String DP

1. Regular Expression Matching

1. Problem Description: Regular Expression Matching

   • Given a string s and a pattern p, implement regular expression matching with support for '.' and '*'.

   • '.' matches any single character.

   • '*' matches zero or more of the preceding element.

   • The matching should cover the entire input string (not partial).

   Leetcode Problem Link: https://leetcode.com/problems/regular-expression-matching/

2. Optimized Java Code with Detailed Comments:

class Solution {
    public boolean isMatch(String s, String p) {
        // Create a 2D boolean array to store the matching results
        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
        
        // Empty string matches empty pattern
        dp[0][0] = true;
        
        // Fill the first row of the DP table
        for (int j = 1; j <= p.length(); j++) {
            if (p.charAt(j - 1) == '*') {
                dp[0][j] = dp[0][j - 2];
            }
        }
        
        // Fill the DP table using bottom-up approach
        for (int i = 1; i <= s.length(); i++) {
            for (int j = 1; j <= p.length(); j++) {
                if (s.charAt(i - 1) == p.charAt(j - 1) || p.charAt(j - 1) == '.') {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (p.charAt(j - 1) == '*') {
                    // Consider the case when '*' matches zero preceding element
                    dp[i][j] = dp[i][j - 2];
                    
                    // Consider the case when '*' matches one or more preceding element
                    if (p.charAt(j - 2) == '.' || p.charAt(j - 2) == s.charAt(i - 1)) {
                        dp[i][j] = dp[i][j] || dp[i - 1][j];
                    }
                }
            }
        }
        
        // Return the final result from DP table
        return dp[s.length()][p.length()];
    }
}

1. Algorithm Description and Interview Process Tips:

• The problem is solved using Dynamic Programming (DP).

• We create a 2D DP table to store the matching results for all possible prefixes of the input string 's' and pattern 'p'.

• DP table has dimensions (s.length() + 1) x (p.length() + 1) to handle empty strings as well.

• The DP table is filled using a bottom-up approach. The base case is an empty string matching an empty pattern, which is set to true.

• We then fill the first row of the DP table, handling cases where '*' matches zero preceding element.

• For each character in 's' and 'p', we check if the characters match or if the pattern character is '.' (matches any character).

• If either of these conditions is true, the result depends on the previous matching result without the current characters.

• If the pattern character is '', we consider two cases: '' matches zero preceding element or '*' matches one or more preceding elements.

• The final result is obtained from the bottom-right cell of the DP table.

During an interview process, follow these steps:

1. Understand the problem statement and requirements.

2. Identify the key features of the regular expression matching.

3. Discuss possible solutions and their time complexities.

4. Choose the optimal solution and implement it.

5. Test the solution with different input cases, including edge cases.

6. Explain the solution to the interviewer, highlighting the algorithm and time complexity analysis.

7. Discuss potential optimizations or alternative approaches if time permits.

Remember to communicate your thoughts and steps clearly during the interview process.

2. Wildcard Matching

1. Problem: The problem is to implement wildcard pattern matching with support for '?' and ''. The matching should cover the entire input string, not just a substring. '?' matches any single character, while '' matches any sequence of characters (including the empty sequence). The matching should be case-sensitive.

Problem link: https://leetcode.com/problems/wildcard-matching/

1. Optimized Java code for the problem "Wildcard Matching":

class Solution {
    public boolean isMatch(String s, String p) {
        int sLen = s.length(), pLen = p.length();
        int sIndex = 0, pIndex = 0, sStar = -1, pStar = -1;

        while (sIndex < sLen) {
            if (pIndex < pLen && (p.charAt(pIndex) == '?' || p.charAt(pIndex) == s.charAt(sIndex))) {
                sIndex++;
                pIndex++;
            } else if (pIndex < pLen && p.charAt(pIndex) == '*') {
                sStar = sIndex;
                pStar = pIndex;
                pIndex++;
            } else if (pStar != -1) {
                sStar++;
                sIndex = sStar;
                pIndex = pStar + 1;
            } else {
                return false;
            }
        }

        while (pIndex < pLen && p.charAt(pIndex) == '*') {
            pIndex++;
        }

        return pIndex == pLen;
    }
}

1. Algorithm and thought process:

The problem can be solved using a greedy approach with two pointers.

• We iterate over the string 's' and pattern 'p'.

• For each character in 's', we check if the corresponding character in 'p' matches. If the character is a '?', or both characters are the same, we increment the pointers correspondingly.

• If we encounter a '*', we keep track of the current positions of the star and the next position after it in the pattern.

• When we encounter a mismatch, we check if we have seen a '' before. If we have, we increment the star pointer and reset the string pointer to the next position after the star. This allows the '' to match any sequence of characters.

• If there is no '*' and we encounter a mismatch, we return false.

• After iterating over 's', we check if there are any remaining '*'s in the pattern. If so, we skip them, as they can match empty sequences.

This algorithm has a time complexity of O(sLen + pLen), where sLen is the length of the string 's' and pLen is the length of the pattern 'p'.

During an interview, it's important to analyze the problem, understand the requirements, and come up with an initial brute-force solution. From there, you can optimize the solution by using different data structures or algorithms. Discuss the time and space complexity of your solution and mention any potential edge cases or optimizations. Also, make sure to write clean and readable code with appropriate variable names and comments.

3. Longest Common Subsequence

1. Problem Description: The problem "Longest Common Subsequence" on LeetCode asks you to find the length of the longest common subsequence (LCS) between two given strings.

Link to the problem on LeetCode: https://leetcode.com/problems/longest-common-subsequence/

1. Optimized Java Code:

public class LongestCommonSubsequence {

    public static int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length();
        int n = text2.length();

        int[][] dp = new int[m + 1][n + 1];

        // Building the DP table
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (text1.charAt(i - 1) == text2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }

        return dp[m][n];
    }

    public static void main(String[] args) {
        String text1 = "abcde";
        String text2 = "ace";
        int longestCommonSubsequenceLength = longestCommonSubsequence(text1, text2);
        System.out.println("Length of the Longest Common Subsequence: " + longestCommonSubsequenceLength);
    }
}

1. Detailed Description:

• The problem requires finding the length of the longest common subsequence (LCS) between two given strings.

• The LCS is a sequence that appears in the same order, but not necessarily contiguous, in both strings.

• The algorithm for finding the LCS can be solved using dynamic programming.

• To solve this problem, we can use a two-dimensional array dp to build a dynamic programming table.

• The value at dp[i][j] represents the length of the LCS of the substrings text1.substring(0, i) and text2.substring(0, j).

• We initialize the DP table with zeroes and iterate through the strings text1 and text2.

• If the characters at the current indices i-1 and j-1 in text1 and text2 are the same, we increment the LCS length by one: dp[i][j] = dp[i - 1][j - 1] + 1.

• If the characters are not the same, we take the maximum of the lengths of the LCS by excluding the characters: dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]).

• After iterating through both strings, the length of the LCS will be stored in dp[m][n], where m and n are the lengths of text1 and text2 respectively.

• Finally, we return the length of the LCS.

• During an interview, it is important to walk through the problem, understand the definitions, and explain your thought process before jumping into coding. Specifically, explain how the problem can be solved efficiently using dynamic programming.

4. Edit Distance

1. Problem Description: The problem "Edit Distance" on LeetCode asks us to find the minimum number of operations required to convert one string into another string. The operations allowed are insertion, deletion, and substitution of a character.

Link to the problem on LeetCode: https://leetcode.com/problems/edit-distance/

1. Optimized Java Code for "Edit Distance":

class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();
        
        int[][] dp = new int[m+1][n+1];
        
        // base case - if one of the strings is empty
        for (int i = 0; i <= m; i++) {
            dp[i][0] = i;
        }
        
        for (int j = 0; j <= n; j++) {
            dp[0][j] = j;
        }
        
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (word1.charAt(i-1) == word2.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1];
                } else {
                    dp[i][j] = 1 + Math.min(dp[i-1][j-1], Math.min(dp[i][j-1], dp[i-1][j]));
                }
            }
        }
        
        return dp[m][n];
    }
}

1. Algorithm and Interview Process: The problem "Edit Distance" can be solved using a dynamic programming approach. The given problem asks us to find the minimum number of operations required to convert one string into another string.

To solve this problem, we can use a 2D array, dp, of size (m+1) x (n+1), where m and n are the lengths of word1 and word2 respectively. dp[i][j] represents the minimum number of operations required to convert the first i characters of word1 to the first j characters of word2.

We can initialize the base cases: if one of the strings is empty, the minimum number of operations required would be equal to the length of the non-empty string.

Then, we iterate through the characters of word1 and word2 and fill in the dp array by considering the three operations:

1. If the characters at i-1 and j-1 are equal, then no operation is needed, and we can copy the value from dp[i-1][j-1].

2. If the characters are not equal, we have three options: a. We can substitute the character at i-1 in word1 with the character at j-1 in word2, so we take dp[i-1][j-1] and add 1 to it. b. We can delete the character at i-1 in word1, so we take dp[i-1][j] and add 1 to it. c. We can insert the character at j-1 in word2 into word1, so we take dp[i][j-1] and add 1 to it.

Finally, the answer will be stored in dp[m][n], which will represent the minimum number of operations required to convert word1 to word2.

During an interview process, it is essential to understand the problem statement and come up with a plan before jumping into code implementation. Here are some key steps to follow during the interview process:

1. Read the problem statement carefully and understand the input and output requirements.

2. Identify any specific constraints or edge cases.

3. Analyze the problem and design an algorithm using an appropriate approach (e.g., in this case, dynamic programming).

4. Break down the problem into smaller subproblems and think about the base cases.

5. Visualize the problem using a diagram or example inputs to get a better understanding.

6. Think about the data structures and variables needed to solve the problem efficiently.

7. Implement the algorithm, test it using various test cases, and check for any edge cases.

By following these steps, you can provide a clear and structured solution to the problem, which will impress the interviewer and showcase your problem-solving skills.

5. Palindrome Partitioning II

1. Problem Description: The problem "Palindrome Partitioning II" asks us to find the minimum number of cuts required to partition a given string such that each partition is a palindrome. We need to return the minimum number of cuts needed.

LeetCode Link: https://leetcode.com/problems/palindrome-partitioning-ii/

1. Optimized Java Code:

public class PalindromePartitioningII {
    public int minCut(String s) {
        int n = s.length();
        int[] dp = new int[n];
        boolean[][] isPalindrome = new boolean[n][n];

        for (int i = 0; i < n; i++) {
            dp[i] = i; // maximum cuts needed will be i
            isPalindrome[i][i] = true;
        }

        // check for each substring length starting from 2
        for (int len = 2; len <= n; len++) {
            for (int start = 0; start <= n - len; start++) {
                int end = start + len - 1;

                if (s.charAt(start) == s.charAt(end)) {
                    if (len == 2 || isPalindrome[start + 1][end - 1]) {
                        isPalindrome[start][end] = true;
                    }
                }
            }
        }

        for (int i = 1; i < n; i++) {
            if (isPalindrome[0][i]) {
                dp[i] = 0;
            } else {
                for (int j = 0; j < i; j++) {
                    if (isPalindrome[j + 1][i]) {
                        dp[i] = Math.min(dp[i], dp[j] + 1);
                    }
                }
            }
        }

        return dp[n - 1];
    }

    public static void main(String[] args) {
        PalindromePartitioningII pp = new PalindromePartitioningII();
        String s = "aab";
        System.out.println(pp.minCut(s)); // Output: 1
    }
}

1. Algorithm Explanation:

The problem can be solved using dynamic programming. We can start by initializing two arrays: dp to keep track of the minimum cuts needed, and isPalindrome to store the information about whether a substring is a palindrome.

First, we initialize dp[i] to i because the maximum number of cuts needed for a string of length i will be i.

Next, we initialize isPalindrome[i][i] as true because a single character is always a palindrome.

We then iterate over each substring length len starting from 2. For each substring, we check if the first and last characters are equal. If they are, we check if the remaining substring (excluding the first and last characters) is also a palindrome. If both conditions are met, we set isPalindrome[start][end] as true.

In the final step, we iterate from 1 to n-1 and check if any substring ending at index i is a palindrome. If it is, then we set dp[i] as 0 because no cuts are needed in this case. Otherwise, we try to find the minimum cuts by iterating from 0 to i-1 and checking if the substring from j+1 to i is a palindrome. If it is, we update dp[i] as min(dp[i], dp[j] + 1), indicating that one additional cut is needed compared to the minimum cuts needed for substring ending at index j.

Finally, we return dp[n-1], the minimum number of cuts needed for the entire string.

During the interview process, it is important to understand the problem statement clearly and identify the repeated subproblems. The problem requires us to find the minimum cuts, which can be broken down into smaller subproblems. It is important to realize that if we can efficiently find whether a substring is a palindrome or not, we can solve the problem using dynamic programming. Hence, the first step is to come up with a way to efficiently determine whether a substring is a palindrome or not. Once we have this information, we can use it to build the solution step by step.