Graph

27. Clone Graph

1. Problem: Clone Graph

   • Leetcode link: https://leetcode.com/problems/clone-graph/

2. Optimized Java code for Clone Graph with detailed comments:

// Definition for a Node.
class Node {
    int val;
    List<Node> neighbors;

    public Node(int val) {
        this.val = val;
        this.neighbors = new ArrayList<Node>();
    }
}

class Solution {
    // Map to store nodes already cloned
    Map<Node, Node> map = new HashMap<>();

    public Node cloneGraph(Node node) {
        // Return null if input node is null
        if (node == null) return null;

        // Return cloned node if it is already present in the map
        if (map.containsKey(node))
            return map.get(node);

        // Create the clone for the current node
        Node cloneNode = new Node(node.val);
        // Put the clone node in the map
        map.put(node, cloneNode);

        // Iterate through the neighbors of the current node
        for (Node neighbor : node.neighbors) {
            // Clone each neighbor and add it to the clone node's neighbors list
            cloneNode.neighbors.add(cloneGraph(neighbor));
        }

        return cloneNode;
    }
}

1. Detailed description of the algorithm and how to think through it during the interview process:

The problem requires cloning a graph, which means creating a deep copy of the given graph such that the structure and values of each node and its neighbors are preserved. The graph is represented by a list of nodes, each containing a value and a list of its neighbors.

To solve this problem, we can use a depth-first search (DFS) approach along with recursion. The basic idea is to traverse the original graph and create a clone for each node, while also adding the cloned nodes to a map to keep track of them. This map will help us avoid duplicating nodes and handle cycles in the graph.

When cloning a node, we first check if it has already been cloned and present in the map. If so, we can directly return the cloned node from the map. If not, we create a new clone node with the same value as the original node, add it to the map, and then recursively clone the neighbors of the current node. The recursive call will create clones for the neighbors and add them to the clone node's neighbors list.

By following this approach, we can recursively clone all the nodes in the graph and maintain the proper connections between them. The map ensures that we don't duplicate nodes and handle cycles properly.

During the interview process, it is important to explain the DFS approach and the concept of cloning nodes and using a map to keep track of cloned nodes. Taking the interviewer through the thought process and reasoning behind the solution shows understanding and problem-solving skills.

1. Problems that use a similar algorithm:

   • Clone Binary Tree (https://leetcode.com/problems/clone-binary-tree/)

   • Clone N-ary Tree (https://leetcode.com/problems/clone-n-ary-tree/)

   • Copy List with Random Pointer (https://leetcode.com/problems/copy-list-with-random-pointer/)

2. Similar problems with detailed Java code and descriptions:

a) Problem: Clone Binary Tree

• Leetcode link: https://leetcode.com/problems/clone-binary-tree/

// Definition for a binary tree node.
class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    public TreeNode(int val) {
        this.val = val;
        this.left = null;
        this.right = null;
    }
}

class Solution {
    // Map to store clones of tree nodes
    Map<TreeNode, TreeNode> map = new HashMap<>();

    public TreeNode cloneTree(TreeNode root) {
        // Return null if root is null
        if (root == null) return null;

        // Return the cloned root if it is already present in the map
        if (map.containsKey(root))
            return map.get(root);

        // Create a clone of the root node
        TreeNode cloneNode = new TreeNode(root.val);
        // Add the clone node to the map
        map.put(root, cloneNode);

        // Recursively clone the left and right subtrees of the root
        cloneNode.left = cloneTree(root.left);
        cloneNode.right = cloneTree(root.right);

        return cloneNode;
    }
}

b) Problem: Clone N-ary Tree

• Leetcode link: https://leetcode.com/problems/clone-n-ary-tree/

// Definition for a Node.
class Node {
    int val;
    List<Node> children;

    public Node(int val) {
        this.val = val;
        this.children = new ArrayList<Node>();
    }
}

class Solution {
    // Map to store clones of tree nodes
    Map<Node, Node> map = new HashMap<>();

    public Node cloneTree(Node root) {
        // Return null if root is null
        if (root == null) return null;

        // Return the cloned root if it is already present in the map
        if (map.containsKey(root))
            return map.get(root);

        // Create a clone of the root node
        Node cloneNode = new Node(root.val);
        // Add the clone node to the map
        map.put(root, cloneNode);

        // Recursively clone each child of the root
        for (Node child : root.children) {
            cloneNode.children.add(cloneTree(child));
        }

        return cloneNode;
    }
}

c) Problem: Copy List with Random Pointer

• Leetcode link: https://leetcode.com/problems/copy-list-with-random-pointer/

// Definition for a Node.
class Node {
    int val;
    Node next;
    Node random;

    public Node(int val) {
        this.val = val;
        this.next = null;
        this.random = null;
    }
}

class Solution {
    // Map to store clones of list nodes
    Map<Node, Node> map = new HashMap<>();

    public Node copyRandomList(Node head) {
        // Return null if head is null
        if (head == null) return null;

        // Return the cloned head if it is already present in the map
        if (map.containsKey(head))
            return map.get(head);

        // Create a clone of the current node
        Node cloneNode = new Node(head.val);
        // Add the clone node to the map
        map.put(head, cloneNode);

        // Recursively clone the next and random pointers
        cloneNode.next = copyRandomList(head.next);
        cloneNode.random = copyRandomList(head.random);

        return cloneNode;
    }
}

In all these problems, the same DFS approach along with recursion and using a map to store clones is used to solve the problem of cloning a given tree or list.

28. Course Schedule

1. Problem Description and Link: Problem: Course Schedule Link: https://leetcode.com/problems/course-schedule/

2. Optimized Java Code for "Course Schedule" with detailed comments:

import java.util.*;

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        // Construct the graph using adjacency list
        List<List<Integer>> adjacencyList = new ArrayList<>(numCourses);
        for (int i = 0; i < numCourses; i++) {
            adjacencyList.add(new ArrayList<Integer>());
        }
        
        // Populate the adjacency list
        for (int[] prerequisite : prerequisites) {
            int course = prerequisite[0];
            int prerequisiteCourse = prerequisite[1];
            adjacencyList.get(course).add(prerequisiteCourse);
        }
        
        // Track the visit status of each course (0: not visited, 1: visiting, 2: visited)
        int[] visitStatus = new int[numCourses];
        
        // Check if there is a cycle in the graph using DFS
        for (int i = 0; i < numCourses; i++) {
            if (hasCycle(adjacencyList, visitStatus, i)) {
                return false;
            }
        }
        
        // If there is no cycle, return true
        return true;
    }
    
    private boolean hasCycle(List<List<Integer>> adjacencyList, int[] visitStatus, int course) {
        // If the current course is being visited, there is a cycle
        if (visitStatus[course] == 1) {
            return true;
        }
        
        // If the current course has already been visited, there is no cycle
        if (visitStatus[course] == 2) {
            return false;
        }
        
        // Mark the current course as visiting
        visitStatus[course] = 1;
        
        // Traverse the prerequisites of the current course
        for (int prerequisite : adjacencyList.get(course)) {
            if (hasCycle(adjacencyList, visitStatus, prerequisite)) {
                return true;
            }
        }
        
        // Mark the current course as visited
        visitStatus[course] = 2;
        
        return false;
    }
}

1. Algorithm Description and Interview Tips: The algorithm solves the problem using a depth-first search (DFS) approach. It constructs a graph using an adjacency list and then checks if there is a cycle in the graph. If there is a cycle, it means that the course schedule is not possible. Otherwise, it returns true.

During an interview, it is important to understand the problem requirements and constraints. The problem requires determining if a given course schedule is possible, given a list of prerequisites. The algorithm constructs a graph using an adjacency list to represent the prerequisites and then checks if there is a cycle using DFS.

To think through the problem during an interview, you could follow these steps:

• Understand the problem requirements and constraints.

• Identify the data structure needed to represent the graph (adjacency list in this case) and construct the graph based on the prerequisites.

• Implement the DFS algorithm to check if there is a cycle in the graph.

• If there is a cycle, return false. Otherwise, return true.

1. Related Problems: The algorithm used in the "Course Schedule" problem can be applied to similar problems that involve detecting cycles in a directed graph. Some examples are:

• Course Schedule II (https://leetcode.com/problems/course-schedule-ii/)

• Course Schedule III (https://leetcode.com/problems/course-schedule-iii/)

1. Similar Problems with Java Code:

a. Course Schedule II:

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        List<List<Integer>> adjacencyList = new ArrayList<>(numCourses);
        for (int i = 0; i < numCourses; i++) {
            adjacencyList.add(new ArrayList<>());
        }
        int[] inDegrees = new int[numCourses];
        
        for (int[] prerequisite : prerequisites) {
            int course = prerequisite[0];
            int prerequisiteCourse = prerequisite[1];
            adjacencyList.get(prerequisiteCourse).add(course);
            inDegrees[course]++;
        }
        
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < numCourses; i++) {
            if (inDegrees[i] == 0) {
                queue.offer(i);
            }
        }
        
        int[] result = new int[numCourses];
        int index = 0;
        
        while (!queue.isEmpty()) {
            int course = queue.poll();
            result[index++] = course;
            for (int adjacentCourse : adjacencyList.get(course)) {
                inDegrees[adjacentCourse]--;
                if (inDegrees[adjacentCourse] == 0) {
                    queue.offer(adjacentCourse);
                }
            }
        }
        
        if (index < numCourses) {
            return new int[0];
        }
        
        return result;
    }
}

b. Course Schedule III:

import java.util.*;

class Solution {
    public int scheduleCourse(int[][] courses) {
        Arrays.sort(courses, (a, b) -> a[1] - b[1]);
        
        PriorityQueue<Integer> courseDurations = new PriorityQueue<>((a, b) -> b - a);
        int currentTime = 0;
        
        for (int[] course : courses) {
            if (currentTime + course[0] <= course[1]) {
                courseDurations.offer(course[0]);
                currentTime += course[0];
            } else if (!courseDurations.isEmpty() && courseDurations.peek() > course[0]) {
                currentTime -= courseDurations.poll() - course[0];
                courseDurations.offer(course[0]);
            }
        }
        
        return courseDurations.size();
    }
}

These solutions demonstrate how the algorithm used in the "Course Schedule" problem can be applied to similar problems by modifying some parts of the code.

29. Pacific Atlantic Water Flow

1. Problem Description: The problem "Pacific Atlantic Water Flow" is a leetcode problem with a unique twist. Given a matrix of heights representing the height of the land at each cell, we need to find all the cells from where water can flow to both the Pacific Ocean and the Atlantic Ocean. Link: https://leetcode.com/problems/pacific-atlantic-water-flow/

2. Optimized Java code with detailed comments:

class Solution {
    public List<List<Integer>> pacificAtlantic(int[][] matrix) {
        // Create a result list to store the cells that can flow to both oceans
        List<List<Integer>> result = new ArrayList<>();
        
        // Check for invalid inputs
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return result;
        }
        
        int rows = matrix.length;
        int cols = matrix[0].length;
        
        // Create two boolean arrays to mark the cells that can flow to each ocean
        boolean[][] pacific = new boolean[rows][cols];
        boolean[][] atlantic = new boolean[rows][cols];
        
        // Start from the boundaries and perform DFS to mark the cells that can flow to the respective oceans
        for (int i = 0; i < rows; i++) {
            dfs(matrix, pacific, Integer.MIN_VALUE, i, 0);
            dfs(matrix, atlantic, Integer.MIN_VALUE, i, cols - 1);
        }
        
        for (int j = 0; j < cols; j++) {
            dfs(matrix, pacific, Integer.MIN_VALUE, 0, j);
            dfs(matrix, atlantic, Integer.MIN_VALUE, rows - 1, j);
        }
        
        // Check for the cells that can flow to both oceans and add them to the result list
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (pacific[i][j] && atlantic[i][j]) {
                    List<Integer> cell = new ArrayList<>();
                    cell.add(i);
                    cell.add(j);
                    result.add(cell);
                }
            }
        }
        
        return result;
    }
    
    private void dfs(int[][] matrix, boolean[][] ocean, int prevHeight, int row, int col) {
        // Check for out of bounds and invalid heights
        if (row < 0 || row >= matrix.length || col < 0 || col >= matrix[0].length || ocean[row][col] || matrix[row][col] < prevHeight) {
            return;
        }
        
        // Mark the cell as reachable to the ocean
        ocean[row][col] = true;
        
        // Perform DFS recursively on the neighboring cells
        dfs(matrix, ocean, matrix[row][col], row - 1, col);
        dfs(matrix, ocean, matrix[row][col], row + 1, col);
        dfs(matrix, ocean, matrix[row][col], row, col - 1);
        dfs(matrix, ocean, matrix[row][col], row, col + 1);
    }
}

1. Algorithm Description: To solve this problem, we can use a Depth First Search (DFS) approach. The idea is to start from the boundaries of the matrix and perform DFS to mark all the cells that can reach the Pacific Ocean and Atlantic Ocean separately. Finally, we find the cells that can reach both oceans and add them to the result list.

During the interview process, it's important to discuss the problem requirements and clarify any doubts. We can communicate our approach to the interviewer and explain the reasons behind choosing DFS as our solution. By giving a step-by-step explanation of the code, including the purpose and functionality of each component, we can showcase our understanding of the problem and our ability to write clean and optimized code.

1. Similar Algorithm Application: The algorithm used in the "Pacific Atlantic Water Flow" problem, which involves determining reachable cells based on certain conditions, can be applied to various other problems with similar requirements. These include:

• Finding all the connected components in a graph.

• Determining the areas affected by a flood or fire.

• Identifying the regions reachable by different sources, such as power distribution or network coverage.

1. Similar Problems with Java Code and Descriptions: Below is a list of similar problems that can be solved using the same algorithm:

• Problem: "Number of Islands" (https://leetcode.com/problems/number-of-islands/) Description: Given a 2D grid map of '1's (land) and '0's (water), count the number of islands.

  class Solution {
      public int numIslands(char[][] grid) {
          // Same DFS approach as "Pacific Atlantic Water Flow"
      }
  }

• Problem: "Flood Fill" (https://leetcode.com/problems/flood-fill/) Description: An image is represented by a 2D integer array. Given a starting pixel (row and column) and a new color, fill the surrounding area until the boundary is reached.

  class Solution {
      public int[][] floodFill(int[][] image, int sr, int sc, int newColor) {
          // Same DFS approach as "Pacific Atlantic Water Flow"
      }
  }

• Problem: "Surrounded Regions" (https://leetcode.com/problems/surrounded-regions/) Description: Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

  class Solution {
      public void solve(char[][] board) {
          // Same DFS approach as "Pacific Atlantic Water Flow"
      }
  }

By practicing and understanding the algorithm used in "Pacific Atlantic Water Flow", we can easily tackle these similar problems during interview discussions or coding challenges.

30. Number of Islands

1. Problem Description: The "Number of Islands" problem on LeetCode can be found at the following link: https://leetcode.com/problems/number-of-islands/

Given a 2D grid 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 surrounded by water.

Example: Input: 11000 11000 00100 00011

Output: 3

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

class Solution {
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }
        
        int numIslands = 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') {
                    numIslands++;
                    dfs(grid, i, j);
                }
            }
        }
        
        return numIslands;
    }
    
    private void dfs(char[][] grid, int row, int col) {
        int rows = grid.length;
        int cols = grid[0].length;
        
        if (row < 0 || col < 0 || row >= rows || col >= cols || grid[row][col] == '0') {
            return;
        }
        
        grid[row][col] = '0'; // mark current cell as visited
        
        // recursively visit all adjacent cells
        dfs(grid, row + 1, col);
        dfs(grid, row - 1, col);
        dfs(grid, row, col + 1);
        dfs(grid, row, col - 1);
    }
}

1. Algorithm Description:

In this problem, we use a depth-first search (DFS) approach to count the number of islands.

1. Iterate over the entire 2D grid.

2. If a cell contains '1', increment the island count and call the DFS function on this cell.

3. In the DFS function, mark the current cell as visited by changing the value to '0'.

4. Recursively explore all adjacent cells (up, down, left, right) and mark them as visited.

5. Keep track of the number of islands visited.

During an interview, it is important to explain the algorithm step by step, emphasizing the DFS approach and explaining how it helps identify connected land cells and count islands.

1. Problem Variations: The algorithm used for the "Number of Islands" problem can also be applied to solve similar problems that involve identifying connected components in graphs or grids. Some of these problems include:

• "Surrounded Regions": The goal is to flip all 'O's in a board surrounded by 'X' into 'X'. (https://leetcode.com/problems/surrounded-regions/)

• "Max Area of Island": Find the maximum area of an island in a grid. (https://leetcode.com/problems/max-area-of-island/)

• "Number of Connected Components in an Undirected Graph": Given an undirected graph with n nodes and edges represented by an edge list, count the number of connected components. (https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/)

1. Similar Problems:

a) "Number of Islands II": Given a 2D grid, add islands one by one and return the number of islands after each addition. (https://leetcode.com/problems/number-of-islands-ii/)

class Solution {
    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        int[] roots = new int[m * n];
        Arrays.fill(roots, -1);
        
        int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        List<Integer> result = new ArrayList<>();
        int count = 0;
        
        for (int[] pos : positions) {
            int row = pos[0];
            int col = pos[1];
            int index = row * n + col;
            
            if (roots[index] == -1) {
                roots[index] = index;
                count++;

                for (int[] dir : dirs) {
                    int newRow = row + dir[0];
                    int newCol = col + dir[1];
                    int newIndex = newRow * n + newCol;
                    
                    if (newRow >= 0 && newRow < m && newCol >= 0 && newCol < n && roots[newIndex] != -1) {
                        int rootIndex = find(root, newIndex);
                        if (index != rootIndex) {
                            roots[rootIndex] = index;
                            count--;
                        }
                    }
                }
            }
            
            result.add(count);
        }
        
        return result;
    }
    
    private int find(int[] roots, int index) {
        while (roots[index] != index) {
            roots[index] = roots[roots[index]];
            index = roots[index];
        }
        
        return index;
    }
}

b) "Number of Distinct Islands": Given a 2D grid of 1's (land) and 0's (water), find the number of distinct islands. Two islands are considered distinct if and only if they have the same shape (not considering rotations or reflections). (https://leetcode.com/problems/number-of-distinct-islands/)

class Solution {
    public int numDistinctIslands(int[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }
        
        Set<String> islands = new HashSet<>();
        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) {
                    StringBuilder path = new StringBuilder();
                    dfs(grid, i, j, path, "o");
                    islands.add(path.toString());
                }
            }
        }
        
        return islands.size();
    }
    
    private void dfs(int[][] grid, int row, int col, StringBuilder path, String dir) {
        int rows = grid.length;
        int cols = grid[0].length;
        
        if (row < 0 || col < 0 || row >= rows || col >= cols || grid[row][col] == 0) {
            return;
        }
        
        grid[row][col] = 0; // mark current cell as visited
        
        path.append(dir);
        
        dfs(grid, row + 1, col, path, "d");
        dfs(grid, row - 1, col, path, "u");
        dfs(grid, row, col + 1, path, "r");
        dfs(grid, row, col - 1, path, "l");
        
        path.append("b"); // backtracking step
    }
}

These are just a few examples of similar problems that follow similar DFS-based approaches.

31. Longest Consecutive Sequence

1. Problem description and problem link: Problem: Longest Consecutive Sequence - LeetCode #128 Given an unsorted array of integers, find the length of the longest consecutive elements sequence. Link: https://leetcode.com/problems/longest-consecutive-sequence/

2. Optimized Java code for "Longest Consecutive Sequence" problem:

import java.util.HashSet;

class Solution {
    public int longestConsecutive(int[] nums) {
        // Create a hash set to hold all the numbers present in the array
        HashSet<Integer> set = new HashSet<>();
        
        // Add all numbers to the set
        for (int num : nums) {
            set.add(num);
        }
        
        int longestStreak = 0;
        
        // Iterate through each number in the set
        for (int num : set) {
            // Check if the number is the start of a sequence
            if (!set.contains(num - 1)) {
                int currentNum = num;
                int currentStreak = 1;
                
                // Check the length of the consecutive sequence starting from the current number
                while (set.contains(currentNum + 1)) {
                    currentNum += 1;
                    currentStreak += 1;
                }
                
                // Update the longest streak if necessary
                longestStreak = Math.max(longestStreak, currentStreak);
            }
        }
        
        return longestStreak;
    }
}

1. Algorithm explanation and interview thoughts: The problem asks us to find the length of the longest consecutive elements sequence in an unsorted array. To solve this problem efficiently, we can use a HashSet to store all the numbers in the array. Then, we iterate through each number in the HashSet. For each number, we check if it is the start of a sequence by verifying if the number minus one is present in the HashSet. If it is not present, we start counting the length of the consecutive sequence by iterating from the current number and increasing it by one until we reach the end of the sequence. We keep track of the longest streak we find and return it at the end.

   During an interview, it is crucial to communicate your approach clearly. You should explain the chosen data structure (HashSet in this case) and elaborate on how it helps in solving the problem efficiently. Mention that using a HashSet, we can quickly check if a number is present in the set or not. Discuss the intuition behind iterating through each number and counting the length of consecutive sequences. Also, be ready to discuss the time and space complexity of your solution.

2. Abstracted problems that use a similar algorithm: The algorithm used in the "Longest Consecutive Sequence" problem can be applied to solve similar problems that require finding consecutive elements.

   Some abstracted problems that can be solved using a similar algorithm are:

   • Maximum Gaps - LeetCode #164 (https://leetcode.com/problems/maximum-gap/) (Find the maximum difference between two sorted adjacent elements in an unsorted array.)

   • Longest Subarray of 1's After Deleting One Element - LeetCode #1493 (Find the length of the longest subarray of consecutive 1's after deleting one element in a binary array.)

   • Binary Tree Longest Consecutive Sequence - LeetCode #298 (Find the length of the longest consecutive sequence in a binary tree.)

3. Similar problems with detailed Java code and descriptions:

   A. Maximum Gaps - LeetCode #164 Problem: Given an unsorted array, find the maximum gap between any two adjacent elements after sorting the array. Link: https://leetcode.com/problems/maximum-gap/

   import java.util.Arrays;
   
   class Solution {
       public int maximumGap(int[] nums) {
           if (nums.length < 2)
               return 0;
   
           Arrays.sort(nums);
           int maxGap = 0;
           
           for (int i = 1; i < nums.length; i++) {
               maxGap = Math.max(maxGap, nums[i] - nums[i - 1]);
           }
           
           return maxGap;
       }
   }

   B. Longest Subarray of 1's After Deleting One Element - LeetCode #1493 Problem: Given a binary array, find the length of the longest subarray of consecutive 1's after deleting one element. Link: https://leetcode.com/problems/longest-subarray-of-1s-after-deleting-one-element/

   class Solution {
       public int longestSubarray(int[] nums) {
           int maxLen = 0;
           int left = 0;
           int right = 0;
           int zeros = 0;
           
           while (right < nums.length) {
               if (nums[right] == 0) {
                   zeros++;
               }
               
               while (zeros > 1) {
                   if (nums[left] == 0) {
                       zeros--;
                   }
                   left++;
               }
               
               maxLen = Math.max(maxLen, right - left);
               right++;
           }
           
           return maxLen;
       }
   }

   C. Binary Tree Longest Consecutive Sequence - LeetCode #298 Problem: Given a binary tree, find the length of the longest consecutive elements sequence. Link: https://leetcode.com/problems/binary-tree-longest-consecutive-sequence/

   class Solution {
       public int longestConsecutive(TreeNode root) {
           return dfs(root, null, 0);
       }
       
       private int dfs(TreeNode node, TreeNode parent, int length) {
           if (node == null)
               return length;
           
           int currentLength = (parent != null && node.val == parent.val + 1) ? length + 1 : 1;
           
           int left = dfs(node.left, node, currentLength);
           int right = dfs(node.right, node, currentLength);
           
           return Math.max(currentLength, Math.max(left, right));
       }
   }

   Note: In the above examples, assume that the necessary data structures and classes are imported and that the TreeNode class is provided.

32. Alien Dictionary (Leetcode Premium)

Sure! Here is the requested information:

1. Problem Description and Link:

   • Problem: Alien Dictionary (Leetcode Premium)

   • Problem Link: Alien Dictionary Problem on Leetcode
2. Optimized Java Code for "Alien Dictionary" Problem:

import java.util.*;

public class AlienDictionary {

    public static String alienOrder(String[] words) {
        // Step 1: Create an adjacency list to represent the graph
        Map<Character, List<Character>> adjList = new HashMap<>();
        Map<Character, Integer> inDegree = new HashMap<>();

        // Initialize inDegree for each character as 0
        for (String word : words) {
            for (char c : word.toCharArray()) {
                inDegree.put(c, 0);
            }
        }

        // Step 2: Build the graph based on the given words
        for (int i = 0; i < words.length - 1; i++) {
            String current = words[i];
            String next = words[i + 1];

            // Check prefix-suffix relationship between current and next words
            if (current.length() > next.length() && current.startsWith(next)) {
                return ""; // Invalid input
            }

            for (int j = 0; j < Math.min(current.length(), next.length()); j++) {
                char c1 = current.charAt(j);
                char c2 = next.charAt(j);

                if (c1 != c2) {
                    // Add edge in the graph
                    adjList.computeIfAbsent(c1, k -> new ArrayList<>()).add(c2);

                    // Increment inDegree of the next character
                    inDegree.put(c2, inDegree.get(c2) + 1);
                    break; // No need to check further characters
                }
            }
        }

        // Step 3: Perform a Topological Sort on the graph
        StringBuilder sb = new StringBuilder();
        Queue<Character> queue = new LinkedList<>();

        for (char c : inDegree.keySet()) {
            if (inDegree.get(c) == 0) {
                queue.offer(c);
            }
        }

        while (!queue.isEmpty()) {
            char c = queue.poll();
            sb.append(c);

            if (adjList.containsKey(c)) {
                for (char neighbor : adjList.get(c)) {
                    inDegree.put(neighbor, inDegree.get(neighbor) - 1);
                    if (inDegree.get(neighbor) == 0) {
                        queue.offer(neighbor);
                    }
                }
            }
        }

        // Check if there is a cyclic dependency in the graph
        if (sb.length() != inDegree.size()) {
            return "";
        }

        return sb.toString();
    }

    public static void main(String[] args) {
        String[] words = {"wrt","wrf","er","ett","rftt"};
        String alienOrder = alienOrder(words);
        System.out.println("Alien Order: " + alienOrder);
    }
}

1. Description of the Algorithm and Interview Thought Process:

   • This problem is a classic example of topological sorting in Directed Acyclic Graphs (DAGs).

   • Step 1: We start by creating an adjacency list to represent the graph and also a map to keep track of each character's inDegree (i.e., the number of inward edges).

   • Step 2: We build the graph by comparing each pair of adjacent words. We check the prefix-suffix relationship between the current and next word to ensure the given words are in a valid order.

   • Step 3: Once the graph is built, we perform a topological sort using a queue. We add the characters with inDegree 0 to the queue, and for each character processed, we reduce the inDegree of its neighboring characters. If the inDegree becomes 0 for a neighboring character, we add it to the queue. We continue this process until the queue is empty.

   • Finally, we check if the length of our resulting string matches the total number of characters. If it does, it means there are no cyclic dependencies in the graph, and we return the resulting string as the valid alien order. Otherwise, we return an empty string.

   • This algorithm has a time complexity of O(n), where n is the total number of characters in all words, and a space complexity of O(1), as the maximum space used is O(26) for storing the characters of the alphabet.

2. Abstract the Problems that Use Similar Algorithm:

   • Any problem that involves finding a valid ordering of elements based on some dependencies or precedence can be solved using a topological sort algorithm. Some examples include:

     • Course Schedule (LeetCode Premium)

     • Task Scheduler (LeetCode)

     • Job Scheduling (Various platforms)

3. Similar Problems with Detailed Java Code and Descriptions:

   • Course Schedule II (LeetCode Premium):

     • Problem Link: Course Schedule II Problem on LeetCode
     • Java Code:

       public int[] findOrder(int numCourses, int[][] prerequisites) {
           // Build adjacency list and in-degree map
           Map<Integer, List<Integer>> adjList = new HashMap<>();
           int[] inDegree = new int[numCourses];
       
           for (int[] prerequisite : prerequisites) {
               int current = prerequisite[0];
               int next = prerequisite[1];
       
               adjList.computeIfAbsent(next, k -> new ArrayList<>()).add(current);
               inDegree[current]++;
           }
       
           // Perform topological sort
           int[] order = new int[numCourses];
           Queue<Integer> queue = new LinkedList<>();
           int count = 0;
       
           for (int i = 0; i < numCourses; i++) {
               if (inDegree[i] == 0) {
                   queue.offer(i);
               }
           }
       
           while (!queue.isEmpty()) {
               int course = queue.poll();
               order[count++] = course;
       
               if (adjList.containsKey(course)) {
                   for (int dependency : adjList.get(course)) {
                       inDegree[dependency]--;
                       if (inDegree[dependency] == 0) {
                           queue.offer(dependency);
                       }
                   }
               }
           }
       
           return count == numCourses ? order : new int[0];
       }

   • Task Scheduler (LeetCode):

     • Problem Link: Task Scheduler Problem on LeetCode
     • Java Code:

       public int leastInterval(char[] tasks, int n) {
           int[] frequencies = new int[26];
           for (char task : tasks) {
               frequencies[task - 'A']++;
           }
           PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
           for (int frequency : frequencies) {
               if (frequency > 0) {
                   maxHeap.offer(frequency);
               }
           }
           int intervals = 0;
           while (!maxHeap.isEmpty()) {
               List<Integer> temp = new ArrayList<>();
               for (int i = 0; i < n + 1; i++) {
                   if (!maxHeap.isEmpty()) {
                       temp.add(maxHeap.poll() - 1);
                   }
               }
               for (int frequency : temp) {
                   if (frequency > 0) {
                       maxHeap.offer(frequency);
                   }
               }
               intervals += maxHeap.isEmpty() ? temp.size() : n + 1;
           }
           return intervals;
       }

33. Graph Valid Tree (Leetcode Premium)

1. Problem Description: 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. The algorithm should satisfy the following conditions:

   • There is exactly one root node.

   • The graph must be fully connected (all nodes are reachable from the root).

   • There are no cycles in the graph.

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

2. Optimized Java Code with detailed comments:

class Solution {
    public boolean validTree(int n, int[][] edges) {
        // Create an adjacency list to represent the graph
        List<List<Integer>> adjacencyList = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adjacencyList.add(new ArrayList<>());
        }
        
        // Populate the adjacency list with the given edges
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adjacencyList.get(u).add(v);
            adjacencyList.get(v).add(u);
        }
        
        // Create a visited array to track visited nodes during DFS
        boolean[] visited = new boolean[n];
        
        // Start DFS from the root node (0)
        if (!dfs(adjacencyList, visited, 0, -1)) {
            return false; // Graph is not fully connected
        }
        
        // Check if there are any unvisited nodes (disconnected components)
        for (boolean nodeVisited : visited) {
            if (!nodeVisited) {
                return false; // Graph is not fully connected
            }
        }
        
        return true;
    }
    
    private boolean dfs(List<List<Integer>> adjacencyList, boolean[] visited, int currentNode, int parentNode) {
        visited[currentNode] = true;
        
        // Traverse all neighbors of the current node
        for (int neighbor : adjacencyList.get(currentNode)) {
            if (visited[neighbor] && neighbor != parentNode) {
                return false; // Cycle detected
            } else if (!visited[neighbor]) {
                if (!dfs(adjacencyList, visited, neighbor, currentNode)) {
                    return false; // DFS returns false, indicating cycle
                }
            }
        }
        
        return true;
    }
}

1. Algorithm Explanation:

The algorithm uses a depth-first search (DFS) approach to check if the given graph is a valid tree.

To solve this problem, we follow these steps:

• Create an adjacency list to represent the undirected graph using the given edges.

• Create a boolean visited array to keep track of visited nodes during DFS traversal.

• Start the DFS traversal from a root node, usually node 0.

  • Mark the current node as visited.

  • Recursively visit all the neighbors of the current node.

  • If a neighbor has been visited before and is not the parent of the current node, a cycle is detected, and the graph is not a valid tree. Return false.

• After the DFS traversal, check if there are any remaining unvisited nodes. If so, the graph is not fully connected, and it's not a valid tree. Return false.

• If the DFS traversal completes without detecting any cycle or disconnected nodes, the graph is a valid tree. Return true.

During an interview, it is essential to explain the reasoning behind using a DFS approach for this problem. DFS is a natural choice for exploring all connected components of the graph. By starting from a root node, we can reach all other nodes if the graph is a valid tree. Additionally, the algorithm avoids revisiting nodes by using the visited array, preventing infinite loops when cycles are present.

1. Similar problems where the same algorithm can be used:

   • Connected Components in an Undirected Graph

   • Cycle detection in an Undirected Graph

   • Tree traversal problems

2. Similar problems with detailed descriptions and Java code:

• Problem: Connected Components in an Undirected Graph

  • Description: Given an undirected graph, find all connected components.

  • Leetcode Link: N/A

  • Java Code:

class Solution {
    public List<List<Integer>> findConnectedComponents(int n, int[][] edges) {
        List<List<Integer>> connectedComponents = new ArrayList<>();
        List<List<Integer>> adjacencyList = new ArrayList<>();
        
        for (int i = 0; i < n; i++) {
            adjacencyList.add(new ArrayList<>());
        }
        
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adjacencyList.get(u).add(v);
            adjacencyList.get(v).add(u);
        }
        
        boolean[] visited = new boolean[n];
        
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                List<Integer> component = new ArrayList<>();
                dfs(adjacencyList, visited, i, component);
                connectedComponents.add(component);
            }
        }
        
        return connectedComponents;
    }
    
    private void dfs(List<List<Integer>> adjacencyList, boolean[] visited, int currentNode, List<Integer> component) {
        visited[currentNode] = true;
        component.add(currentNode);
        
        for (int neighbor : adjacencyList.get(currentNode)) {
            if (!visited[neighbor]) {
                dfs(adjacencyList, visited, neighbor, component);
            }
        }
    }
}

• Problem: Cycle Detection in an Undirected Graph

  • Description: Given an undirected graph, determine if it contains any cycle.

  • Leetcode Link: N/A

  • Java Code:

class Solution {
    public boolean hasCycle(int n, int[][] edges) {
        List<List<Integer>> adjacencyList = new ArrayList<>();
        
        for (int i = 0; i < n; i++) {
            adjacencyList.add(new ArrayList<>());
        }
        
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adjacencyList.get(u).add(v);
            adjacencyList.get(v).add(u);
        }
        
        boolean[] visited = new boolean[n];
        
        if (dfs(adjacencyList, visited, 0, -1)) {
            return true;
        }
        
        return false;
    }
    
    private boolean dfs(List<List<Integer>> adjacencyList, boolean[] visited, int currentNode, int parentNode) {
        visited[currentNode] = true;
        
        for (int neighbor : adjacencyList.get(currentNode)) {
            if (visited[neighbor] && neighbor != parentNode) {
                return true; // Cycle detected
            } else if (!visited[neighbor]) {
                if (dfs(adjacencyList, visited, neighbor, currentNode)) {
                    return true; // DFS returns true, indicating cycle
                }
            }
        }
        
        return false;
    }
}

Note: The above code snippets are simplified and may not have full error checking or edge cases handling. They are intended to demonstrate the algorithm and its usage.

34. Number of Connected Components in an Undirected Graph (Leetcode Premium)

1. Problem Description: You are given an undirected graph consisting of n nodes labeled from 0 to n - 1. You are also given an array edges, where each edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the graph. A connected component is a set of nodes where there is a path between every pair of nodes in the set.

Return the number of connected components in the graph.

Leetcode Link: https://leetcode.com/problems/number-of-connected-components-in-an-undirected-graph/

1. Optimized Java Code:

class Solution {
    public int countComponents(int n, int[][] edges) {
        int[] parent = new int[n];
        int count = n;

        // Initialize each node as a separate component
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }

        // Union the nodes from the given edges
        for (int[] edge : edges) {
            int a = find(parent, edge[0]);
            int b = find(parent, edge[1]);
            if (a != b) {
                parent[a] = b; // Union
                count--;
            }
        }

        return count;
    }

    private int find(int[] parent, int node) {
        // Find the root of the node
        if (parent[node] == node) {
            return node;
        }

        // Path compression for optimization
        return parent[node] = find(parent, parent[node]);
    }
}

1. Algorithm Explanation:

The problem can be solved using the Union-Find algorithm. Here's a step-by-step explanation:

• Create an array parent with size n. Each index in the array represents a node, and the value at each index represents the parent of that node.

• Initialize each node as a separate component by setting the value of each index in parent array to itself (i.e., parent[i] = i).

• Traverse through the given edges array and for each edge, find the parent of both nodes using the find function.

  • The find function performs path compression by recursively finding the root of the node and updating the parent of each visited node to the root node. This optimization improves the efficiency of subsequent find operations.

• If the parents of the two nodes are different, it means they belong to different components. In this case, union them by setting the parent of one node to be the parent of the other node (parent[a] = b).

• After processing all the edges, the number of connected components will be equal to the number of distinct parent nodes in the parent array.

• Return the count of connected components.

During an interview, you can discuss the Union-Find algorithm as a possible solution to the problem. Explain the concept of disjoint sets, where each node initially belongs to its own set. By merging sets (i.e., union operation) based on the given edges, we can determine the number of connected components in the graph.

1. Abstracted Problems:

The algorithm used here can be applied to solve similar problems involving graphs and finding connected components. It is useful in various scenarios, including:

• Network analysis and clustering

• Social networks and friendship connections

• Geographic connectivity analysis

• Component separation in image processing algorithms

1. Similar Problems:

Here are some similar problems that can be solved using the Union-Find algorithm:

a) Number of Islands (https://leetcode.com/problems/number-of-islands/)

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

        int[] parent = new int[rows * cols];
        int[] rank = new int[rows * cols];

        for (int i = 0; i < rows * cols; i++) {
            parent[i] = i;
        }

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

                    int index = i * cols + j;

                    if (i > 0 && grid[i - 1][j] == '1') {
                        union(parent, rank, index, (i - 1) * cols + j);
                    }
                    if (i < rows - 1 && grid[i + 1][j] == '1') {
                        union(parent, rank, index, (i + 1) * cols + j);
                    }
                    if (j > 0 && grid[i][j - 1] == '1') {
                        union(parent, rank, index, i * cols + (j - 1));
                    }
                    if (j < cols - 1 && grid[i][j + 1] == '1') {
                        union(parent, rank, index, i * cols + (j + 1));
                    }
                }
            }
        }

        return count;
    }
    
    private int find(int[] parent, int node) {
        if (parent[node] == node) {
            return node;
        }

        return parent[node] = find(parent, parent[node]);
    }
    
    private void union(int[] parent, int[] rank, int a, int b) {
        int rootA = find(parent, a);
        int rootB = find(parent, b);

        if (rootA == rootB) {
            return;
        }

        if (rank[rootA] < rank[rootB]) {
            parent[rootA] = rootB;
        } else if (rank[rootA] > rank[rootB]) {
            parent[rootB] = rootA;
        } else {
            parent[rootA] = rootB;
            rank[rootB]++;
        }
    }
}

b) Evaluate Division (https://leetcode.com/problems/evaluate-division/)

class Solution {
    public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {
        Map<String, String> parent = new HashMap<>();
        Map<String, Double> weight = new HashMap<>();

        for (int i = 0; i < equations.size(); i++) {
            String dividend = equations.get(i).get(0);
            String divisor = equations.get(i).get(1);
            double quotient = values[i];

            union(parent, weight, dividend, divisor, quotient);
        }

        double[] result = new double[queries.size()];

        for (int i = 0; i < queries.size(); i++) {
            String dividend = queries.get(i).get(0);
            String divisor = queries.get(i).get(1);

            if (!parent.containsKey(dividend) || !parent.containsKey(divisor)) {
                result[i] = -1.0;
            } else {
                String rootDividend = find(parent, weight, dividend);
                String rootDivisor = find(parent, weight, divisor);

                if (!rootDividend.equals(rootDivisor)) {
                    result[i] = -1.0;
                } else {
                    result[i] = weight.get(dividend) / weight.get(divisor);
                }
            }
        }

        return result;
    }
    
    private String find(Map<String, String> parent, Map<String, Double> weight, String node) {
        String parentNode = parent.getOrDefault(node, node);

        if (!parentNode.equals(node)) {
            double nodeWeight = weight.getOrDefault(node, 1.0);
            double parentWeight = weight.getOrDefault(parentNode, 1.0);

            weight.put(node, nodeWeight * parentWeight);
            parent.put(node, find(parent, weight, parentNode));
        }

        return parent.getOrDefault(node, node);
    }
    
    private void union(Map<String, String> parent, Map<String, Double> weight, String dividend, String divisor, double quotient) {
        String rootDividend = find(parent, weight, dividend);
        String rootDivisor = find(parent, weight, divisor);

        double divisorWeight = weight.getOrDefault(dividend, 1.0) * quotient / weight.getOrDefault(divisor, 1.0);

        parent.put(rootDividend, rootDivisor);
        weight.put(rootDividend, divisorWeight);
    }
}

These examples demonstrate how the Union-Find algorithm can be used in different problem domains. The specific implementation details vary based on the problem requirements.