What is the time complexity of Find Closest Node to Given Two Nodes?

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

What data structures are used to solve Find Closest Node to Given Two Nodes?

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

What are the key insights for solving Find Closest Node to Given Two Nodes?

Insight 1: We need to calculate the distances from both `node1` and `node2` to all other nodes in the graph. This can be efficiently done using a simple iterative approach, as each node has at most one outgoing edge. Insight 2: Since cycles are allowed, we need to detect them when calculating distances to avoid infinite loops. This is done by checking if a node has already been visited (distance is not -1) before visiting it again.

Which companies ask LeetCode 2359. Find Closest Node to Given Two Nodes in interviews?

This problem is frequently asked at Juspay.

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Find Closest Node to Given Two Nodes - LeetCode 2359 Solution

Find Closest Node to Given Two Nodes - Complete Solution Guide

Find Closest Node to Given Two Nodes is LeetCode problem 2359, 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

You are given a directed graph of n nodes numbered from 0 to n - 1 , where each node has at most one outgoing edge. The graph is represented with a given 0-indexed array edges of size n , indicating that there is a directed edge from node i to node edges[i] . If there is no outgoing edge from i , then edges[i] == -1 . You are also given two integers node1 and node2 . Return the index of the node that can be reached from both node1 and node2 , such that the maximum between the distance from node1

Detailed Explanation

The problem asks us to find the node that is reachable from both `node1` and `node2` in a directed graph, such that the maximum of the distances from `node1` and `node2` to that node is minimized. The graph is represented by an array `edges`, where `edges[i]` indicates the node that node `i` points to. If `edges[i]` is -1, it means node `i` has no outgoing edge. We need to return the index of the node that satisfies the minimum maximum distance condition. If multiple nodes satisfy this condition, return the one with the smallest index. If no such node exists (no common reachable node), return -1.

Solution Approach

The solution calculates the distances from `node1` and `node2` to all other nodes separately using the `get_distances` function (or equivalent in other languages). Then, it iterates through all the nodes and checks if the node is reachable from both `node1` and `node2` (i.e., both `dist1[i]` and `dist2[i]` are not -1). If a node is reachable from both, the maximum of the two distances is calculated. The solution maintains the minimum of these maximum distances and the corresponding node index. Finally, it returns the node index with the minimum maximum distance. If no such node exists, it returns -1.

Step-by-Step Algorithm

Step 1: Create a function `get_distances(start_node)` to calculate the distances from a given `start_node` to all other nodes in the graph.
Step 2: In `get_distances`, initialize a `distances` array of size `n` with all elements set to -1. This array stores the distances from `start_node` to each node.
Step 3: Start from the `start_node` and iterate through the graph, updating the `distances` array and incrementing the distance at each step.
Step 4: Detect cycles by checking if the current node has already been visited (i.e., its distance in the `distances` array is not -1). If a cycle is detected, stop the iteration.
Step 5: Call `get_distances` twice, once for `node1` and once for `node2`, to get the `dist1` and `dist2` arrays, respectively.
Step 6: Initialize `min_max_dist` to infinity and `result_node` to -1.
Step 7: Iterate through all the nodes in the graph.
Step 8: For each node `i`, check if it is reachable from both `node1` and `node2` (i.e., `dist1[i] != -1` and `dist2[i] != -1`).
Step 9: If the node is reachable from both, calculate the maximum of the two distances: `max_dist = max(dist1[i], dist2[i])`.
Step 10: If `max_dist` is less than `min_max_dist`, update `min_max_dist` to `max_dist` and `result_node` to `i`.
Step 11: After iterating through all the nodes, return `result_node`.

Key Insights

Insight 1: We need to calculate the distances from both `node1` and `node2` to all other nodes in the graph. This can be efficiently done using a simple iterative approach, as each node has at most one outgoing edge.
Insight 2: Since cycles are allowed, we need to detect them when calculating distances to avoid infinite loops. This is done by checking if a node has already been visited (distance is not -1) before visiting it again.
Insight 3: The optimal solution will be a node `i` where `dist1[i] != -1` and `dist2[i] != -1`, implying it's reachable from both `node1` and `node2`. We need to minimize `max(dist1[i], dist2[i])`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Depth-First Search, Graph.

Companies

Asked at: Juspay.

Find Closest Node to Given Two Nodes | LeetCode 2359

Medium

Depth-First Search Graph

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int closestMeetingNode(int[] edges, int node1, int node2) {
        int n = edges.length;

        class Helper {
            public int[] getDistances(int startNode) {
                int[] distances = new int[n];
                Arrays.fill(distances, -1);
                int dist = 0;
                int currNode = startNode;

                while (currNode != -1 && distances[currNode] == -1) {
                    distances[currNode] = dist;
                    dist += 1;
                    currNode = edges[currNode];
                }

                return distances;
            }
        }
        
        Helper helper = new Helper();
        int[] dist1 = helper.getDistances(node1);
        int[] dist2 = helper.getDistances(node2);

        int minMaxDist = Integer.MAX_VALUE;
        int resultNode = -1;

        for (int i = 0; i < n; i++) {
            if (dist1[i] != -1 && dist2[i] != -1) {
                int maxDist = Math.max(dist1[i], dist2[i]);

                if (maxDist < minMaxDist) {
                    minMaxDist = maxDist;
                    resultNode = i;
                }
            }
        }

        return resultNode;
    }
}

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Problem Info

Description

Problem 2359: Find Closest Node to Given Two Nodes - Detailed explanation and solution approaches.

Topics

Depth-First Search Graph

Companies

Juspay

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2359

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: Create a function `get_distances(start_node)` to calculate the distances from a given `start_node` to all other nodes in the graph.
2
Step 2: In `get_distances`, initialize a `distances` array of size `n` with all elements set to -1. This array stores the distances from `start_node` to each node.
3
Step 3: Start from the `start_node` and iterate through the graph, updating the `distances` array and incrementing the distance at each step.
4
Step 4: Detect cycles by checking if the current node has already been visited (i.e., its distance in the `distances` array is not -1). If a cycle is detected, stop the iteration.
5
Step 5: Call `get_distances` twice, once for `node1` and once for `node2`, to get the `dist1` and `dist2` arrays, respectively.
6
Step 6: Initialize `min_max_dist` to infinity and `result_node` to -1.
7
Step 7: Iterate through all the nodes in the graph.
8
Step 8: For each node `i`, check if it is reachable from both `node1` and `node2` (i.e., `dist1[i] != -1` and `dist2[i] != -1`).
9
Step 9: If the node is reachable from both, calculate the maximum of the two distances: `max_dist = max(dist1[i], dist2[i])`.
10
Step 10: If `max_dist` is less than `min_max_dist`, update `min_max_dist` to `max_dist` and `result_node` to `i`.
11
Step 11: After iterating through all the nodes, return `result_node`.

💡

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 time complexity is linear because we traverse each node at most a constant number of times (maximum 3). Space complexity is linear since we use distance arrays that store information for each node.

📖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

Breadth-First Search (BFS)

Instead of the iterative approach used in the provided code, we could use Breadth-First Search (BFS) to calculate the distances from node1 and node2. The underlying principle remains the same.

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

Trade-offs: BFS might be slightly more intuitive for some developers, but in this specific problem, the iterative approach is equally efficient and potentially simpler to implement since each node has at most one outgoing edge.

✨

Key Insights

1Insight 1: We need to calculate the distances from both `node1` and `node2` to all other nodes in the graph. This can be efficiently done using a simple iterative approach, as each node has at most one outgoing edge.
2Insight 2: Since cycles are allowed, we need to detect them when calculating distances to avoid infinite loops. This is done by checking if a node has already been visited (distance is not -1) before visiting it again.
3Insight 3: The optimal solution will be a node `i` where `dist1[i] != -1` and `dist2[i] != -1`, implying it's reachable from both `node1` and `node2`. We need to minimize `max(dist1[i], dist2[i])`.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: `edges[i] == -1` for all `i`. This means no node has any outgoing edge. If `node1 == node2`, the answer is `node1`. If `node1 != node2`, the answer is -1 because no node is reachable from any other node. The code handles this correctly as `dist1` and `dist2` will be all -1 except for dist[node1] and dist[node2] at index node1 and node2 respectively.
Edge case 2: There is no common reachable node from `node1` and `node2`. In this case, the loop will not find any node where both `dist1[i]` and `dist2[i]` are not -1. The `result_node` will remain -1, which is the correct output.
Edge case 3: There are cycles in the graph. The `get_distances` function correctly handles cycles by checking if a node has already been visited before visiting it again. This prevents infinite loops.
Edge case 4: node1 and node2 are the same. The algorithm handles this correctly; the distances will be calculated from the same starting point, and the closest meeting node would be node1 itself.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling cycles correctly. If cycles are not detected, the `get_distances` function might run into an infinite loop, leading to a Time Limit Exceeded (TLE) error.

✗Common mistake 2: Not initializing the `distances` array correctly. If the array is not initialized to -1, the algorithm might produce incorrect results because it will not be able to distinguish between unvisited nodes and nodes with a distance of -1.

✗Common mistake 3: Forgetting the condition that if multiple nodes have the same minimum maximum distance, the node with the smallest index should be returned. The given solution implicitly takes care of this because the loop iterates from 0 to n-1.

✅

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: Social Network Analysis: Finding the closest person who is mutually connected to two different individuals in a social network.

🏦Real-world scenario 2: Network Routing: In a communication network, finding the closest server that can be reached from two different client machines, minimizing the maximum latency between each client and the server.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 20, 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 BFS vs DFS for graph problems?

Use BFS for shortest path in unweighted graphs, level-order processing, or when exploring neighbors systematically. Use DFS for exhaustive search, detecting cycles, topological sort, or when you need to explore complete paths. DFS uses less memory for very wide graphs.

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 "Find Closest Node to Given Two Nodes"?

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

🤖 Need help with this problem?