes back at its starting vertex.
  • Score each distinct visited vertex once while allowing repeated visits and traversals.
  • Justify that the search or optimization covers every feasible return walk under the supplied bounds.

Optional extensions

What additional information could rule out a partial walk that cannot return to the start in time?