String

6. Longest Substring Without Repeating Characters

1. Problem Description: The problem "Longest Substring Without Repeating Characters" asks us to find the length of the longest substring without repeating characters in a given string.

LeetCode Link: https://leetcode.com/problems/longest-substring-without-repeating-characters/

1. Optimized Java Code:

Here's the optimized Java code for the problem "Longest Substring Without Repeating Characters" with detailed comments:

class Solution {
    public int lengthOfLongestSubstring(String s) {
        // Create a HashMap to store the last occurrence index of each character
        HashMap<Character, Integer> map = new HashMap<>();
        int maxLength = 0;
        int startIndex = 0;
        
        // Traverse the string
        for (int endIndex = 0; endIndex < s.length(); endIndex++) {
            char c = s.charAt(endIndex);
            
            // If the character is already present in the map and its last occurrence
            // index is greater than or equal to the startIndex, update the startIndex
            if (map.containsKey(c) && map.get(c) >= startIndex) {
                startIndex = map.get(c) + 1;
            }
            
            // Update the last occurrence index of the character in the map
            map.put(c, endIndex);
            
            // Update the maximum length of the substring
            maxLength = Math.max(maxLength, endIndex - startIndex + 1);
        }
        
        return maxLength;
    }
}

1. Algorithm Explanation:

To solve this problem efficiently, we can use the sliding window technique along with a HashMap to store the last occurrence index of each character.

The algorithm starts with two pointers, startIndex and endIndex, both initialized to 0. The sliding window consists of the substring between these two pointers.

We traverse the string from left to right, continuously expanding the window by moving the endIndex. If a character is already present in the window (i.e., it is a repeating character), we need to adjust the window's start by updating the startIndex to the next index of the repeating character.

To keep track of the last occurrence index of each character, we use a HashMap. If a character is not present in the map, we insert it along with its current index. If a character is already present in the map, we update its last occurrence index to the current index.

During the traversal, at each step, we update the maxLength by taking the maximum of the current maxLength and the length of the current window (i.e., endIndex - startIndex + 1).

This algorithm has a time complexity of O(n), where n is the length of the input string, as we traverse the string once using the sliding window technique.

7. Valid Palindrome

1. Problem Description:

The problem "Valid Palindrome" is to determine if a given string is a valid palindrome. A palindrome is a word, phrase, number, or other sequence of characters that reads the same forward and backward. The given string may contain alphanumeric characters and ignore the case sensitivity.

You can find the problem on LeetCode using the following link: https://leetcode.com/problems/valid-palindrome/

1. Optimized Java Code:

Here is the optimized Java code for the "Valid Palindrome" problem with detailed comments:

public class Solution {
    public boolean isPalindrome(String s) {
        // Convert the given string to lowercase
        s = s.toLowerCase();
        
        // Initialize two pointers, one at the start and one at the end of the string
        int left = 0;
        int right = s.length() - 1;
        
        // Traverse the string from both ends towards the middle
        while (left < right) {
            // Skip non-alphanumeric characters from the left pointer
            while (left < right && !Character.isLetterOrDigit(s.charAt(left))) {
                left++;
            }
            
            // Skip non-alphanumeric characters from the right pointer
            while (left < right && !Character.isLetterOrDigit(s.charAt(right))) {
                right--;
            }
            
            // Compare the characters at the left and right pointers
            if (s.charAt(left) != s.charAt(right)) {
                return false;
            }
            
            // Move the pointers towards the middle
            left++;
            right--;
        }
        
        // All characters have been compared and no mismatches found
        return true;
    }
}

1. Algorithm and Approach:

To check if a given string is a palindrome, we can make use of two pointers approach. We initialize two pointers, one at the start and one at the end of the string. We then traverse the string from both ends towards the middle.

While traversing, we skip any non-alphanumeric characters by using the Character.isLetterOrDigit() method. We compare the characters at the left and right pointers, and if they are not equal, we return false as it is not a palindrome.

If all the characters have been compared and no mismatches are found, we return true as it is a palindrome.

During the interview process, approach the problem by breaking it down into smaller steps:

1. Convert the string to lowercase to ignore case sensitivity.

2. Use two pointers approach to compare characters from both ends.

3. Skip non-alphanumeric characters using character checking methods.

4. Return the result based on the comparison of characters.

Take into consideration that the optimized code uses constant space and runs in linear time complexity O(n), where n is the length of the input string.

8. Reverse Vowels in a String

1. Problem Description:

Given a string, reverse only the vowels of the string and return it. The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both lowercase and uppercase.

Example: Input: "leetcode" Output: "leotcede"

Leetcode Link: Reverse Vowels in a String

1. Java code for the problem:

public class Solution {
    public String reverseVowels(String s) {
        char[] chars = s.toCharArray();
        int left = 0;
        int right = chars.length - 1;
        
        while (left < right) {
            if (isVowel(chars[left]) && isVowel(chars[right])) {
                // Swap the vowels
                char temp = chars[left];
                chars[left] = chars[right];
                chars[right] = temp;
                left++;
                right--;
            } else if (!isVowel(chars[left])) {
                left++;
            } else if (!isVowel(chars[right])) {
                right--;
            }
        }
        
        return new String(chars);
    }
    
    private boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || 
               c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U';
    }
}

1. Algorithm Explanation:

The algorithm uses the two-pointer approach to reverse the vowels in the string. We start with two pointers, left at the beginning of the string and right at the end of the string.

While the left pointer is less than the right pointer, we iterate through the string. If both left and right pointers point to vowels, we swap them. If the left pointer does not point to a vowel, we increment it. If the right pointer does not point to a vowel, we decrement it. This way, we keep swapping vowels until we reach the middle of the string.

The isVowel function is used to check if a character is a vowel. It checks if the character is equal to any of the vowel letters (both lowercase and uppercase).

The solution has a time complexity of O(n), where n is the length of the string. Since we are using a char array to perform the operations, the overall space complexity is O(n) as well.

During the interview process, it is important to explain the approach and the reasoning behind it. You can also mention alternative approaches and discuss their pros and cons. Additionally, it is recommended to write the code correctly, handling edge cases and explaining the code logic using comments.

9. Longest Palindromic Substring

1. Problem Description and LeetCode Link: The problem "Longest Palindromic Substring" requires finding the longest substring that is a palindrome in a given string. A palindrome is a word or phrase that is read the same backward as forward.

LeetCode Link: https://leetcode.com/problems/longest-palindromic-substring/

1. Optimized Java Code:

public class Solution {
    public String longestPalindrome(String s) {
        if (s == null || s.length() == 0)
            return "";
        
        int start = 0, end = 0;
        
        for (int i = 0; i < s.length(); i++) {
            int len1 = expandAroundCenter(s, i, i);
            int len2 = expandAroundCenter(s, i, i + 1);
            int len = Math.max(len1, len2);
            if (len > end - start) {
                start = i - (len - 1) / 2;
                end = i + len / 2;
            }
        }
        
        return s.substring(start, end + 1);
    }
    
    private int expandAroundCenter(String s, int left, int right) {
        while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
            left--;
            right++;
        }
        return right - left - 1;
    }
}

1. Detailed Description and Algorithm: To approach this problem during an interview, we can use the concept of expanding around centers. The idea is to check each character in the string as a center and expand outward to find the longest palindromic substring.

The algorithm starts by initializing two pointers, start and end, which represent the indices of the longest palindromic substring found so far. We traverse the given string and at each index i, we expand around i as well as around i and i + 1 (to consider both even and odd-length palindromes).

The expandAroundCenter() method takes a center character(s) and checks if expanding left and right from that center forms a palindrome. It starts with initializing left and right pointers at the center and keeps expanding until the characters at left and right are no longer equal or reach the boundaries of the string. The length of the palindrome is then calculated as right - left - 1 (since right and left have been already incremented/decremented respectively after exiting the while loop).

In the main method, we iterate over each character in the string and use the expandAroundCenter() method to get the length of the palindromic substring starting from that character as a center. We update the start and end indices if the length of the current palindromic substring is greater than the length of the previously found longest substring.

Finally, we return the substring using the start and end indices.

The time complexity of this algorithm is O(n^2) where n is the length of the input string. This is because for every character, we expand around it to check if a palindrome is formed. The space complexity is O(1) as we are only using a constant amount of extra space.

In an interview, it is important to discuss the approach, explain the algorithm, and write clean and optimized code with proper comments.

10. Minimum Window Substring

1. The problem is called "Minimum Window Substring" and can be found on LeetCode at the following link: https://leetcode.com/problems/minimum-window-substring/

2. Here is the optimized Java code for the problem "Minimum Window Substring" with detailed comments:

class Solution {
    public String minWindow(String s, String t) {
        // Initialize two arrays to store the frequency of characters in string s and t
        int[] targetFreq = new int[128]; // Assume ASCII characters
        int[] windowFreq = new int[128];
        
        // Calculate the frequency of characters in string t
        for (char c : t.toCharArray()) {
            targetFreq[c]++;
        }
        
        int left = 0; // Left pointer of the sliding window
        int right = 0; // Right pointer of the sliding window
        
        int count = 0; // Counter to keep track of the characters in string t found in string s
        int minLen = Integer.MAX_VALUE; // Minimum window length
        int startIndex = 0; // Start index of the minimum window substring
        
        // Start expanding the sliding window
        while (right < s.length()) {
            char currentChar = s.charAt(right);
            
            // If the current character in string s exists in string t, increase the count
            if (windowFreq[currentChar] < targetFreq[currentChar]) {
                count++;
            }
            
            windowFreq[currentChar]++;
            
            // If all characters in string t are found in string s
            while (count == t.length()) {
                // Update the minimum window substring if it is smaller than the previous minimum
                if (right - left + 1 < minLen) {
                    minLen = right - left + 1;
                    startIndex = left;
                }
                
                char leftChar = s.charAt(left);
                
                // Shrink the sliding window by moving the left pointer
                windowFreq[leftChar]--;
                
                // If the character at left pointer is part of string t, decrease the count
                if (windowFreq[leftChar] < targetFreq[leftChar]) {
                    count--;
                }
                
                left++;
            }
            
            right++; // Expand the sliding window by moving the right pointer
        }
        
        return minLen == Integer.MAX_VALUE ? "" : s.substring(startIndex, startIndex + minLen);
    }
}

1. Algorithm and thought process:

• The problem asks to find the minimum window substring in string s that contains all the characters from string t in any order.

• We can solve this problem using the sliding window technique.

• First, we calculate the frequency of characters in string t using an array called targetFreq.

• Then, we initialize two pointers, left and right, and move the right pointer to expand the sliding window.

• While expanding the window, we check if the current character at the right pointer exists in string t. If it does, we increase the count, which indicates the number of characters from string t found in string s.

• Once all the characters from string t are found in the current window, we shrink the window from the left pointer until the count decreases.

• While shrinking the window, we keep track of the minimum window length and its start index.

• We continue expanding and shrinking the window until we reach the end of string s.

• Finally, we return the minimum window substring based on the minimum window length and start index.

During an interview, it is crucial to communicate your thought process clearly. Here is a step-by-step breakdown of how to approach the problem:

1. Understand the problem requirements and constraints.

2. Identify possible approaches to solve the problem and choose the most suitable one.

3. Clearly define the input and output formats.

4. Write down the algorithm in pseudocode or explain it verbally.

5. Start coding by initializing variables and data structures.

6. Implement each step of the algorithm using appropriate data structures and control flow.

7. Test the code against various test cases, including edge cases, to ensure correctness.

8. Optimize the code if possible by reducing time or space complexity.

9. Add comments to explain the code logic and improve code readability.

10. Verify that the code is optimized, easy to understand, and meets the problem requirements.

Reiterating the steps and discussing any optimizations or improvements after coding demonstrates a solid problem-solving approach during an interview.