Union Find

1. Number of Islands (LeetCode #200)

1. Problem Description:

The problem "Number of Islands" is described as follows:

Given a 2D grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example: Input: 11110 11010 11000 00000

Output: 1

Input: 11000 11000 00100 00011

Output: 3

Problem Link: https://leetcode.com/problems/number-of-islands/

1. Optimized Java Code for "Number of Islands":

public class Solution {
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }

        int islandCount = 0;
        int rows = grid.length;
        int cols = grid[0].length;

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == '1') {
                    islandCount++;
                    dfs(grid, i, j);
                }
            }
        }

        return islandCount;
    }

    private void dfs(char[][] grid, int row, int col) {
        if (row < 0 || col < 0 || row >= grid.length || col >= grid[0].length || grid[row][col] != '1') {
            return;
        }

        grid[row][col] = '0';

        dfs(grid, row - 1, col);
        dfs(grid, row + 1, col);
        dfs(grid, row, col - 1);
        dfs(grid, row, col + 1);
    }
}

1. Algorithm Explanation:

The algorithm uses Depth-First Search (DFS) to solve the problem.

1. First, we check if the input grid is empty or null. If it is, we return 0 as there are no islands.

2. We maintain a count variable, initially set to 0, to store the number of islands.

3. We iterate through each cell in the grid using two nested loops.

4. For each cell, if it is a land cell ('1'), we increment the island count by 1 and call the dfs function.

5. The dfs function is a recursive function that performs DFS on the grid. It marks the current cell as visited by changing its value to '0' (to avoid counting it again), and then recursively calls dfs on its adjacent land cells (up, down, left, and right).

6. The base case for the dfs function is when the current cell is out of bounds or is a water cell ('0').

7. Finally, we return the total island count.

During the interview process, it's important to approach this problem in a systematic way:

1. Understand the problem: Read the problem statement carefully and make sure you understand the requirements and constraints. Take note of any examples provided.

2. Analyze the problem: Break down the problem into smaller sub-problems and think about how you can solve each sub-problem. Identify any patterns or algorithms that can be used.

3. Brainstorm solutions: Come up with possible solutions for the problem. Think about different approaches, such as using a graph traversal algorithm like DFS or BFS.

4. Choose an approach: Evaluate the pros and cons of each approach and choose the one that seems most suitable for the problem at hand. Consider efficiency, time complexity, and any specific requirements of the problem.

5. Implement the solution: Write code that implements your chosen approach. Make sure to handle edge cases and test your code on various test cases, including the given examples.

6. Optimize the solution: Once you have a working solution, analyze its time complexity and identify any potential bottlenecks. Look for opportunities to optimize your code and reduce unnecessary operations.

7. Test the solution: Verify that your code is working correctly by testing it on additional test cases and edge cases. Make sure to cover various scenarios and check for any possible edge cases.

2. Redundant Connection (LeetCode #684)

1. Problem Description:

The problem "Redundant Connection" states that given a set of directed edges connecting n nodes in a graph, one edge is redundant, meaning it can be removed without affecting the connectivity of any other edges. Our task is to find this redundant edge and return it as an array of two integers.

Problem Link: Redundant Connection (LeetCode #684)

1. Optimized Java Code:

class Solution {
    private int[] parent;
    
    public int[] findRedundantConnection(int[][] edges) {
        int n = edges.length;
        parent = new int[n + 1];
        
        // Initially, all nodes are parent of themselves
        for (int i = 1; i <= n; i++) {
            parent[i] = i;
        }
        
        int[] redundantEdge = new int[2];
        
        // Traverse the edges and find the redundant edge
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            
            // If u and v have same parent, then the edge is redundant
            if (find(u) == find(v)) {
                redundantEdge[0] = u;
                redundantEdge[1] = v;
            } else {
                // Merge the sets of u and v
                union(u, v);
            }
        }
        
        return redundantEdge;
    }
    
    private int find(int x) {
        if (parent[x] != x) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
    }
    
    private void union(int x, int y) {
        parent[find(x)] = find(y);
    }
}

1. Algorithm Explanation and Interview Strategy:

To solve this problem efficiently during an interview, we can make use of the Union-Find algorithm. Union-Find is a data structure that allows efficient union and find operations on disjoint sets.

In this problem, we need to find the redundant edge, which means we need to find the edge that forms a cycle in the given graph. The Union-Find algorithm helps us to find cycles efficiently.

The algorithm works as follows:

• Create a parent array to keep track of the parent node for each node.

• Initially, all nodes are parent of themselves, so we set parent[i] = i for each node i.

• Traverse through the given edges in the graph.

  • For each edge (u, v), check if u and v have the same parent. If they do, this edge is redundant since it forms a cycle in the graph.

  • If u and v have different parents, merge the sets of u and v by setting parent[find(u)] = find(v).

• Return the redundant edge found in the above traversal.

During an interview, it is important to explain the concept of the Union-Find algorithm and how it can be used to solve the problem efficiently. Make sure to discuss the time and space complexity of the solution and address any possible edge cases.

3. Graph Valid Tree (LeetCode #261)

1. The problem "Graph Valid Tree" is described as follows:

Given n nodes labeled from 0 to n-1 and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.

Example 1: Input: n = 5, edges = [[0,1], [0,2], [0,3], [1,4]] Output: true

Example 2: Input: n = 5, edges = [[0,1], [1,2], [2,3], [1,3], [1,4]] Output: false

Note:

• You can assume that no duplicate edges will appear in edges

• Since all edges are undirected, [0,1] is the same as [1,0] and thus will not appear together in edges.

Problem Link: https://leetcode.com/problems/graph-valid-tree/

1. Here is the optimized Java code for the problem "Graph Valid Tree" (LeetCode #261) with detailed comments:

public class Solution { public boolean validTree(int n, int[][] edges) { // Create an array to store the parent of each node int[] parent = new int[n];

    // Initially, set the parent of each node to itself
    for(int i = 0; i < n; i++)
        parent[i] = i;
    
    // Iterate through each edge
    for(int[] edge : edges) {
        int x = find(parent, edge[0]);
        int y = find(parent, edge[1]);
        
        // If the two nodes already have the same parent, there is a cycle
        if(x == y)
            return false;
        
        // Union the two nodes by setting the parent of one to the other
        parent[x] = y;
    }
    
    // Check if the graph contains a single connected component
    int connectedComponents = 0;
    for(int i = 0; i < n; i++) {
        if(parent[i] == i)
            connectedComponents++;
    }
    
    // If there is only one connected component, it is a valid tree
    return connectedComponents == 1;
}

// Find the parent of a node using union-find algorithm
private int find(int[] parent, int node) {
    if(parent[node] != node)
        parent[node] = find(parent, parent[node]);
    return parent[node];
}

}

1. During the interview, you can approach the problem using the following steps:

• The problem asks us to determine whether a graph is a valid tree or not. A valid tree is a connected graph with no cycles.

• To solve this problem, we can use the union-find algorithm, which is commonly used for solving connectivity problems.

• First, we create an array to store the parent of each node. Initially, we set the parent of each node to itself.

• Then, we iterate through the edges and find the parent of each node using the union-find algorithm.

• If two nodes already have the same parent, it means there is a cycle in the graph, and the graph is not a valid tree. Return false.

• Otherwise, we union the two nodes by setting the parent of one node to the other.

• After iterating through all the edges, we check if the graph contains a single connected component. We count the number of connected components by checking the parent of each node. If there is only one connected component, it means the graph is a valid tree. Return true.

• The time complexity of this solution is O(n + m), where n is the number of nodes and m is the number of edges. We iterate through all the edges once, and finding the parent of each node takes O(1) time on average.

• Overall, this solution provides an efficient way to determine whether a graph is a valid tree or not.

4. Accounts Merge (LeetCode #721)

1. Problem Description:

The problem "Accounts Merge" (LeetCode #721) can be found at the following link: https://leetcode.com/problems/accounts-merge/

You are given a list of accounts where each element accounts[i] is a list of strings, representing an email address and a name. You need to merge the accounts if they are linked by a common email address.

After merging the accounts, return the accounts in the following format: the first element of each merged account is the name, and the rest of the elements are emails sorted lexicographically. The accounts themselves can be returned in any order.

1. Optimized Java Code:

import java.util.*;

class Solution {
    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        Map<String, String> emailToName = new HashMap<>();
        Map<String, List<String>> graph = new HashMap<>();
        List<List<String>> result = new ArrayList<>();
        
        // Populate the email to name mapping and the graph
        for (List<String> account : accounts) {
            String name = account.get(0);
            for (int i = 1; i < account.size(); i++) {
                String email = account.get(i);
                // Add email to name mapping
                emailToName.put(email, name);
                
                // Create an edge between emails in the same account
                if (!graph.containsKey(email)) {
                    graph.put(email, new ArrayList<>());
                }
                if (i > 1) {
                    String prevEmail = account.get(i - 1);
                    graph.get(email).add(prevEmail);
                    graph.get(prevEmail).add(email);
                }
            }
        }
        
        Set<String> visited = new HashSet<>();
        // Perform DFS to find connected components
        for (String email : emailToName.keySet()) {
            if (!visited.contains(email)) {
                List<String> mergedAccount = new ArrayList<>();
                dfs(email, graph, visited, mergedAccount);
                Collections.sort(mergedAccount);
                mergedAccount.add(0, emailToName.get(email));
                result.add(mergedAccount);
            }
        }
        
        return result;
    }
    
    private void dfs(String email, Map<String, List<String>> graph, Set<String> visited, List<String> mergedAccount) {
        visited.add(email);
        mergedAccount.add(email);
        
        if (graph.containsKey(email)) {
            for (String neighbor : graph.get(email)) {
                if (!visited.contains(neighbor)) {
                    dfs(neighbor, graph, visited, mergedAccount);
                }
            }
        }
    }
}

1. Algorithm Description:

To solve this problem, we can use the concept of a graph and perform a Depth-First Search (DFS) to find connected components. Here is a step-by-step algorithm to solve the problem:

• Create two data structures: emailToName (a map to store email to name mapping) and graph (a map to store the graph representation).

• Iterate through each account in the given list of accounts.

  • Extract the name of the account from the first element of the account list.

  • For each email in the account (starting from the second element):

    • Add the email to the emailToName map with its corresponding name.

    • Create an edge between the current email and the previous email (if any) in the same account.

    • If a graph node for the email does not exist, create one and store an empty list as its value.

• Create an empty set called visited to keep track of visited emails.

• Iterate through each email in the emailToName map.

  • If the email has not been visited:

    • Create an empty list called mergedAccount to store the merged account.

    • Perform a depth-first search (DFS) starting from the current email, passing in the graph, visited, and mergedAccount.

    • Sort the mergedAccount list lexicographically.

    • Add the corresponding name to the first position in the mergedAccount list.

    • Add the mergedAccount list to the result list.

• Finally, return the result list containing the merged accounts.

During an interview, it is important to communicate the algorithm step-by-step, highlighting the data structures used and their purposes. It is also beneficial to discuss the time complexity of the solution, which in this case is O(N log N), where N is the total number of emails, due to the sorting process.

5. Evaluate Division (LeetCode #399)

1. The problem: Evaluate Division (LeetCode #399) Link: https://leetcode.com/problems/evaluate-division/

Problem Description: You are given an array of equations "equations" and an array of real numbers "values", where equations[i] = [A[i], B[i]] and values[i] represent the equation A[i] / B[i] = values[i]. Each A[i] and B[i] is a string that represents a single variable.

You are also given some queries, where queries[j] = [C[j], D[j]] represents the jth query where you must find the answer for C[j] / D[j] = ?.

You are given the array equations and the array values. Return an array of all the answers. If the answer does not exist, return -1.0.

Note: The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.

Example: Input: equations = [["a","b"],["b","c"]], values = [2.0,3.0], queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]] Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]

1. Optimized Java code for Evaluate Division:

import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        
        // Creating adjacency matrix to store the values of equations
        Map<String, Map<String, Double>> adjacencyMatrix = buildAdjacencyMatrix(equations, values);
        
        int n = queries.size();
        double[] results = new double[n];
        
        for (int i = 0; i < n; i++) {
            
            List<String> query = queries.get(i);
            String source = query.get(0);
            String target = query.get(1);
            
            if (!adjacencyMatrix.containsKey(source) || !adjacencyMatrix.containsKey(target)) {
                // If either the source or target variable is not in the equation list, answer does not exist.
                results[i] = -1.0;
            } else if (source.equals(target)) {
                // If source and target are the same variable, result is 1.0
                results[i] = 1.0;
            } else {
                // Perform DFS on the adjacency matrix to find the answer
                results[i] = dfs(adjacencyMatrix, source, target, new HashSet<String>(), 1.0);
            }
        }
        
        return results;
    }
    
    // Function to build the adjacency matrix
    private Map<String, Map<String, Double>> buildAdjacencyMatrix(List<List<String>> equations, double[] values) {
        
        Map<String, Map<String, Double>> adjacencyMatrix = new HashMap<>();
        
        for (int i = 0; i < equations.size(); i++) {
            List<String> equation = equations.get(i);
            String source = equation.get(0);
            String target = equation.get(1);
            double value = values[i];
            
            // Adding the equation value to the adjacency matrix in both forward and backward direction
            adjacencyMatrix.computeIfAbsent(source, k -> new HashMap<String, Double>()).put(target, value);
            adjacencyMatrix.computeIfAbsent(target, k -> new HashMap<String, Double>()).put(source, 1.0 / value);
        }
        
        return adjacencyMatrix;
    }
    
    // DFS function to find the answer using the adjacency matrix
    private double dfs(Map<String, Map<String, Double>> adjacencyMatrix, String source, String target, Set<String> visited, double value) {
        
        // Adding the current variable to visited set
        visited.add(source);
        
        if (!adjacencyMatrix.containsKey(source)) {
            // If source variable is not in the adjacency matrix, answer does not exist.
            return -1.0;
        }
        
        if (source.equals(target)) {
            // If source and target are the same variable, return the accumulated value
            return value;
        }
        
        Map<String, Double> neighbors = adjacencyMatrix.get(source);
        for (Map.Entry<String, Double> neighbor : neighbors.entrySet()) {
            String next = neighbor.getKey();
            double val = neighbor.getValue();
            
            if (!visited.contains(next)) {
                // Recursively call the DFS function on the next variable
                double result = dfs(adjacencyMatrix, next, target, visited, value * val);
                if (result != -1.0) {
                    // If the answer is found in the next recursion, return the accumulated value
                    return result;
                }
            }
        }
        
        // If no answer is found, return -1.0
        return -1.0;
    }
}

1. Algorithm Explanation and Interview Approach: The given problem can be solved using the Graph data structure. In this problem, we need to evaluate the division between variables. We can represent the variables and their division values as an adjacency matrix, where each variable is a node and the division value is the weighted edge between two nodes.

To solve this problem, we can follow the steps below:

1. Build the adjacency matrix using the given equations and values. Create an empty adjacency matrix as a HashMap of HashMaps, where each inner HashMap represents the weighted edges for a node (variable).

2. Loop through the equations and values to populate the adjacency matrix. For each equation, add the division values to the adjacency matrix in both forward and backward directions.

3. For each query, perform a Depth-First Search (DFS) on the adjacency matrix to find the answer. If either the source or target variable is not in the adjacency matrix, return -1.0. If the source and target variables are the same, return 1.0. Otherwise, recursively call the DFS function on the neighbors of the source variable until the target variable is reached or no more neighbors are available.

4. Return the results for all the queries.

During an interview, it is essential to communicate the approach clearly and demonstrate your understanding of graph traversal algorithms like Depth-First Search. You can discuss the time and space complexity and potential optimization techniques like memoization to optimize repeated DFS calls. Additionally, you can also analyze other possible approaches like Breadth-First Search or the Floyd-Warshall algorithm for solving this problem.

6. Regions Cut By Slashes (LeetCode #959)

1. Problem Description: The problem "Regions Cut By Slashes" is to find the number of regions formed by slashes in a given grid. The grid is represented as a 2D string array, where each cell can be empty (' '), a forward slash ('/'), or a backward slash (''). Slashes divide the grid into regions. The goal is to count the number of regions. You can find the problem on LeetCode using the following link: https://leetcode.com/problems/regions-cut-by-slashes/

2. Java Code:

class Solution {
    public int regionsBySlashes(String[] grid) {
        int n = grid.length;
        int size = n * n * 4; // maximum number of nodes
        
        // Initialize the DSU (Disjoint Set Union) data structure
        int[] parent = new int[size];
        int[] rank = new int[size];
        for (int i = 0; i < size; i++) {
            parent[i] = i;
            rank[i] = 1;
        }
        
        int regions = size;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                char cell = grid[i].charAt(j);
                int root = 4 * (i * n + j);
                
                // Connect the nodes based on the slashes
                if (cell == '/') {
                    int union1 = find(parent, root + 0);
                    int union2 = find(parent, root + 1);
                    if (union1 != union2) {
                        union(parent, rank, union1, union2);
                        regions--;
                    }
                    
                    union1 = find(parent, root + 2);
                    union2 = find(parent, root + 3);
                    if (union1 != union2) {
                        union(parent, rank, union1, union2);
                        regions--;
                    }
                    
                    union1 = find(parent, root + 0);
                    union2 = find(parent, root + 3);
                    if (union1 != union2) {
                        union(parent, rank, union1, union2);
                        regions--;
                    }
                    
                } else if (cell == '\\') {
                    int union1 = find(parent, root + 0);
                    int union2 = find(parent, root + 3);
                    if (union1 != union2) {
                        union(parent, rank, union1, union2);
                        regions--;
                    }
                    
                    union1 = find(parent, root + 1);
                    union2 = find(parent, root + 2);
                    if (union1 != union2) {
                        union(parent, rank, union1, union2);
                        regions--;
                    }
                    
                } else {
                    // cell is empty, connect all sides
                    union(parent, rank, root + 0, root + 1);
                    union(parent, rank, root + 1, root + 2);
                    union(parent, rank, root + 2, root + 3);
                    regions -= 3; // decrement regions by 3 because all sides are connected
                }
            }
        }
        
        return regions;
    }
    
    // Find the root of a node using path compression
    private int find(int[] parent, int i) {
        if (parent[i] != i) {
            parent[i] = find(parent, parent[i]);
        }
        return parent[i];
    }
    
    // Union the nodes based on rank to optimize the DSU
    private void union(int[] parent, int[] rank, int i, int j) {
        if (rank[i] >= rank[j]) {
            parent[j] = i;
            rank[i] += rank[j];
        } else {
            parent[i] = j;
            rank[j] += rank[i];
        }
    }
}

1. Algorithm Description:

• The problem can be solved using the concept of Disjoint Set Union (DSU).

• We initialize each cell in the grid as a separate node and maintain their parent and rank arrays.

• We iterate through each cell in the grid and handle the three cases: '/', '', and empty.

• For a '/', we connect the appropriate nodes if they are not already connected and decrement the regions count.

• For a '', we connect the appropriate nodes if they are not already connected and decrement the regions count.

• For an empty cell, we connect all the sides of the cell and decrement the regions count by 3.

• Finally, we return the regions count as the result.

During the interview process, it is important to explain the problem clearly, present and explain your algorithm, and provide a well-commented and optimized code solution. Additionally, you can discuss the runtime and space complexity of the solution and any possible further optimizations or alternative approaches.

7. Friend Circles (LeetCode #547)

1. Problem Description: The problem "Friend Circles" is a commonly asked interview question on LeetCode. You can find the problem description and test cases on LeetCode's website using the following link: https://leetcode.com/problems/friend-circles/

The problem statement is as follows:

You are given a 2D array, representing an M x M matrix, where each cell contains either 0 or 1. If matrix[i][j] == 1, then person i and person j are friends. This friendship is transitive, which means if A is a friend of B, and B is a friend of C, then A is also a friend of C.

You need to return the total number of friend circles in the given matrix.

1. Optimized Java Code with Detailed Comments:

class Solution {
    public int findCircleNum(int[][] M) {
        int n = M.length;
        boolean[] visited = new boolean[n];
        int count = 0;
        
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                // Start a new friend circle
                dfs(M, i, visited);
                count++;
            }
        }
        
        return count;
    }
    
    // Depth-first search to find all friends in a circle
    private void dfs(int[][] M, int i, boolean[] visited) {
        visited[i] = true;
        
        for (int j = 0; j < M.length; j++) {
            if (M[i][j] == 1 && !visited[j]) {
                dfs(M, j, visited);
            }
        }
    }
}

1. Detailed Description:

The problem can be solved using a depth-first search (DFS) algorithm. We start with the first person, and for each person, we perform a DFS to find all other people in the same friend circle.

In the main function, we initialize a boolean array visited to keep track of whether a person has been visited or not. We also initialize a count variable to keep track of the number of friend circles.

We then iterate through each person in the given matrix. If the person has not been visited yet, it means we have found a new friend circle. We call the dfs function to perform a DFS on that person, marking all visited persons in the same circle.

The dfs function takes the matrix, the current person index, and the visited array as parameters. We mark the current person as visited and then iterate through all other persons. If two persons are friends (i.e., the matrix value is 1) and the friend has not been visited yet, we recursively call the dfs function for that friend. This process continues until we have visited all possible friends in the circle.

By counting the number of times we call the dfs function, we can determine the total number of friend circles.

During the interview process, it is important to clearly understand the problem requirements and constraints. It is also crucial to explain your approach to the interviewer and discuss the time and space complexity of your solution. Additionally, mentioning the chosen algorithm and explaining how it works can showcase your understanding of the problem and your problem-solving skills.

8. Most Stones Removed with Same Row or Column (LeetCode #947)

1. Problem Description:

The problem "Most Stones Removed with Same Row or Column" (LeetCode #947) can be found at the following link: https://leetcode.com/problems/most-stones-removed-with-same-row-or-column/

In this problem, we are given a list of stones represented by their coordinates on a 2D plane. Each stone is represented by a pair of integers (x, y), representing its x-coordinate and y-coordinate respectively. A stone can be removed if and only if there is another stone in the same row or the same column.

The goal is to determine the maximum number of stones that can be removed.

1. Optimized Java Code:

Here is the optimized Java code for the given problem:

class Solution {
    public int removeStones(int[][] stones) {
        int n = stones.length;
        int maxStones = 0;
        
        // Create a graph to represent the stones
        HashMap<Integer, ArrayList<Integer>> graph = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int x = stones[i][0];
            int y = stones[i][1] + 10000; // To differentiate x and y coordinates
            
            if (!graph.containsKey(x)) {
                graph.put(x, new ArrayList<>());
            }
            if (!graph.containsKey(y)) {
                graph.put(y, new ArrayList<>());
            }
            
            // Connect stones with same x-coordinate and y-coordinate
            graph.get(x).add(y);
            graph.get(y).add(x);
        }
        
        // Perform DFS on all stones to find connected stones
        HashSet<Integer> visited = new HashSet<>();
        for (int[] stone : stones) {
            maxStones += dfs(graph, stone[0], visited) - 1;
        }
        
        return maxStones;
    }
    
    private int dfs(HashMap<Integer, ArrayList<Integer>> graph, int stone, HashSet<Integer> visited) {
        if (visited.contains(stone)) return 0;
        
        visited.add(stone);
        int count = 1;
        
        for (int neighbor : graph.get(stone)) {
            count += dfs(graph, neighbor, visited);
        }
        
        return count;
    }
}

1. Algorithm and Approach:

The problem can be solved using a graph-based approach. We can create a graph where each stone is represented by a node, and an edge connects two stones if they are in the same row or the same column.

Based on the graph, we need to find the connected components or clusters of stones. A connected component is a group of stones that are connected by edges. In each connected component, we can remove all stones except one, as removing any stone will still keep the remaining stones connected.

To find the connected components, we can perform a Depth-First Search (DFS) starting from each stone. We keep track of the number of stones in each connected component and add them to the result. Finally, we return the total number of stones that can be removed.

During an interview, it is important to explain the algorithm and approach in a structured manner. Start by describing the problem, providing the problem link, and then explain the algorithm step by step, highlighting the key components like graph creation, DFS, and counting the stones. It is also recommended to discuss the time and space complexity of the solution.

Remember to write clean and well-commented code, mentioning the purpose of each segment and explaining the logic wherever necessary.

9. Number of Connected Components in an Undirected Graph (LeetCode #323)

1. Problem Description and Link: The problem "Number of Connected Components in an Undirected Graph" can be found on LeetCode here: https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/

2. Java Solution with Detailed Comments:

class Solution {
    // Method to count the number of connected components in an undirected graph
    public int countComponents(int n, int[][] edges) {
        // Create an array to keep track of visited nodes in the graph
        boolean[] visited = new boolean[n];
        
        // Create an adjacency list representation of the graph
        List<List<Integer>> adjList = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adjList.add(new ArrayList<>());
        }
        
        // Build the adjacency list by adding edges to the respective nodes
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adjList.get(u).add(v);
            adjList.get(v).add(u);
        }
        
        // Variable to store the count of connected components
        int count = 0;
        
        // Iterate through each node in the graph
        for (int i = 0; i < n; i++) {
            // If the node is not visited, perform DFS to find connected components
            if (!visited[i]) {
                dfs(i, adjList, visited);
                count++;
            }
        }
        
        // Return the count of connected components
        return count;
    }
    
    // Depth First Search (DFS) method to traverse the graph and mark nodes as visited
    private void dfs(int node, List<List<Integer>> adjList, boolean[] visited) {
        // Mark the current node as visited
        visited[node] = true;
        
        // Iterate through the neighboring nodes of the current node
        for (int neighbor : adjList.get(node)) {
            // If the neighbor is not visited, recursively call DFS on it
            if (!visited[neighbor]) {
                dfs(neighbor, adjList, visited);
            }
        }
    }
}

1. Detailed Description and Approach: During an interview, follow the following approach to solve the problem:

• Understand the problem statement and ask clarifying questions if needed.

• Recognize that the problem can be solved using Depth First Search (DFS) algorithm.

• Create an array to keep track of visited nodes in the graph.

• Build an adjacency list representation of the graph using the given edges.

• Iterate through each node in the graph.

• If the node is not visited, perform DFS to find connected components.

• In the DFS method, mark the current node as visited and recursively call DFS on its neighboring nodes.

• Count the number of times DFS is called, as it indicates the number of connected components.

• Return the count of connected components.

This solution has a time complexity of O(n + m), where n is the number of nodes and m is the number of edges. It uses DFS to traverse the graph and mark nodes as visited, ensuring that each node is visited exactly once.

10. Satisfiability of Equality Equations (LeetCode #990)

1. Problem Description and LeetCode Link: The problem "Satisfiability of Equality Equations" is about determining if a given set of equations can be satisfied or not. The problem statement can be found on LeetCode at the following link: https://leetcode.com/problems/satisfiability-of-equality-equations/

2. Optimized Java Code with Detailed Comments:

class Solution {
    // Helper function to find the root of a character in the parent array
    private int findRoot(int[] parent, int i) {
        if (parent[i] == i) {
            return i;
        }
        return findRoot(parent, parent[i]);
    }
    
    public boolean equationsPossible(String[] equations) {
        // Create a parent array to track the root of each character
        int[] parent = new int[26];
        for (int i = 0; i < 26; i++) {
            parent[i] = i; // Initially, each character is its own root
        }
        
        // Union operation: If two characters are equal, union their roots
        for (String equation : equations) {
            if (equation.charAt(1) == '=') {
                int x = equation.charAt(0) - 'a';
                int y = equation.charAt(3) - 'a';
                int xRoot = findRoot(parent, x);
                int yRoot = findRoot(parent, y);
                if (xRoot != yRoot) {
                    parent[xRoot] = yRoot;
                }
            }
        }
        
        // Check if any inequality equations violate the union-find structure
        for (String equation : equations) {
            if (equation.charAt(1) == '!') {
                int x = equation.charAt(0) - 'a';
                int y = equation.charAt(3) - 'a';
                int xRoot = findRoot(parent, x);
                int yRoot = findRoot(parent, y);
                if (xRoot == yRoot) {
                    return false;
                }
            }
        }
        
        return true;
    }
}

1. Algorithm and Thought Process during an Interview:

During the interview, you can approach this problem using the Union-Find algorithm.

Basically, the idea is to first create a parent array where each character is initially its own root. Then, we iterate through the given equations and perform the union operation when we encounter equality equations (=). This union operation involves finding the root of each character and merging the roots if they are different.

After performing the union operation, we then iterate through the equations again and check if there are any inequality equations (!) that violate the union-find structure. We do this by finding the roots of the characters in each inequality equation and checking if they are the same.

If we find any inequality equation that violates the union-find structure, we return false, indicating that the set of equations cannot be satisfied. Otherwise, if all inequality equations are satisfied, we return true.

During the interview, it's important to explain the logic behind the algorithm, the union-find data structure, and how the code achieves the desired functionality. You can also mention the time and space complexity of the solution, which is O(n) where n is the total number of equations.