Deep Copy, Deep Dive – Clone That Graph!

Clone Graph – Making a Perfect Copy

Today’s problem is a classic graph challenge: deep copying a graph given the reference to one of its nodes.

At first glance, it may seem like a straightforward copy-paste job—but in a connected undirected graph, cycles and shared references make it trickier than it looks!

Let’s break it down!

Best Data Structure for Solving It

To tackle this, we use a HashMap (or Dictionary) to keep track of already cloned nodes. Why?

  • Graphs can have cycles.

  • Nodes can be shared across paths.

  • Without tracking, we’d create duplicate nodes or go into infinite recursion.

Different Approaches – Brute Force to Optimized

1. DFS Approach (Recursive)

We use Depth-First Search to traverse and clone nodes recursively.

class Solution {
    private Map<Node, Node> visited = new HashMap<>();

    public Node cloneGraph(Node node) {
        if (node == null) return null;

        if (visited.containsKey(node)) return visited.get(node);

        Node cloneNode = new Node(node.val, new ArrayList<>());
        visited.put(node, cloneNode);

        for (Node neighbor : node.neighbors) {
            cloneNode.neighbors.add(cloneGraph(neighbor));
        }

        return cloneNode;
    }
}

2. BFS Approach (Iterative)

class Solution {
    public Node cloneGraph(Node node) {
        if (node == null) return null;

        Map<Node, Node> visited = new HashMap<>();
        Queue<Node> queue = new LinkedList<>();

        queue.add(node);
        visited.put(node, new Node(node.val, new ArrayList<>()));

        while (!queue.isEmpty()) {
            Node current = queue.poll();
            for (Node neighbor : current.neighbors) {
                if (!visited.containsKey(neighbor)) {
                    visited.put(neighbor, new Node(neighbor.val, new ArrayList<>()));
                    queue.add(neighbor);
                }
                visited.get(current).neighbors.add(visited.get(neighbor));
            }
        }

        return visited.get(node);
    }
}

⚠️ Common Mistakes & Pitfalls

  • Not tracking visited nodes: This leads to infinite recursion or duplicated nodes.

  • Returning new nodes without connecting neighbors properly.

  • Modifying original nodes instead of creating deep copies.

📊 Time & Space Complexity Analysis

  • Time Complexity: O(N + E)
    Where N is the number of nodes and E is the number of edges (we visit each node and edge once).

  • Space Complexity: O(N)
    Due to the visited map and recursion stack or queue (depending on DFS or BFS).

🌍 Real-World Applications

  • Cloning social networks (like a subgraph of friends).

  • Simulating virtual environments (like game maps or state trees).

  • Versioning and backup of complex reference-linked data structures.


I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

Comments

Post a Comment

Popular posts from this blog

Trailing Zeros in Factorial – How Many and Why?

Trailing Zeros – Counting Zeros in Factorial Factorial-related problems are a fundamental part of Data Structures and Algorithms (DSA). One such classic problem involves counting the number of trailing zeros in N! . In this article, we will explore different approaches to solving this problem, optimize our solution, and examine its real-world applications. Optimal Data Structure for Solving the Problem For this problem, no additional data structures are required. Since factorials grow exponentially, storing the actual factorial value is impractical. Instead, we leverage mathematical properties and division operations to efficiently determine the number of trailing zeros. This approach ensures an optimal solution in terms of both time and space complexity. Approach – From Brute Force to an Optimized Solution Brute Force Approach – Compute and Count A naive approach involves computing the factorial of N and counting the number of trailing zeros. However, this method is highly ineff...
Read more

Divisible Drama — Who’s In, Who’s Out?

 Divisible Drama — Who’s In, Who’s Out? Problem Statement You’re given two positive integers n and m . num1 = sum of all numbers from 1 to n not divisible by m num2 = sum of all numbers from 1 to n divisible by m Your mission, should you choose to accept it: Return num1 - num2 Example Input: n = 10 , m = 3 Output: 19 Breakdown: Not divisible by 3 → [ 1 , 2 , 4 , 5 , 7 , 8 , 10 ] → num1 = 37 Divisible by 3 → [ 3 , 6 , 9 ] → num2 = 18 Answer = 37 - 18 = 19 Best Data Structure for Solving the Problem This one’s a math.  Why overthink it when Gauss already gave us the formula for sum of 1 to n? Different Approaches (from Brute Force to Optimized) Approach 1: Brute Force (aka List & Sum Vibes) Loop from 1 to n , split the nums based on whether they’re divisible by m , and sum them up. class Solution { public int differenceOfSums ( int n, int m) { int num1 = 0 , num2 = 0 ; for ( int i = 1 ; i <= n; i++)...
Read more

Zero Array After Queries

Zero Array After Queries You are given an integer array nums and a list of queries, where each query is a pair [li, ri] representing a range. Each query allows you to decrement any subset of elements in the range [li, ri] by 1. Your task is to determine if it is possible to make every element in nums equal to zero after applying all queries in order . Example Input: nums = [ 1 , 0 , 1 ] queries = [[0, 2]] Output: true Explanation: We can select indices 0 and 2 and decrement both by 1 to obtain [0, 0, 0] . Best Data Structure for the Job We are working with multiple range updates and need to evaluate coverage over indices. The best choice here is the difference array combined with a prefix sum . It allows efficient range additions and helps track how many times each index is affected by the queries. Different Approaches Approach 1: Brute Force This naive method iterates over every index for every query and performs the decrement operations directly. class Solution {...
Read more