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Mastering Leetcode Problem 2359: The Closest Meeting Node Explained

By Talent Navigator

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Published Jun 6, 2025

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4 min read

https://www.youtube.com/watch?v=i4WLvAVVlw0&ab_channel=TalentNavigator

Understanding graph problems can be challenging, especially when they involve various paths and nodes. One such intriguing problem is Leetcode Problem 2359, which focuses on finding the closest meeting node between two nodes in a graph. This article will delve into how to approach this problem effectively, providing detailed explanations of the concepts involved and the coding implementation.

Problem Overview

Leetcode Problem 2359 presents a graph represented by edges, with two specified nodes. The objective is to identify the closest node where both nodes "meet" based on the shortest path calculated through the edges. This translates into finding the node that has the minimum distance from both given nodes. The challenge lies in efficiently traversing and analyzing the graph.

Key Terms

Before we dive into the solution, let’s clarify some crucial terms:

• Graph: A collection of nodes (or vertices) connected by edges.
• Node: A fundamental part of a graph representing an entity.
• Depth-First Search (DFS): An algorithm for traversing or searching tree or graph data structures, going as deep as possible down one path before backing up.
• Distances: The shortest paths from the specified nodes to all other nodes in the graph.

Solution Approach

To solve Problem 2359, we follow these steps:

1. Initialize Distance Vectors: We create two vectors to store the distances for both nodes (let's call them distance1 and distance2). These vectors are initialized with infinity to signify that initially, the distance to every node is unknown.
2. Depth-First Search (DFS): We will employ DFS to populate these distance vectors. Starting from each of the two given nodes, we traverse the graph and update the distances accordingly.
3. Finding the Closest Meeting Node: After filling in the distance vectors, we will iterate through the nodes to find the one with the minimum maximum distance from both vectors. This node is our closest meeting point.

Step-by-Step Implementation

Now, let's break down the implementation in detail. Below is how the coding is structured:

int closestMeetingNode(vector<int>& edges, int node1, int node2) {
    int n = edges.size();
    vector<int> distance1(n, INT_MAX);
    vector<int> distance2(n, INT_MAX);

    // Initialize visited arrays
    vector<bool> visited1(n, false);
    vector<bool> visited2(n, false);
    distance1[node1] = 0;
    distance2[node2] = 0;

    // Perform DFS for both nodes
    dfs(edges, node1, distance1, visited1);
    dfs(edges, node2, distance2, visited2);

    int minDistanceNode = -1;
    int minMaxDistance = INT_MAX;

    // Find the closest meeting node
    for (int i = 0; i < n; ++i) {
        int maxDist = max(distance1[i], distance2[i]);
        if (maxDist < minMaxDistance) {
            minMaxDistance = maxDist;
            minDistanceNode = i;
        }
    }
    return minDistanceNode;
}

void dfs(vector<int>& edges, int node, vector<int>& distance, vector<bool>& visited) {
    visited[node] = true;
    int nextNode = edges[node];
    if (nextNode != -1 && !visited[nextNode]) {
        distance[nextNode] = distance[node] + 1;
        dfs(edges, nextNode, distance, visited);
    }
}

Explanation of the Code

• Graph Edges: The function takes the vector edges representing the graph along with the two nodes.
• Distance Initialization: The distances are initialized to "infinity" to signify that initially all distances are unknown.
• DFS Function: This function marks nodes as visited and updates the distances as it traverses the graph recursively. It checks for connections based on the edges provided.
• Finding the Minimum: Lastly, the algorithm traverses all nodes to compute the maximum distance from the two sources and finds the minimum among them, identifying the closest meeting node.

Complexity Analysis

The time complexity of this approach is O(n), where n is the number of edges. This is due to the requirement to traverse each node once. The space complexity is also O(n) due to the storage used by the distance arrays and visited flags.

Conclusion

Leetcode Problem 2359 poses a unique challenge in algorithmic design and programming logic. By breaking down the problem and employing a systematic approach that combines depth-first search with strategic distance management, you can arrive at an efficient solution.

Learning how to process complex graph-based problems significantly enhances your coding skills and prepares you for competitive programming and technical interviews. For more tips and in-depth guides on solving similar coding challenges, consider exploring various algorithm topics further. Happy coding!