Stack

1. Decode String (LeetCode #394)

1. Problem Description: The problem "Decode String" is a string manipulation problem where we are given a string encoded in a specific format and we need to decode it. The input string can have digits, square brackets, and lowercase English letters. The encoding rules are as follows:

• Digits represent the number of times the following enclosed characters should be repeated.

• Square brackets '[ ]' represent a group of enclosed characters that should be repeated. The task is to decode the given string and return the decoded version.

LeetCode Link: https://leetcode.com/problems/decode-string/

1. Optimized Java Code with Detailed Comments:

public class DecodeString {
    public String decodeString(String s) {
        Stack<Integer> countStack = new Stack<Integer>(); // Stack to store the counts
        Stack<String> stringStack = new Stack<String>(); // Stack to store the encoded strings
        StringBuilder decodedString = new StringBuilder();
        int count = 0; // Keeps track of the count for repetition
        String currentString = ""; // Current string being built

        // Iterate through each character in the input string
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) { // If the character is a digit
                count = count * 10 + (c - '0'); // Calculate the count
            } else if (c == '[') { // If the character is an opening bracket
                countStack.push(count); // Push the current count to the stack
                stringStack.push(currentString); // Push the current string to the stack
                count = 0; // Reset the count
                currentString = ""; // Reset the current string
            } else if (c == ']') { // If the character is a closing bracket
                StringBuilder temp = new StringBuilder(); // StringBuilder to store repeated string
                int repetitions = countStack.pop(); // Pop the count from stack (number of repetitions)
                for (int j = 0; j < repetitions; j++) {
                    temp.append(currentString); // Append the current string to temp for repetitions
                }
                currentString = stringStack.pop() + temp.toString(); // Append the previous string to current with repetitions
            } else { // If the character is a letter
                currentString += c; // Append the character to the current string
            }
        }
        
        return currentString; // Return the final decoded string
    }
}

1. Detailed Description of Algorithm and Approach:

• Initialize two stacks: countStack to store the counts of repetitions and stringStack to store the encoded strings.

• Initialize decodedString as a StringBuilder to store the final decoded string.

• Initialize count as 0 to keep track of the count for repetitions and currentString as an empty string to store the current encoded string being built.

• Iterate through each character in the input string s.

• If the character is a digit, convert it to a number and update the count variable by multiplying it with 10 and adding the current digit.

• If the character is an opening bracket [, push the current count to the countStack and push the current string to the stringStack. Reset both count and currentString.

• If the character is a closing bracket ], pop the previous count from the countStack. Create a StringBuilder temp to store the repeated string.

• Repeat the current string repetitions number of times and append it to temp.

• Pop the previous string from stringStack and append it with temp.toString() to update the currentString.

• If the character is a letter, append it to the currentString.

• After iterating through all characters, return the currentString as the final decoded string.

During an interview, it is important to explain the approach step by step and mention the time and space complexity of the solution. In this problem, the time complexity of the solution is O(N), where N is the length of the input string, as we iterate through each character only once. The space complexity is O(M), where M is the maximum number of recursive calls made to the stack, which depends on the nesting of brackets in the input string.

2. Valid Parentheses (LeetCode #20)

1. Problem Description: The problem "Valid Parentheses" states that given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid. An input string is valid if:

• Open brackets must be closed by the same type of brackets.

• Open brackets must be closed in the correct order.

We need to implement a function that returns a boolean indicating if the input string is valid or not.

Problem Link: https://leetcode.com/problems/valid-parentheses/

1. Optimized Java Code:

import java.util.Stack;

class Solution {
    public boolean isValid(String s) {
        // Creating a stack to store the opening brackets
        Stack<Character> stack = new Stack<>();
        
        // Iterating through each character in the string
        for (char c : s.toCharArray()) {
            // If the current character is an opening bracket, push it to the stack
            if (c == '(' || c == '[' || c == '{') {
                stack.push(c);
            }
            // If the current character is a closing bracket
            else if (c == ')' || c == ']' || c == '}') {
                // If the stack is empty, there is no opening bracket for the current closing bracket
                if (stack.isEmpty()) {
                    return false;
                }
                
                // Checking if the current closing bracket matches the top of the stack
                char top = stack.pop();
                if ((c == ')' && top != '(') || (c == ']' && top != '[') || (c == '}' && top != '{')) {
                    return false;
                }
            }
        }
        
        // If there are still opening brackets left in the stack, the string is invalid
        return stack.isEmpty();
    }
}

1. Algorithm and Approach:

The algorithm for solving this problem can be summarized as follows:

• We initialize an empty stack to store the opening brackets.

• We iterate through each character in the input string.

• If the current character is an opening bracket ('(', '[', or '{'), we push it onto the stack.

• If the current character is a closing bracket (')', ']', or '}'), we check if the stack is empty.

  • If the stack is empty, it means there is no opening bracket for the current closing bracket, so we return false.

  • Otherwise, we pop the top element from the stack and check if it matches the current closing bracket. If not, we return false.

• After iterating through all the characters in the string, we check if there are any remaining opening brackets in the stack. If there are, it means the string is invalid, so we return false.

• If there are no remaining opening brackets and we have not returned false at any point, it means the string is valid, so we return true.

During the interview process, it is important to carefully read and understand the problem statement. In this particular problem, understanding the concept of valid parentheses and the rules for their arrangement is crucial. Breaking down the problem into smaller steps and identifying the proper data structure and algorithm is also important.

The use of a stack to store the opening brackets and the iteration through each character in the string are common strategies in solving problems related to parentheses combinations. By using a stack, we can efficiently check for the valid arrangement of parentheses. Additionally, by iterating through each character in the string, we can process and validate the parentheses in a sequential manner.

It is also a good practice to test the code with different inputs, including edge cases, to ensure its correctness and efficiency.

3. Remove Duplicate Letters (LeetCode #316)

1. Problem Description:

   • Given a string s, remove duplicate letters such that every letter appears once and only once in the final result. You must make sure your result is the smallest in lexicographical order among all possible results.

   • Link to the problem on LeetCode: Remove Duplicate Letters
2. Optimized Java Code:

import java.util.Stack;

class Solution {
    public String removeDuplicateLetters(String s) {
        int[] lastOccurrence = new int[26];
        boolean[] used = new boolean[26];
        Stack<Character> stack = new Stack<>();

        // Track the last occurrence of each character in the string
        for (int i = 0; i < s.length(); i++) {
            lastOccurrence[s.charAt(i) - 'a'] = i;
        }

        // Iterate through each character in the string
        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);

            // Skip if the character is already in the result
            if (used[ch - 'a']) {
                continue;
            }

            // Remove any character from the stack that is greater than current character
            while (!stack.isEmpty() && stack.peek() > ch && lastOccurrence[stack.peek() - 'a'] > i) {
                used[stack.pop() - 'a'] = false;
            }

            stack.push(ch);
            used[ch - 'a'] = true;
        }

        StringBuilder result = new StringBuilder();
        while (!stack.isEmpty()) {
            result.insert(0, stack.pop());
        }

        return result.toString();
    }
}

1. Algorithm and Interview Process:

   • This problem can be solved using a greedy algorithm approach with the help of a stack.

   • The idea is to maintain a stack that stores the characters of the result string in lexicographical order.

   • We iterate through each character in the string and check if it has already been used in the result. If so, we skip it.

   • Otherwise, we compare it with the top of the stack and pop any characters that are greater than the current character and have further occurrences in the remaining string. This ensures that the result string is the smallest lexicographically.

   • Finally, we build the result using a StringBuilder by inserting characters in reverse order from the stack.

   • During an interview, it's important to explain the problem clearly to the interviewer and analyze the given problem statement. Understanding the problem is the first step in solving it effectively.

   • Next, you can explain your approach, discussing the data structures, algorithms, and optimizations you plan to use.

   • During the implementation, make sure to add useful comments to explain your thought process and steps taken to solve the problem. This highlights your understanding and clarity of the code.

   • Additionally, you can discuss the time and space complexity of your solution, ensuring it meets the requirements of the problem constraints.

   • Finally, it's important to test your code with different test cases to ensure correctness and handle edge cases. Discussing the test cases and their expected outputs can further demonstrate your problem-solving skills.

4. Evaluate Reverse Polish Notation (LeetCode #150)

Problem: Evaluate Reverse Polish Notation (LeetCode #150)

Problem Description: The problem is to evaluate the given arithmetic expression in Reverse Polish Notation (RPN). RPN is a mathematical notation in which every operator follows all its operands.

In Reverse Polish Notation, operators are written after their operands. For example, the expression "2 1 + 3 *" is equivalent to "((2+1)*3)".

LeetCode Link: https://leetcode.com/problems/evaluate-reverse-polish-notation/

Algorithm Explanation: To evaluate the expression in Reverse Polish Notation, we can use a stack.

1. Iterate through each element in the given input tokens.

2. If the current element is a number, push it onto the stack.

3. If the current element is an operator, pop the top two elements from the stack, apply the operator, and push the result back onto the stack.

4. After iterating through all the elements, the final result will be the top element of the stack.

Java Code (with detailed comments):

import java.util.Stack;

public class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer> stack = new Stack<>();
        
        for (String token : tokens) {
            if (isOperator(token)) {
                int operand2 = stack.pop();
                int operand1 = stack.pop();
                int result = performOperation(token, operand1, operand2);
                stack.push(result);
            } else {
                // Current token is a number, convert it to int and push onto the stack
                stack.push(Integer.parseInt(token));
            }
        }
        
        // The final result will be the top element of the stack
        return stack.pop();
    }
    
    private boolean isOperator(String token) {
        return token.equals("+") || token.equals("-") || token.equals("*") || token.equals("/");
    }
    
    private int performOperation(String operator, int operand1, int operand2) {
        switch (operator) {
            case "+":
                return operand1 + operand2;
            case "-":
                return operand1 - operand2;
            case "*":
                return operand1 * operand2;
            case "/":
                return operand1 / operand2;
            default:
                return 0;
        }
    }
}

During an interview process:

1. Understand the problem: Read the problem statement carefully and ensure you understand the requirements and constraints.

2. Identify the approach: Recognize that a stack can be used to evaluate the expression in RPN.

3. Plan the algorithm: Break down the problem into smaller steps, such as iterating through the tokens, handling operators, and performing calculations.

4. Write code: Implement the algorithm in code, ensuring you handle all edge cases and write clean and readable code.

5. Test thoroughly: Test the code with various test cases, including different arithmetic expressions in RPN, to verify its correctness.

6. Optimize if needed: Analyze the time and space complexity of the algorithm. If necessary, optimize the code for better performance.

7. Explain your solution: During the interview, explain your algorithm step by step, justify its correctness, and discuss its time and space complexity.

5. Next Greater Element II (LeetCode #503)

1. Problem Description: The problem "Next Greater Element II" is a LeetCode problem (#503). The problem statement can be found at the following link: Next Greater Element II - LeetCode #503
2. Optimized Java Code with Detailed Comments:

import java.util.Arrays;
import java.util.Stack;

class Solution {
    public int[] nextGreaterElements(int[] nums) {
        int n = nums.length;
        int[] res = new int[n];
        Arrays.fill(res, -1); // Initialize result array with -1

        Stack<Integer> stack = new Stack<>(); // Stack to store indices of decreasing elements

        // Iterate the array twice to handle circular nature of the array
        for (int i = 0; i < 2 * n; i++) {
            int num = nums[i % n]; // Use modulo operator to get the correct index

            while (!stack.isEmpty() && nums[stack.peek()] < num) {
                res[stack.pop()] = num; // Set the next greater element for the popped index
            }

            if (i < n) {
                stack.push(i); // Store indices of the elements in the first round
            }
        }

        return res;
    }
}

1. Algorithm Description and Interview Process: When approaching this problem during an interview, it is important to understand the problem requirements and constraints. The problem asks to find the next greater element for every element in a non-empty circular array. If there is no such greater element, the result should be -1.

To solve this problem, an optimized approach is to iterate the array twice using a single loop. This handles the circular nature of the array. We initialize a result array with -1, indicating that there is no next greater element by default. We also use a stack to store the indices of the decreasing elements.

While iterating the array, we compare the current element with the top element in the stack. If the current element is greater, we update the result array at the corresponding index with the current element. We continue this comparison until we cannot find a greater element or the stack becomes empty. By doing this, we ensure that the top element in the stack points to the index of the largest element seen so far.

In the first round of iteration, we store the indices of the elements in the stack as we haven't visited them before. In the second round, we do not store any new indices in the stack; we only use the stored indices to update the result array with the next greater elements.

The time complexity of this algorithm is O(n), as we iterate the array twice. The space complexity is O(n), as we use a stack to store indices.

During an interview, you can mention the efficient use of stack and the optimization of solving the problem in a single loop. You can also discuss the time and space complexity of the algorithm to show your knowledge of analyzing algorithms. Additionally, it's important to test your solution on various test cases to verify its correctness and handle edge cases.

6. Simplify Path (LeetCode #71)

Sure, I can help you with that!

1. Problem Description and Link: The problem "Simplify Path" is described as follows:

Given an absolute path for a file (Unix-style), simplify it. In other words, convert it to the canonical path.

In a UNIX-style file system:

• A period '.' refers to the current directory.

• A double period '..' denotes going up one level to the parent directory.

More details about the problem can be found here: https://leetcode.com/problems/simplify-path/

1. Optimized Java Code with Detailed Comments:

import java.util.*;

class Solution {
    public String simplifyPath(String path) {
        Deque<String> stack = new LinkedList<>();

        // Split the input path by '/' and process each component
        String[] components = path.split("/");
        for (String component : components) {
            if (component.equals(".") || component.isEmpty()) {
                // Ignore '.' and empty strings (caused by consecutive '/' characters)
                continue;
            } else if (component.equals("..")) {
                // If we encounter '..', pop the last directory from stack (if it exists)
                if (!stack.isEmpty()) {
                    stack.pop();
                }
            } else {
                // Otherwise, add the directory to stack
                stack.push(component);
            }
        }
        
        // Build the simplified path by joining the components from stack in reverse order
        StringBuilder result = new StringBuilder();
        while (!stack.isEmpty()) {
            result.insert(0, "/" + stack.pop());
        }
        
        // If the result is empty, the simplified path should be "/"
        return result.length() == 0 ? "/" : result.toString();
    }
}

1. Algorithm Description and Thinking Process:

During the interview process, it's important to understand the problem and come up with an efficient algorithm. Here is one way to approach the "Simplify Path" problem:

• We can split the input path by '/' to get the components of the path.

• We can use a stack to keep track of the directories in the canonical path.

• For each component in the path:

  • If it is '.', we ignore it as it refers to the current directory.

  • If it is '..', we pop the last directory from the stack (if it exists), as it refers to going up one level to the parent directory.

  • Otherwise, it is a valid directory name and we add it to the stack.

• Once we have processed all the components, we can build the simplified path by joining the directories from the stack in reverse order.

• If the stack is empty, the simplified path should be just '/'.

• Otherwise, we can pop directories from the stack and prepend them to the result string along with '/' to get the simplified path.

This algorithm takes O(N) time and O(N) space, where N is the length of the input path.

7. Basic Calculator (LeetCode #224)

1. Problem Description: The problem "Basic Calculator" is a LeetCode problem (#224) that requires you to implement a calculator that can evaluate simple arithmetic expressions that include parentheses and the operators '+', '-', '*', and '/'. The expressions provided will be valid and only contain non-negative integers and spaces.

LeetCode link: Basic Calculator

1. Optimized Java Code with Detailed Comments:

import java.util.Stack;

public class BasicCalculator {
    public int calculate(String s) {
        // Remove all whitespace characters from the expression
        s = s.replaceAll("\\s", "");
        
        Stack<Integer> stack = new Stack<>();
        int result = 0;   // Final result of the evaluation
        int num = 0;      // Tracks the current number being evaluated
        int sign = 1;     // Tracks the sign of the current expression (+1 for positive, -1 for negative)
        
        for (char c : s.toCharArray()) {
            if (Character.isDigit(c)) {
                num = num * 10 + c - '0';
            } else if (c == '+') {
                result += sign * num;
                num = 0;
                sign = 1;
            } else if (c == '-') {
                result += sign * num;
                num = 0;
                sign = -1;
            } else if (c == '(') {
                stack.push(result);
                stack.push(sign);
                result = 0;
                sign = 1;
            } else if (c == ')') {
                result += sign * num;
                num = 0;
                result *= stack.pop();    // Multiply the inner result by the sign from the stack
                result += stack.pop();    // Add the outer result from the stack
            }
        }
        
        // Add the last number and the final result to handle the case where there are no closing parentheses
        result += sign * num;

        return result;
    }

    // Testing the implementation
    public static void main(String[] args) {
        BasicCalculator calculator = new BasicCalculator();

        // Example test cases
        System.out.println(calculator.calculate("1 + 1"));              // Expected output: 2
        System.out.println(calculator.calculate(" 2-1 + 2 "));          // Expected output: 3
        System.out.println(calculator.calculate("(1+(4+5+2)-3)+(6+8)"));// Expected output: 23
    }
}

1. Detailed Description of the Algorithm and Interview Process:

• The algorithm solves the problem using a stack to keep track of intermediate results during the evaluation of the expression.

• It iterates over each character in the input string and performs different operations depending on the character encountered.

• If a digit (0-9) is encountered, it forms the current number by multiplying the existing number by 10 and adding the current digit.

• If a '+' or '-' operator is encountered, it adds the current number to the result (with the correct sign) and resets the current number and sign.

• If an opening bracket '(' is encountered, it pushes the current result and sign onto the stack and resets the result and sign.

• If a closing bracket ')' is encountered, it adds the current number to the result (with the correct sign), multiplies the result by the sign on top of the stack, and adds the outer result from the stack.

• Once all characters in the input string are processed, the final number (if any) and the result are added together to handle the case where there are no closing brackets.

• The final result is returned.

During an interview, it is crucial to understand the problem requirements and constraints. Start by clarifying any ambiguities and discussing potential edge cases. In this problem, it is important to handle negative numbers, whitespace characters, and valid input expressions.

To walk through the algorithm during an interview, you can explain and implement the code step by step, discussing time and space complexity analysis. Emphasize the usage of the stack to track intermediate results and handle nested parentheses properly. Additionally, you can mention the importance of regular expression and character manipulation techniques for efficient processing of the input string.

Overall, the goal is to showcase your problem-solving skills and ability to translate the problem statement into a working algorithm, along with highlighting your understanding of data structures (e.g., stack) and proper implementation techniques.

8. Expression Add Operators (LeetCode #282)

1. Problem Description: The problem "Expression Add Operators" is given on LeetCode. The problem link is https://leetcode.com/problems/expression-add-operators/.

Given a string of digits, and a target value, we need to add binary operators (+, -, or *) between the digits to evaluate the expression and obtain the target value. The expression must be formed in such a way that it follows the operator precedence, and any multiplication is done before addition and subtraction.

Example: Input: num = "123", target = 6 Output: ["1+2+3", "123"]

1. Optimized Java code with detailed comments:

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

class Solution {
    public List<String> addOperators(String num, int target) {
        List<String> res = new ArrayList<>();
        backtrack(res, "", num, target, 0, 0, 0);
        return res;
    }
    
    private void backtrack(List<String> res, String expr, String num, int target, int index, long eval, long prevNum) {
        if (index == num.length()) {
            if (eval == target) {
                res.add(expr);
            }
            return;
        }
        
        for (int i = index; i < num.length(); i++) {
            if (i > index && num.charAt(index) == '0') {
                // To avoid adding numbers with leading zeros
                break;
            }
            
            long currNum = Long.parseLong(num.substring(index, i + 1));
            
            if (index == 0) {
                // Starting point in the expression
                backtrack(res, expr + currNum, num, target, i + 1, currNum, currNum);
            } else {
                // Adding operator '+'
                backtrack(res, expr + "+" + currNum, num, target, i + 1, eval + currNum, currNum);
                
                // Adding operator '-'
                backtrack(res, expr + "-" + currNum, num, target, i + 1, eval - currNum, -currNum);
                
                // Adding operator '*'
                backtrack(res, expr + "*" + currNum, num, target, i + 1, eval - prevNum + prevNum * currNum, prevNum * currNum);
            }
        }
    }
}

1. Detailed Description of Algorithm and Interview Process:

The given problem can be solved using backtracking. At each step, we can either add '+', '-', or '*' as an operator between the numbers. We will recursively try out all possible combinations of operators and evaluate the expression.

The backtracking algorithm can be summarized as follows:

1. Start with an empty expression and initialize the evaluation and previous number to 0.

2. Iterate through the digits of the given input number.

3. At each digit, check if adding an operator with the current digit is possible (avoiding numbers with leading zeros).

4. If it's the starting point (index = 0), go to the next digit recursively.

5. If it's not the starting point, try adding '+', '-', and '*' operators and go to the next digit recursively.

6. Keep track of the evaluation and the previous number at each step.

7. If we reach the end of the input number, check if the evaluation matches the target.

8. If the evaluation matches the target, add the expression to the result list.

9. Backtrack by removing the last operator and digit to explore other possibilities.

During an interview, it is important to understand the problem completely before jumping into the implementation. Break down the problem into smaller subproblems and identify patterns or problem-solving techniques that can be applied.

Here are some tips to tackle the "Expression Add Operators" problem or similar problems:

1. Understand the problem requirements and constraints: Understand the problem statement, input format, and output format. Identify any specific constraints or edge cases that need to be considered.

2. Break down the problem into smaller subproblems: Identify the subproblems involved in solving the main problem. In this case, the main problem involves finding all possible expressions by adding operators, while the subproblems involve evaluating expressions recursively.

3. Think of recursive/backtracking approach: For problems involving combinations or permutations, a recursive or backtracking approach can be useful. Break down the problem into smaller subproblems and iteratively solve them.

4. Identify the base case and recursive case: Identify the base case(s) where the recursion terminates, and the recursive case(s) where the function calls itself recursively with updated parameters.

5. Keep track of intermediate results: Think about what information needs to be tracked at each step of the recursion. In this problem, we need to track the current expression, evaluation, previous number, and index.

6. Experiment with small inputs: Before implementing the code, try out the algorithm on small inputs manually. This will help validate the approach and catch any logical errors.

7. Optimize the solution if possible: Once the basic solution is working, think about potential optimizations. In this problem, avoiding numbers with leading zeros is an important optimization.

8. Test thoroughly: Test the code with different test cases, including edge cases and corner cases. Ensure that the code produces the correct output and handles all possible scenarios.

Remember, during an interview, the interviewer is not only interested in seeing the final code but also in how you approach the problem, think through different scenarios, and optimize the solution.

9. Valid Parenthesis String (LeetCode #678)

1. Problem Description: The problem "Valid Parenthesis String" is described as follows: Given a string containing only three types of characters: '(', ')', and '*', write a function to check whether this string is valid. We define the validity of a string by these rules:

2. Any left parenthesis '(' must have a corresponding right parenthesis ')'.

3. Any right parenthesis ')' must have a corresponding left parenthesis '('.

4. Left parenthesis '(' must go before the corresponding right parenthesis ')'.

5. '*' could be treated as a single right parenthesis ')' or a single left parenthesis '(' or an empty string.

6. An empty string is also valid.

You can find the problem on LeetCode here: https://leetcode.com/problems/valid-parenthesis-string/

1. Optimized Java Code with Detailed Comments:

public class Solution {
    public boolean checkValidString(String s) {
        int minOpen = 0;  // minimum number of open parentheses
        int maxOpen = 0;  // maximum number of open parentheses
        
        for (char c : s.toCharArray()) {
            if (c == '(') {
                minOpen++;
                maxOpen++;
            } else if (c == ')') {
                minOpen--;  // consider '*' as ')'
                maxOpen--;  // consider '*' as '('
            } else {  // c == '*'
                minOpen--;  // consider '*' as ')'
                maxOpen++;  // consider '*' as '('
            }
            
            minOpen = Math.max(minOpen, 0);  // ensure minOpen is non-negative
            
            if (maxOpen < 0) {
                return false;  // more ')' than available '(' and '*'s
            }
        }
        
        return minOpen == 0;  // check if all open parentheses are closed
    }
}

1. Algorithm and Thinking Process: To solve this problem, we can use a greedy approach and keep track of the minimum and maximum number of open parentheses encountered so far while iterating through the string.

The intuition behind this algorithm is as follows:

• Every open parenthesis '(' can be considered as an increment of both minOpen and maxOpen.

• Every closing parenthesis ')' can be considered as a decrement of both minOpen and maxOpen.

• Every asterisk '*' can be considered as a decrement of minOpen (considering it as ')') and an increment of maxOpen (considering it as '(').

At any point in the iteration, if the count of ')' exceeds the available count of '(' and '*', then the string is invalid and we return false. We keep track of the minimum and maximum number of open parentheses encountered to handle cases where there are more closing parentheses than the available opening parentheses.

After the full iteration, we check if there are no open parentheses left, i.e., the minOpen is 0. If it is, then all open parentheses have been closed, making the string valid. Otherwise, some open parentheses are left unclosed, making the string invalid.

This algorithm has a time complexity of O(n), where n is the length of the input string 's'. It only requires a single pass through the string.

10. Decode Ways (LeetCode #91)

1. Problem Description: The problem "Decode Ways" is a dynamic programming problem from LeetCode, specifically problem number 91. The problem can be found on LeetCode using the following link: Decode Ways (LeetCode #91)

Given a non-empty string containing only digits, we need to determine the total number of ways to decode it. The mapping of alphabets to digits is given by "A" = 1, "B" = 2, ..., "Z" = 26.

Examples:

• Input: "12" Output: 2 Explanation: It could be decoded as "AB" (1 2) or "L" (12).

• Input: "226" Output: 3 Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).

1. Optimized Java Code: Below is the optimized Java code for the "Decode Ways" problem with detailed comments:

class Solution {
    public int numDecodings(String s) {
        // Store the number of ways to decode substrings of s
        int[] dp = new int[s.length() + 1];

        // Initialize base cases
        dp[0] = 1; // Empty string has 1 decoding
        dp[1] = s.charAt(0) != '0' ? 1 : 0; // Single character can be decoded only if it is not '0'

        // Build up the dynamic programming table
        for (int i = 2; i < dp.length; i++) {
            // Check if single-digit decoding is possible
            int oneDigit = Integer.parseInt(s.substring(i - 1, i));
            if (oneDigit >= 1 && oneDigit <= 9) {
                dp[i] += dp[i - 1];
            }
            
            // Check if two-digit decoding is possible
            int twoDigits = Integer.parseInt(s.substring(i - 2, i));
            if (twoDigits >= 10 && twoDigits <= 26) {
                dp[i] += dp[i - 2];
            }
        }

        // Return the number of ways to decode the entire string
        return dp[s.length()];
    }
}

1. Algorithm and Approach: To solve this problem efficiently, we can utilize dynamic programming with an array dp to store the number of ways to decode substrings of s. The approach follows these steps:

• Initialize dp[0] and dp[1] as base cases:

  • dp[0] is set to 1 as the empty string has one decoding.

  • dp[1] is set to 1 if the single character is not '0' (indicating that it can be decoded as a letter).

• Build up the dynamic programming table dp from index 2 to s.length():

  • For each index i, we check if single-digit decoding is possible by parsing the substring s.substring(i - 1, i) into an integer. If the integer value is between 1 and 9 (inclusive), we can add the current value of dp[i - 1] to dp[i] to calculate the number of ways to decode the current substring.

  • Similarly, we check if two-digit decoding is possible by parsing the substring s.substring(i - 2, i) into an integer. If the integer value is between 10 and 26 (inclusive), we can add the current value of dp[i - 2] to dp[i].

• Finally, we return the value of dp[s.length()], which represents the number of ways to decode the entire string s.

During an interview process, it is crucial to explain the approach to the interviewer, followed by implementing the solution in code. The above code provides a concise and optimized solution to the problem, while the detailed comments help in understanding the logic behind it.

11. Backspace String Compare (LeetCode #844)

1. Problem Description: The problem "Backspace String Compare" is described as follows: Given two strings s and t, determine if they are equal after processing the backspaces. A backspace character '#' means deleting the previous character. Return true if the two strings are equal after backspace processing, and false otherwise.

Problem Link: https://leetcode.com/problems/backspace-string-compare/

1. Optimized Java Code with Detailed Comments:

class Solution {
    public boolean backspaceCompare(String s, String t) {
        // Convert the strings to char arrays for easy manipulation
        char[] sArr = s.toCharArray();
        char[] tArr = t.toCharArray();
        
        // Apply the backspace processing for both strings
        processBackspaces(sArr);
        processBackspaces(tArr);
        
        // Compare the processed strings
        return Arrays.equals(sArr, tArr);
    }
    
    private void processBackspaces(char[] arr) {
        int i = 0; // Pointer to keep track of the current index
        
        for (char c : arr) {
            if (c == '#') {
                if (i > 0) {
                    i--; // decrement the pointer if there are valid characters to delete
                    
                    // Set the character at pointer index to a null character
                    arr[i] = '\0'; 
                }
            } else {
                // Set the current character at pointer index
                arr[i] = c;
                i++; // increment the pointer
            }
        }
    }
}

1. Detailed Description:

The backspace string compare problem can be solved using an optimized approach with O(n) time complexity and O(1) space complexity. The algorithm follows these steps:

• Step 1: Convert the input strings 's' and 't' into character arrays for easy manipulation.

• Step 2: Apply the backspace processing for both strings using the processBackspaces() helper method.

• Step 3: The processBackspaces() method iterates over each character in the array and performs the following operations:

  • If the current character is a backspace '#' and there are valid characters to delete (i > 0), the pointer is decremented (i--), and the character at the pointer index is set to a null character ('\0').

  • If the current character is not a backspace, it is set at the current index (arr[i]), and the pointer is incremented (i++).

• Step 4: After backspace processing, both strings are compared using the Arrays.equals() method. If they are equal, return true; otherwise, return false.

During an interview, it is important to think through the problem step by step and communicate your thought process with the interviewer. Here's a breakdown of how to approach this problem:

1. Understand the problem requirements: Start by carefully reading and understanding the problem statement. Identify the input and output requirements.

2. Identify the key steps: Analyze the problem and break it down into smaller steps. In this case, the key steps are applying the backspace processing and comparing the processed strings.

3. Define the data structure: Determine the appropriate data structure to use. In this problem, converting the strings to character arrays simplifies the manipulation process.

4. Develop the algorithm: Determine the algorithmic approach to solve the problem. The optimized approach for this problem involves processing the backspaces in both strings using a single loop and then comparing the processed strings.

5. Implement the solution: Transform the algorithm into code, ensuring that it is correct and efficient. Add necessary comments to explain the code logic and make it more readable.

6. Test the solution: Verify the correctness of the code by running various test cases, including edge cases and corner cases.

Remember to communicate with the interviewer throughout the process, explaining your thought process, assumptions, and any optimizations you made.

12. Minimum Remove to Make Valid Parentheses (LeetCode #1249)

1. Problem Description and Link: The problem "Minimum Remove to Make Valid Parentheses" is available on LeetCode at the following link: https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/

2. Optimized Java Code with Detailed Comments:

import java.util.*;

class Solution {
    public String minRemoveToMakeValid(String s) {
        StringBuilder sb = new StringBuilder(s);
        Stack<Integer> stack = new Stack<>();
        
        // First Pass: Remove all ')'
        for (int i = 0; i < sb.length(); i++) {
            if (sb.charAt(i) == '(') {
                stack.push(i);
            } else if (sb.charAt(i) == ')') {
                if (!stack.isEmpty()) {
                    stack.pop();
                } else {
                    sb.setCharAt(i, '*'); // Mark ')' for removal
                }
            }
        }
        
        // Second Pass: Remove extra '('
        while (!stack.isEmpty()) {
            sb.setCharAt(stack.pop(), '*'); // Mark '(' for removal
        }
        
        // Construct the resulting string
        StringBuilder res = new StringBuilder();
        for (int i = 0; i < sb.length(); i++) {
            if (sb.charAt(i) != '*') {
                res.append(sb.charAt(i));
            }
        }
        
        return res.toString();
    }
}

1. Algorithm and Thought Process:

The algorithm is based on the idea of using a stack to keep track of opening parentheses '(' and removing unnecessary closing parentheses ')' from the input string.

Here's how the algorithm works:

• We use a StringBuilder to make string modification easier.

• We iterate through the string from left to right in two passes.

• In the first pass, we remove all unnecessary closing parentheses ')' by maintaining a stack. Whenever we encounter a closing parentheses and the stack is empty, we mark it for removal by replacing it with '*'. If the stack is not empty, we pop an opening parentheses '(' from the stack, indicating a valid pair.

• In the second pass, we remove all remaining opening parentheses '(' from the stack. Since there is no corresponding closing parentheses, we mark them for removal by replacing them with '*'.

• Finally, we construct the resulting string by appending all characters that are not '*'.

This algorithm has a time complexity of O(n) and a space complexity of O(n), where 'n' is the length of the input string.

13. Daily Temperatures (LeetCode #739)

1. Problem Description: The problem "Daily Temperatures" can be found on LeetCode using the following link: https://leetcode.com/problems/daily-temperatures/

Given a list of daily temperatures, return a list such that, for each day in the input, tells you how many days you would have to wait until a warmer temperature. If there is no future day for which this is possible, put 0 instead.

Example: Input: temperatures = [73, 74, 75, 71, 69, 72, 76, 73] Output: [1, 1, 4, 2, 1, 1, 0, 0]

1. Optimized Java Code with Detailed Comments:

import java.util.*;

public class DailyTemperatures {
    public int[] dailyTemperatures(int[] temperatures) {
        int[] result = new int[temperatures.length]; // Initialize the result array
        Stack<Integer> stack = new Stack<>(); // Initialize a stack to track indices

        for (int i = 0; i < temperatures.length; i++) {
            while (!stack.isEmpty() && temperatures[i] > temperatures[stack.peek()]) {
                int index = stack.pop();
                result[index] = i - index; // Store the difference in indices as the result
            }
            stack.push(i);
        }

        return result;
    }
}

1. Algorithm Description and Interview Process Insight:

The provided solution uses a stack to solve the problem efficiently in linear time complexity (O(N)).

In this problem, we need to find the number of days we have to wait until a warmer temperature is encountered for each day in the input list.

To solve the problem, we iterate through the temperatures from left to right and push their indices onto the stack. For each temperature, we compare it with the temperature at the top of the stack. If the current temperature is warmer than the temperature at the top of the stack, it means we have found a warmer temperature, and we can calculate the difference in indices to determine the number of days we have to wait. We then update the result for the index at the top of the stack and pop it from the stack.

By using a stack, we only keep track of the indices of the temperatures which have not found a warmer day yet. As soon as a warmer temperature is encountered, we can directly update the result for the indices popped from the stack. This approach avoids unnecessary comparisons and ensures that we find the number of days to wait in linear time complexity.

During an interview, explaining the algorithm can showcase your understanding of stack data structure and efficient problem-solving techniques. Additionally, focusing on the time complexity analysis and the reasoning behind the chosen approach can demonstrate your ability to think critically and optimize solutions.

14. Longest Valid Parentheses (LeetCode #32)

1. The problem "Longest Valid Parentheses" on LeetCode can be found at the following link: https://leetcode.com/problems/longest-valid-parentheses/

2. Here is an optimized Java code for the problem "Longest Valid Parentheses" with detailed comments:

class Solution {
    public int longestValidParentheses(String s) {
        int n = s.length();
        int maxLength = 0;
        
        // We use a stack to store the indices of opening parentheses
        Stack<Integer> stack = new Stack<>();
        
        // We initialize the stack with -1 to handle cases where the first character is a closing parentheses
        stack.push(-1);
        
        for (int i = 0; i < n; i++) {
            // If we encounter an opening parentheses, we push its index onto the stack
            if (s.charAt(i) == '(') {
                stack.push(i);
            } else { // If we encounter a closing parentheses
                stack.pop(); // Pop the corresponding opening parentheses index
                
                // If the stack becomes empty, we push the current closing parentheses index onto the stack to mark the start of a new valid string
                if (stack.isEmpty()) {
                    stack.push(i);
                } else { // If the stack is not empty, we calculate the length of the current valid string
                    maxLength = Math.max(maxLength, i - stack.peek());
                }
            }
        }
        
        return maxLength;
    }
}

1. Detailed description of the algorithm for the "Longest Valid Parentheses" problem:

This problem can be solved efficiently using a stack. The idea is to push the index of every opening parentheses onto the stack. For every closing parentheses encountered, we need to check if it matches with the opening parentheses at the top of the stack.

To handle cases where the first character is a closing parentheses, we can start by pushing -1 onto the stack. This acts as a marker that signifies the start of a new valid string.

For each closing parentheses, we pop the corresponding opening parentheses from the stack. If the stack becomes empty, we push the current closing parentheses index onto the stack to mark the start of a new valid string.

To calculate the length of the current valid string, we take the difference between the current index and the top of the stack (which represents the index of the last unmatched opening parentheses). We update the maxLength if the current length is greater.

By following this approach, we can obtain the longest valid parentheses substring in linear time complexity, which is efficient for most cases.

15. Remove All Adjacent Duplicates In String (LeetCode #1047)

1. Problem Description: The problem is to remove all adjacent duplicates in a given string. We need to keep removing duplicates until no adjacent duplicates are present. The final string without any duplicates should be returned.

Problem Link: Remove All Adjacent Duplicates In String

1. Optimized Java Code with Detailed Comments:

class Solution {
    public String removeDuplicates(String S) {
        // Create a stack to store characters
        Stack<Character> stack = new Stack<>();
        
        // Iterate over each character in the string
        for (char c : S.toCharArray()) {
            // Check if the current character is equal to the top of the stack
            // If it is, pop the top of the stack to remove the adjacent duplicate
            // If it is not, push the current character onto the stack
            if (!stack.isEmpty() && stack.peek() == c) {
                stack.pop();
            } else {
                stack.push(c);
            }
        }
        
        // Convert the stack to a string
        StringBuilder sb = new StringBuilder();
        while (!stack.isEmpty()) {
            sb.append(stack.pop());
        }
        
        // Reverse the string and return the result
        return sb.reverse().toString();
    }
}

1. Description of the Algorithm and Interview Thought Process:

The problem can be approached using a stack data structure. The idea is to iterate over each character in the string and maintain a stack.

During the iteration, we compare the current character with the top of the stack. If they are equal, it means we have found adjacent duplicates. In this case, we remove the top of the stack to ensure that the duplicates are removed. If they are not equal, we simply push the current character onto the stack.

After iterating over the entire string, we convert the stack to a string by popping each element from the stack and appending it to a StringBuilder. Finally, we reverse the string and return the result.

During an interview, it is important to demonstrate a clear understanding of the problem requirements and constraints. Additionally, explaining the chosen algorithm and the reasoning behind it can showcase problem-solving skills and logical thinking.

Here are some key points to discuss during an interview:

• Identify that a stack data structure can be used to solve the problem efficiently.

• Understand the concept of adjacent duplicates and how to remove them using a stack.

• Explain the approach of iterating over the string and comparing characters with the top of the stack.

• Discuss the stack operations involved, such as pushing, popping, and checking for emptiness.

• Clarify how to convert the stack to a string and reverse it to obtain the final result.

• Address any time or space complexity considerations related to the chosen algorithm. In this case, the time complexity is O(n) and the space complexity is O(n) due to the stack and the StringBuilder.

Remember to communicate your thoughts clearly and provide concise and efficient code solutions.

16. Valid Number (LeetCode #65)

1. Problem Description: The problem "Valid Number" on LeetCode (problem #65) asks us to determine if a given string represents a valid number or not. The string may contain spaces at the beginning or end, an optional plus or minus sign, digits, a dot (.), and/or an exponent part (e) with an optional sign. A valid number must satisfy the following criteria:

• It can contain leading or trailing spaces, but not in between numbers or signs.

• It may contain an optional plus or minus sign before the number.

• It may contain digits (0-9), representing the integer or decimal part of the number.

• It may contain a dot (.) to represent the decimal part of the number.

• It may contain an exponent part (e or E) to represent the exponent part of the number.

• The integer part, if present, should not be empty.

• The decimal part, if present, should not be empty.

• The number should not contain any leading zeros except for the integer zero.

• The number should not contain any invalid characters except for spaces, plus or minus sign, digits (0-9), dot, and exponent part.

Link to the problem: Valid Number - LeetCode #65

1. Optimized Java Code with detailed comments:

class Solution {
    public boolean isNumber(String s) {
        // Variable to track if we have seen a digit
        boolean seenDigit = false;
        // Variable to track if we have seen a dot
        boolean seenDot = false;
        // Variable to track if we have seen an exponent
        boolean seenExponent = false;
        
        // Remove any leading or trailing spaces
        s = s.trim();
        
        // Iterate through each character in the string
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            
            // If current character is a digit
            if (Character.isDigit(c)) {
                seenDigit = true;
            }
            // If current character is a dot
            else if (c == '.') {
                // Dot should not have been seen before and there should not be an exponent already
                if (seenDot || seenExponent) {
                    return false;
                }
                seenDot = true;
            }
            // If current character is an exponent
            else if (c == 'e' || c == 'E') {
                // Exponent should not have been seen before or there should not be any digit before the exponent
                if (seenExponent || !seenDigit) {
                    return false;
                }
                seenExponent = true;
                // Reset seenDigit to false as we need to get digits after the exponent
                seenDigit = false;
            }
            // If current character is a sign (+ or -)
            else if (c == '+' || c == '-') {
                // Sign should not be in the middle of the number or there should not be a sign right before the exponent
                if (i != 0 && s.charAt(i-1) != 'e' && s.charAt(i-1) != 'E') {
                    return false;
                }
            }
            // If current character is not a valid character
            else {
                return false;
            }
        }
        
        // We should have at least seen a digit in the number
        return seenDigit;
    }
}

1. Algorithm and Thinking Process during Interviews: The problem asks us to determine if a given string represents a valid number or not. To solve this problem, we can use a simple iterative approach, where we check each character of the string and apply the necessary conditions to validate the number.

• We initialize three boolean variables: seenDigit, seenDot, and seenExponent, to keep track of whether we have seen a digit, a dot, and an exponent, respectively.

• We start by removing any leading or trailing spaces from the input string using the trim() method.

• Then, we iterate through each character of the string using a for loop.

• For each character, we check the following conditions:

  • If the character is a digit, we set seenDigit to true.

  • If the character is a dot, we check if we have already seen a dot or an exponent. If so, the number is not a valid number, and we return false.

  • If the character is an exponent, we check if we have already seen an exponent or if we have not seen any digits before the exponent. If so, the number is not a valid number, and we return false. Otherwise, we set seenExponent to true and reset seenDigit to false as we need to get digits after the exponent.

  • If the character is a sign (+ or -), we check if it appears in the middle of the number or right before the exponent. If so, the number is not a valid number, and we return false.

  • If the character is not a valid character (other than spaces, plus or minus sign, digits, dot, and exponent), we return false.

• After iterating through all characters, we check if we have seen at least one digit in the number. If not, the number is not a valid number, and we return false.

• If we reach the end without any invalid conditions, then the number is a valid number, and we return true.

During an interview, it's important to approach the problem methodically and explain each step as you write the code. You should also test your code with various test cases to ensure its correctness. Additionally, try to optimize your code and discuss the time and space complexity of your solution.