Path Finding

6. Word Search (79)

1. The problem "Word Search" on LeetCode can be found at the following link: https://leetcode.com/problems/word-search/

2. Here is the optimized Java code for the "Word Search" problem (79) on LeetCode:

class Solution {
    public boolean exist(char[][] board, String word) {
        int m = board.length;
        int n = board[0].length;
        
        boolean[][] visited = new boolean[m][n];
        
        // Iterate through each cell of the board
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == word.charAt(0) && backtrack(board, word, i, j, 0, visited)) {
                    return true;
                }
            }
        }
        return false;
    }
    
    private boolean backtrack(char[][] board, String word, int i, int j, int index, boolean[][] visited) {
        // Check if the word has been found
        if (index == word.length()) {
            return true;
        }
        
        // Check if the indices are valid and the cell is not visited
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || visited[i][j]) {
            return false;
        }
        
        // Check if the current character matches the character in the word
        if (board[i][j] != word.charAt(index)) {
            return false;
        }
        
        // Mark the cell as visited
        visited[i][j] = true;
        
        // Recursively check adjacent cells
        boolean result = backtrack(board, word, i + 1, j, index + 1, visited) ||
                         backtrack(board, word, i - 1, j, index + 1, visited) ||
                         backtrack(board, word, i, j + 1, index + 1, visited) ||
                         backtrack(board, word, i, j - 1, index + 1, visited);
        
        // Mark the cell as unvisited after backtracking
        visited[i][j] = false;
        
        return result;
    }
}

1. The algorithm for solving the "Word Search" problem is based on backtracking.

During an interview, you would start by understanding the problem statement and constraints. The problem requires finding if a given word can be formed by traversing adjacent cells in a 2D board, following specific rules.

The algorithm starts by iterating through each cell of the board to search for the first character of the word. Once found, a backtracking function is invoked to explore the adjacent cells and check if the remaining characters of the word can be matched. The backtracking function is called recursively to explore all possible paths.

To optimize the solution, we use a boolean array called visited to keep track of already visited cells during the backtracking process. This is essential to prevent visiting a cell more than once in the same path. Before and after each call to the backtracking function, we mark and unmark the current cell as visited respectively.

Additionally, we also perform multiple checks to ensure that we do not go out of bounds while exploring neighboring cells and to validate if the current character matches the character in the word.

By using backtracking, we are able to explore all possible paths in the board and find if the given word can be formed. If the word is found, the function returns true; otherwise, it returns false.

During an interview, it would be important to explain the approach, the reasons for using backtracking, and how the backtracking function works. It would also be valuable to discuss the time and space complexity of the solution (which is O(mn4^k), where m and n are the dimensions of the board, and k is the length of the word).

7. Path Sum (112)

1. Problem Description: The problem "Path Sum" on LeetCode (Problem #112) asks us to determine if there exists a root-to-leaf path in a binary tree, such that the sum of all the values along the path equals a given target sum.

Link to problem on LeetCode: https://leetcode.com/problems/path-sum/

1. Optimized Java Code with Comments:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        // Check if the root is null
        if (root == null) {
            return false;
        }
        
        // Check if the current node is a leaf node
        if (root.left == null && root.right == null) {
            return root.val == targetSum;
        }
        
        // Recursively check if there is a path with the remaining sum in the left or right subtree
        return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val);
    }
}

1. Detailed Description:

The problem can be solved using a recursive approach. The algorithm is as follows:

1. Check if the root of the binary tree is null. If it is null, return false, as there are no nodes to process.

2. Check if the current node is a leaf node (i.e., it has no left or right child). If so, check if the sum of values along the path from the root to the current node equals the target sum. If it does, return true; otherwise, return false.

3. Recursively check if there is a path with the remaining sum in the left subtree or right subtree. Subtract the current node's value from the target sum and call the recursive function on the left and right child nodes.

4. If either of the recursive calls returns true, it means a path with the target sum exists in either the left or right subtree. In this case, return true.

5. If both recursive calls return false, it means there is no path with the target sum in either the left or right subtree. In this case, return false.

During an interview, the interviewer may want to assess your ability to think recursively and solve tree-related problems. To approach this problem, start by understanding the problem statement and requirements. Identify any edge cases and handle them appropriately. Then, think about how a recursive solution can be implemented, considering the base case, recursive function calls, and combining the results.

During the coding process, make sure to write clean and readable code, following good coding practices. Add comments to explain the steps and logic behind the code. Finally, test the implementation with different test cases to validate its correctness.

Remember to communicate your thought process and approach clearly to the interviewer, as they are interested not only in the final solution but also in understanding your problem-solving skills.

8. Minimum Depth of Binary Tree (111)

1. Problem Description and Link: The problem is called "Minimum Depth of Binary Tree" and can be found on LeetCode here: https://leetcode.com/problems/minimum-depth-of-binary-tree/

2. Optimized Java Code: Here is the optimized solution in Java for the "Minimum Depth of Binary Tree" problem:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        // Base case: if the root is null, return 0
        if (root == null) {
            return 0;
        }
        
        // If both left and right child are null, return 1
        if (root.left == null && root.right == null) {
            return 1;
        }
        
        // If left child is null, recursively call minDepth on right child
        if (root.left == null) {
            return minDepth(root.right) + 1;
        }
        
        // If right child is null, recursively call minDepth on left child
        if (root.right == null) {
            return minDepth(root.left) + 1;
        }
        
        // If both left and right child exist, return the minimum depth of both
        return Math.min(minDepth(root.left), minDepth(root.right)) + 1;
    }
}

1. Algorithm Explanation and Thought Process: The problem asks us to find the minimum depth of a binary tree, which is the shortest path from the root to the nearest leaf node. The leaf nodes are the nodes that do not have any children.

To solve this problem, we can use recursion. We can define the base cases as follows:

• If the root is null, the depth is 0.

• If both the left and right child of a node are null, the depth is 1 (since it is a leaf node).

For any other node, we need to consider the following scenarios:

• If the left child is null, we can recursively call minDepth() on the right child and add 1 to the result.

• If the right child is null, we can recursively call minDepth() on the left child and add 1 to the result.

• If both the left and right child exist, we need to return the minimum depth of both subtrees by taking the minimum of minDepth() of each subtree and adding 1 to the result.

By using recursion and considering these scenarios, we can effectively find the minimum depth of the binary tree.

During an interview, it is important to communicate the thought process clearly and concisely. You can mention the recursive approach and explain how the base cases are handled. It is also important to mention that the code provided is the optimized solution with a time complexity of O(n), where n is the number of nodes in the binary tree.

9. Path Sum II (113)

1. Problem Description: The problem "Path Sum II" on LeetCode asks us to find all root-to-leaf paths in a binary tree where each path's sum equals a given target sum.

LeetCode Link: https://leetcode.com/problems/path-sum-ii/

1. Java Code: Here is the optimized Java code for the problem "Path Sum II" with detailed comments:

import java.util.ArrayList;
import java.util.List;

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int val) {
        this.val = val;
    }
}

public class PathSumII {
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> path = new ArrayList<>();
        findPaths(root, targetSum, path, result);
        return result;
    }

    private void findPaths(TreeNode node, int target, List<Integer> path, List<List<Integer>> result) {
        if (node == null)
            return;

        path.add(node.val); // Add the current node value to the path
        target -= node.val; // Update the target sum

        if (node.left == null && node.right == null) { // Leaf node reached
            if (target == 0) {
                result.add(new ArrayList<>(path)); // Add a new path to the result list
            }
        } else { // Recursively find paths in the left and right subtrees
            findPaths(node.left, target, path, result);
            findPaths(node.right, target, path, result);
        }

        // Backtracking: Remove the current node from the path before returning
        path.remove(path.size() - 1);
    }
}

1. Algorithm and Interview Tips: The algorithm uses a recursive approach to explore all root-to-leaf paths in a binary tree. We start with an empty path and traverse the tree in a depth-first manner.

During the traversal, we keep track of the current path and the remaining target sum. At each node, we add its value to the path and subtract it from the target sum. If we reach a leaf node and the target sum becomes zero, we have found a valid path.

In the recursive function findPaths(), we check if the current node is a leaf node. If so, we verify if the remaining target sum is zero. If it is, we add the current path to the result list. Otherwise, we continue the traversal by recursively calling the function on the left and right subtrees.

To optimize the memory usage, we pass the current path by reference instead of creating a new path at each node. This saves space and allows us to backtrack efficiently after exploring a path.

During an interview, it is crucial to understand the problem requirements and constraints before jumping into the code. It is also important to communicate your thought process and explain your approach step by step.

Some possible discussion points during the interview process can be:

• Understanding the problem requirements and constraints.

• Discussing the expected input and output formats.

• Identifying the base cases and reasoning about the traversal strategy.

• Analyzing the time and space complexity of the solution and possible optimizations.

• Discussing edge cases and potential error handling.

By demonstrating a clear understanding of the problem and explaining your approach and thought process, you can showcase your problem-solving skills and impress the interviewer.

10. Word Ladder II (126)

1. The problem "Word Ladder II" on LeetCode can be found at the following link: https://leetcode.com/problems/word-ladder-ii/

2. The optimized Java code for the problem "Word Ladder II" (126) with detailed comments is as follows:

import java.util.*;

public class Solution {
    // Use a map to store the adjacent words for each word
    private Map<String, List<String>> adjacentWords;

    public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
        // Initialize the adjacentWords map
        adjacentWords = new HashMap<>();
        for (String word : wordList) {
            generateAdjacencyList(word);
        }

        // Use a queue for BFS
        Queue<List<String>> queue = new LinkedList<>();
        queue.add(new ArrayList<>(Arrays.asList(beginWord)));

        // Use a set to store the visited words
        Set<String> visited = new HashSet<>();
        visited.add(beginWord);

        // Use a set to store the words in the current level
        Set<String> currentLevel = new HashSet<>();
        currentLevel.add(beginWord);

        // Use a flag to indicate if the end word is found
        boolean found = false;

        // Use a list to store the valid ladders
        List<List<String>> ladders = new ArrayList<>();

        while (!queue.isEmpty() && !found) {
            int size = queue.size();
            Set<String> tempLevel = new HashSet<>();

            for (int i = 0; i < size; i++) {
                List<String> ladder = queue.poll();
                String word = ladder.get(ladder.size() - 1);

                for (String neighbor : adjacentWords.get(word)) {
                    if (neighbor.equals(endWord)) {
                        ladder.add(neighbor);
                        ladders.add(new ArrayList<>(ladder));
                        found = true;
                    }

                    if (!visited.contains(neighbor)) {
                        ladder.add(neighbor);
                        queue.add(new ArrayList<>(ladder));
                        ladder.remove(ladder.size() - 1);
                        tempLevel.add(neighbor);
                    }
                }
            }

            visited.addAll(tempLevel);
            currentLevel = tempLevel;
        }

        return ladders;
    }

    private void generateAdjacencyList(String word) {
        List<String> neighbors = new ArrayList<>();

        for (int i = 0; i < word.length(); i++) {
            char[] chars = word.toCharArray();
            for (char c = 'a'; c <= 'z'; c++) {
                chars[i] = c;
                String newWord = new String(chars);
                neighbors.add(newWord);
            }
        }

        neighbors.remove(word);
        adjacentWords.put(word, neighbors);
    }
}

1. Algorithm and Thought Process:

The problem requires finding the shortest transformation sequence from a given start word to a given end word using a set of words from a given word list, where only one letter can be changed at a time. Additionally, the intermediate word in the sequence must have the same length as the start and end words.

To solve this problem, the algorithm uses the concept of bidirectional breadth-first search (BFS), which starts from the beginWord and endWord simultaneously, expanding the search space until the two searches meet. By using both BFS and bidirectional search, we can significantly reduce the time complexity compared to a traditional BFS approach.

The implementation of the algorithm involves several steps:

1. Generate an adjacency list for each word in the wordList. This is done by iterating through each word and replacing each character with all possible lowercase alphabets. The resulting words are added to the adjacency list.

2. Maintain a queue for BFS. Initialize the queue with a list containing the beginWord.

3. Use a set to store the visited words and add the beginWord to the set.

4. Use a set to store the words in the current level and add the beginWord to the set.

5. Use a boolean flag to track if the endWord is found.

6. Use a list to store the valid ladders that reach the endWord.

7. While the queue is not empty and the endWord is not found, do the following:

   • Get the size of the current level (number of words).

   • Create a temporary set for the next level.

   • For each word in the current level, do the following:

     • Get the last word in the ladder.

     • Iterate through each neighbor word (adjacent word) by accessing the adjacentWords map.

       • If the neighbor word is the endWord, add it to the ladder, add the ladder to the valid ladders, and set the found flag to true.

       • If the neighbor word is not visited, add it to the ladder, add the ladder to the queue, remove the last word from the ladder, and add the neighbor word to the temporary set.

   • Add all the words in the temporary set to the visited set.

   • Update the current level with the temporary set.

8. Return the valid ladders.

During an interview, it is important to communicate and explain the algorithm step by step, highlighting the use of bidirectional BFS to optimize the search process. Make sure to emphasize the key data structures used (map, queue, sets, and lists) and explain the purpose of each step in the algorithm. Additionally, discussing the time and space complexity of the algorithm will showcase your problem-solving skills.