Others

21. Valid Parentheses

1. Problem Description: You are given a string containing only parentheses '(' and ')'. Determine if the parentheses are 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.

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

1. Optimized Java Code with Comments:

import java.util.Stack;

class Solution {
    public boolean isValid(String s) {
        // Create a stack to store the opening brackets
        Stack<Character> stack = new Stack<>();

        // Traverse through each character of 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);
            } else { // If the current character is a closing bracket
                // Check if the stack is empty (no opening brackets left)
                // or if the current closing bracket does not match the top of the stack
                if (stack.isEmpty() || !isMatchingBracket(stack.peek(), c)) {
                    return false;
                }
                // If the current closing bracket matches the top of the stack, pop it
                stack.pop();
            }
        }
        
        // After traversing the entire string
        // The stack should be empty for the string to be valid
        return stack.isEmpty();
    }
    
    // Function to check if two brackets are matching
    private boolean isMatchingBracket(char open, char close) {
        return (open == '(' && close == ')') ||
               (open == '{' && close == '}') ||
               (open == '[' && close == ']');
    }
}

1. Algorithm and Thinking Process:

• The problem can be solved using a stack.

• We traverse through each character of the string.

• If the current character is an opening bracket, we push it to the stack.

• If the current character is a closing bracket, we check if it matches the top of the stack (the corresponding opening bracket).

• If it does not match or if the stack is empty (no opening brackets left), the string is not valid and we return false.

• If it matches, we pop the top of the stack.

• After traversing the entire string, the stack should be empty for the string to be valid. If the stack is not empty, we return false.

• In the end, we return true if the stack is empty, indicating a valid string.

• The time complexity of this solution is O(n), where n is the length of the string.

During the interview process, it is important to explain the approach in a clear and concise manner. It is also helpful to discuss the time complexity and potential edge cases. Additionally, you can improve the explanation by providing examples and going through the code step by step.

22. Move Zeroes

1. Problem Description: The problem "Move Zeroes" on LeetCode asks us to move all the zeroes in an input array to the end while maintaining the relative order of the non-zero elements.

LeetCode Link: https://leetcode.com/problems/move-zeroes/

1. Optimized Java Code with Comments:

class Solution {
    public void moveZeroes(int[] nums) {
        // Initialize a pointer to keep track of the position to place the next non-zero element
        int insertPos = 0;
        
        // Traverse the array
        for (int num : nums) {
            // If the current element is non-zero, we move it to its appropriate position
            if (num != 0) {
                nums[insertPos] = num; // Place the non-zero element at the next available position
                insertPos++; // Increment the insert position
            }
        }
        
        // Fill the remaining positions with zeroes
        while (insertPos < nums.length) {
            nums[insertPos] = 0;
            insertPos++;
        }
    }
}

1. Algorithm and Approach: To solve this problem, we can use the two-pointer approach. One pointer (insertPos) keeps track of the position to place the next non-zero element, while the other pointer traverses the input array.

Initially, we set insertPos to 0. Then, for each element in the array:

• If the element is non-zero, we move it to nums[insertPos] (i.e., placing the non-zero element at the next available position) and increment insertPos by 1.

• If the element is zero, we do nothing and move on to the next element.

After traversing the entire array, we know that all non-zero elements have been placed at their respective positions up to insertPos. To complete the task, we fill the remaining positions in nums with zeroes, starting from insertPos until the end of the array.

This approach has a linear runtime complexity of O(n), where n is the length of the input array, and does not require creating a new array or using extra space. It also maintains the relative order of the non-zero elements, as required by the problem statement.

During an interview, it is crucial to explain the algorithmic approach clearly and discuss the trade-offs. Additionally, mentioning the time and space complexity helps demonstrate your understanding of the problem and your problem-solving skills.

23. Longest Mountain in Array

1. Problem Description: The problem "Longest Mountain in Array" asks us to find the length of the longest subarray that forms a mountain. In this problem, a mountain is defined as an array with three sections: an increasing section, a peak, and a decreasing section. The increasing section comes before the peak, and the decreasing section comes after the peak. The peak cannot be at either end of the array.

LeetCode Problem Link: https://leetcode.com/problems/longest-mountain-in-array/

1. Java Code:

class Solution {
    public int longestMountain(int[] A) {
        int maxLength = 0;
        int i = 1;
        
        while (i < A.length) {
            // Skip non-increasing elements
            while (i < A.length && A[i] <= A[i - 1]) {
                i++;
            }
            
            // Start of a potential mountain
            int start = i - 1;
            
            // Find the peak
            while (i < A.length && A[i] > A[i - 1]) {
                i++;
            }
            
            // Check if there was a peak
            if (start != i - 1) {
                // Find the end of the mountain
                while (i < A.length && A[i] < A[i - 1]) {
                    i++;
                }
                
                // Calculate the length of the mountain
                int end = i - 1;
                maxLength = Math.max(maxLength, end - start + 1);
            }
        }
        
        return maxLength;
    }
}

1. Algorithm Explanation:

• We start with a maxLength variable initialized to 0.

• We iterate through the array using an index, starting from 1.

• In each iteration, we skip any non-increasing elements to find the start of a potential mountain.

• After finding the start, we continue iterating to find the peak of the mountain.

• If we find a peak, we then continue iterating to find the end of the mountain (the start of the decreasing section).

• If there was a peak, we calculate the length of the current mountain (end - start + 1) and update maxLength if necessary.

• We repeat this process until we've iterated through the entire array.

• Finally, we return the maxLength.

During an interview, the following thought process can help in solving this problem:

• Understand the problem statement and ask clarifying questions if needed.

• Analyze the problem and come up with a clear plan of action.

• Break down the problem into smaller tasks, if applicable.

• Implement the solution step by step, testing your code at each iteration to catch any mistakes early on.

• Optimize the solution by identifying any unnecessary steps or redundant operations.

• Ensure your code is clean, readable, and follows best practices.

• Test your code on different test cases, including edge cases, to verify its correctness.

• Discuss the time and space complexity of your solution and potential trade-offs.

• Communicate your thoughts and approach clearly to the interviewer.

• Be prepared to discuss alternative solutions or optimizations if asked by the interviewer.

24. Subarray Product Less Than K

1. Problem Description: Subarray Product Less Than K - Given an array of positive integers nums and an integer k, return the number of contiguous subarrays where the product of all the elements in the subarray is less than k. Link: https://leetcode.com/problems/subarray-product-less-than-k/

2. Optimized Java Code:

public class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        if (k <= 1) return 0;
        
        int count = 0;
        int left = 0;
        int product = 1;

        for (int right = 0; right < nums.length; right++) {
            product *= nums[right];
            
            while (product >= k) {
                product /= nums[left];
                left++;
            }
            
            count += right - left + 1;
        }
        
        return count;
    }
}

1. Algorithm Description: To solve this problem, we can use a sliding window approach. We will maintain a window using two pointers - left and right. The product of all elements within this window should be less than k. We start expanding the window by moving the right pointer to the right.

As we move the right pointer, we update the product by multiplying the new element. If the product becomes greater than or equal to k, we shrink the window by moving the left pointer and dividing the product by the element at that index. This shrinking step ensures that all subarrays with smaller product are counted.

At each step, we count the number of subarrays ending at the current index (right pointer). The number of subarrays ending at any index is equal to right - left + 1, since we have a contiguous range from the left pointer to the current index.

Finally, we return the total count of subarrays.

During the interview process, it is important to explain the thought process and explain the algorithm step by step. It is also helpful to run through some example test cases to validate the approach. Additionally, highlighting the time and space complexity of the solution can be beneficial.

25. Max Consecutive Ones III

1. Problem Description: Given an array consisting of only 0s and 1s, you have a certain number of flips that can be performed. By flipping a 0 to 1, you can extend the length of the consecutive subarray of 1s. Return the maximum length of a consecutive subarray of 1s after performing at most k flips.

Problem Link: https://leetcode.com/problems/max-consecutive-ones-iii/

1. Optimized Java Code with Comments:

class Solution {
    public int longestOnes(int[] nums, int k) {
        int left = 0; // left pointer to track the start of the subarray
        int right = 0; // right pointer to track the end of the subarray
        int maxLen = 0; // variable to store the maximum length of consecutive ones
        int zeroCount = 0; // variable to track the count of zeros in the current subarray
        
        while (right < nums.length) {
            if (nums[right] == 0) {
                zeroCount++; // increment zero count when encountering a 0
                
                // Check if the number of flips have exceeded k
                // If yes, move the left pointer to the right until we have enough flips
                while (zeroCount > k) {
                    if (nums[left] == 0) {
                        zeroCount--; // decrement zero count when moving left pointer and encountering a 0
                    }
                    left++; // move left pointer to the right
                }
            }
            
            int currLen = right - left + 1; // current length of the subarray
            maxLen = Math.max(maxLen, currLen); // update maxLen if necessary
            right++; // move right pointer to the right
        }
        
        return maxLen;
    }
}

1. Algorithm Explanation and Interview Approach:

The problem is asking us to find the maximum length of a consecutive subarray of 1s after performing at most k flips. One flip means we can change a 0 to a 1.

To solve this problem, we can use a sliding window approach. We maintain two pointers, left and right, to represent the start and end of the current subarray. We also keep track of the number of zeros in the current subarray using the zeroCount variable.

The idea is to move the right pointer to the right until the number of zeros in the subarray is greater than k. When we encounter a zero, we increment the zeroCount. If the zeroCount exceeds k, we move the left pointer to the right until we have enough flips (zeros) to make the subarray valid again.

By maintaining the maxLen variable, we can update it whenever we find a longer valid subarray. Finally, we return the maxLen as the result.

During an interview, you can approach this problem by explaining the idea of a sliding window and how we can use it to solve this problem efficiently. Start by identifying the variables needed, such as the left and right pointers, maxLen, and zeroCount. Walk through the code and explain how each variable is used and why the algorithm works. Finally, analyze the time and space complexity of the solution.