Category: Java Coding Test

Hello! Today, we will learn about an algorithm technique called Sliding Window. This technique is suitable for problems that involve finding partial sums or maximum/minimum values in continuous data. Additionally, this technique significantly reduces time complexity, making it a frequently tested topic in coding assessments.

What is Sliding Window?

Sliding Window is a technique that creates a window of a specific size in linear data structures like arrays or strings and manipulates the contents of that window. This technique is primarily useful in situations such as:

• When calculating the sum or average of consecutive elements
• When finding the maximum/minimum value that satisfies specific conditions
• When looking for values within defined intervals in various optimization problems

Problem Description

Problem: Maximum Length of Subarray

Given an integer array nums and an integer k, find the length of the longest subarray whose sum is less than or equal to k.

Input Example

nums = [1, 2, 3, 4, 5]
k = 5

Output Example

2

The maximum length of a subarray such as [2, 3] or [1, 2] with a sum less than or equal to 5 is 2.

Solution Process

1. Problem Analysis

This problem is about finding the longest subarray with a sum less than or equal to k from the given array. If solved using brute force, the time complexity would be O(n^2), so we can improve it using the sliding window technique.

2. Applying the Sliding Window Technique

The sliding window technique maintains the current sum of the window using two pointers pointing to the start and end of the array. While adjusting the size of the window, we need to find the maximum length. The basic approach is as follows:

Algorithm

1. Use two pointers start and end to point to the beginning and end of the window.
2. Move the end pointer to the end of the array while calculating the current window’s sum.
3. If the sum exceeds k, move the start pointer to the right to decrease the current window’s sum.
4. Calculate the current window’s length for each case and update the maximum length.

3. Java Code Implementation

Now, let’s write Java code based on the algorithm above:

public class MaxLengthSubarray {
    public static int maxLengthSubarray(int[] nums, int k) {
        int start = 0, end = 0;
        int maxLength = 0;
        int sum = 0;

        while (end < nums.length) {
            sum += nums[end];

            while (sum > k) {
                sum -= nums[start];
                start++;
            }

            maxLength = Math.max(maxLength, end - start + 1);
            end++;
        }

        return maxLength;
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 3, 4, 5};
        int k = 5;
        System.out.println("Maximum length of subarray: " + maxLengthSubarray(nums, k));
    }
}

4. Code Explanation

The code above performs the following tasks:

• The maxLengthSubarray function takes the input array nums and integer k as arguments.
• It initializes the pointers start and end, and uses the sum variable to maintain the current window’s sum.
• In the while loop, the end pointer is moved, adding nums[end] to the sum.
• If the current sum exceeds k, the start pointer is moved to the right to update the sum.
• The maximum length is updated at each step, ultimately returning the maximum length.

Conclusion

Sliding Window is one of the very useful techniques in coding assessments. Through this problem, we learned how to solve algorithm problems more efficiently. Utilizing this technique increases the likelihood of solving various problems more quickly.

If you found this blog post helpful, try out other problems as well. We hope you will learn more about various algorithms and problem-solving techniques!

Problem Description

Write a program that arranges the numbers from 1 to N in ascending order using a stack for the given integer N. You can push numbers onto the stack if needed before outputting the sequence, and you can pop numbers to output them. However, if you cannot output the numbers in the given order, you must print ‘NO’.

Input Format

• The first line contains the integer N. (1 ≤ N ≤ 100,000)
• The next N lines each contain a number that needs to be output. This number is an integer between 1 and N.

Output Format

• If it is possible to output the given numbers in ascending order, print each number in sequence.
• If it is not possible, print ‘NO’.

Example Input

    4
    4
    3
    2
    1
    

Example Output

    YES
    PUSH
    PUSH
    PUSH
    PUSH
    POP
    POP
    POP
    POP
    

Explanation

To solve the problem, you need to manipulate the numbers using a stack data structure appropriately. The basic idea is as follows:

1. Push the numbers from 1 to N onto the stack sequentially.
2. Pop each number from the sequence to output. At this time, the top value of the stack must match the desired output value.
3. If the top value of the stack does not match, it becomes impossible to output any number, and you must print ‘NO’.

Implementation Steps

Below is a Java code snippet to solve the problem:

    import java.util.*;

    public class StackSequence {
        public static void main(String[] args) {
            Scanner scanner = new Scanner(System.in);
            int N = scanner.nextInt();
            int[] sequence = new int[N];
            for (int i = 0; i < N; i++) {
                sequence[i] = scanner.nextInt();
            }

            Stack stack = new Stack<>();
            StringBuilder output = new StringBuilder();
            int current = 1;
            boolean isPossible = true;

            for (int i = 0; i < N; i++) {
                while (current <= sequence[i]) {
                    stack.push(current++);
                    output.append("PUSH\n");
                }
                if (stack.isEmpty() || stack.peek() != sequence[i]) {
                    isPossible = false;
                    break;
                }
                stack.pop();
                output.append("POP\n");
            }

            if (!isPossible) {
                System.out.println("NO");
            } else {
                System.out.println("YES");
                System.out.print(output);
            }

            scanner.close();
        }
    }
    

Code Explanation

The above Java code operates in the following way:

1. Input Reading: It reads N and stores the next N integers in an array.
2. Stack Initialization: It initializes a stack.
3. Pushing from 1 to N: It pushes each number onto the stack. During operations, it outputs the string PUSH.
4. Pop Operation: It checks if the current sequence number matches the top of the stack, and if not, it prints 'NO'.
5. Result Output: If all numbers are successfully popped, it prints 'YES' along with the PUSH and POP operations.

Time Complexity Analysis

In this problem, since each number is pushed and popped from the stack only once, the overall time complexity is O(N). This is the time required to process the input through the sequence.

Conclusion

In this lecture, we learned how to use the stack data structure to output the given numbers in ascending order. In coding tests, stacks can be a useful tool for solving various problems. They can particularly contribute to the design of efficient algorithms considering memory usage and time complexity.

Through this problem, I hope you understand the characteristics of stacks and how to utilize them in solving algorithmic problems. Furthermore, I encourage you to practice similar problems on your own to discover how to utilize stacks effectively.

Problem 1: Valid Parentheses

This problem is to determine if a given string is a valid parentheses string.
A valid parentheses string means that every open parenthesis is properly closed by a corresponding closing parenthesis and follows the correct order.
For example, “()”, “()[]{}”, “{[()]}” are valid, but “(]”, “([)]”, “{)” are not valid.

Problem Description

Implement a function to determine if the given input string s is a valid parentheses string.
A valid parentheses string must satisfy the following conditions:

• Every open parenthesis must have a corresponding closing parenthesis.
• Open parentheses must be properly ordered.

Input Examples

“()” -> true
“()[]{}” -> true
“{[()]}” -> true
“(]” -> false
“([)]” -> false
“{)” -> false

Solution Method

This problem can be solved using a stack.
The stack (FILO: First In Last Out) structure is very useful for operations involving parentheses. The process is as follows.

1. Iterate through the string from left to right.
2. When encountering an open parenthesis ((, {, [), push it onto the stack.
3. When encountering a closing parenthesis (), }, ]), pop from the stack and check if it forms a pair with the most recently opened parenthesis.
4. After processing the entire string, if the stack is empty, it is considered a valid parentheses string.

Java Code Implementation

            
            import java.util.Stack;

            public class ValidParentheses {
                public static boolean isValid(String s) {
                    Stack stack = new Stack<>();
                    for (char c : s.toCharArray()) {
                        if (c == '(' || c == '{' || c == '[') {
                            stack.push(c);
                        } else {
                            if (stack.isEmpty()) return false;
                            char top = stack.pop();
                            if ((c == ')' && top != '(') || 
                                (c == '}' && top != '{') || 
                                (c == ']' && top != '[')) {
                                return false;
                            }
                        }
                    }
                    return stack.isEmpty();
                }

                public static void main(String[] args) {
                    System.out.println(isValid("()")); // true
                    System.out.println(isValid("()[]{}")); // true
                    System.out.println(isValid("{[()]}")); // true
                    System.out.println(isValid("(]")); // false
                    System.out.println(isValid("([)]")); // false
                    System.out.println(isValid("{)")); // false
                }
            }
            
        

Time Complexity

The time complexity of this algorithm is O(n).
Here, n is the length of the string. Since we are using a stack, in the worst case, when there are n open parentheses, there can be n pushes and n pops.

Space Complexity

The space complexity is also O(n).
As the stack can hold a maximum of n open parentheses, the worst-case space complexity is O(n).

Summary of the Problem-Solving Process

In order to solve this problem, we used a stack to track open parentheses and checked each closing parenthesis by popping from the stack.
This method is very efficient for checking the validity of parentheses, and the same principle can be applied to check the validity of other types of parentheses.

Tips for Valid Coding Tests

Stacks and queues are frequently used data structures in coding tests.
It is important to become familiar with how to use stacks and queues through various problems.
Learn common patterns and tricks, and practice various variations to enhance your problem-solving skills.

Problem Description

Write a program that takes a given sequence of integers as input and calculates their sum.
The input is provided in the form of a string, separated by spaces or commas.
Examples of input data that the program should handle include:
"1, 2, 3, 4, 5", "10 20 30", which contain arbitrary numbers.

Input Format

The input is given in the form of a string, with each number separated by spaces or commas.

Example input: "5, 10, 15, 20"

Output Format

The sum of the numbers should be printed as an integer.

Example output: 50

Problem-Solving Process

Step 1: Reading Input Data

Read the string entered by the user and use a
String type variable to handle it.

Step 2: Parsing the String

Split the input string based on spaces or commas to extract each number.
To do this, you can use the String.split() method.

Step 3: Converting Strings to Integers

Convert the split string numbers to integers using the Integer.parseInt() method.

Step 4: Calculating the Sum

Use a loop to calculate the sum of the converted integer array.
Sum each number using a for loop.

Java Code Example

import java.util.Arrays;

public class SumOfNumbers {
    public static void main(String[] args) {
        String input = "5, 10, 15, 20";
        int sum = sumOfNumbers(input);
        System.out.println("Sum of numbers: " + sum);
    }

    public static int sumOfNumbers(String input) {
        // Splitting the string
        String[] numbers = input.split("[,\\s]+");
        // Sum variable
        int sum = 0;
        // Summing numbers
        for (String number : numbers) {
            sum += Integer.parseInt(number);
        }
        return sum;
    }
}

    

Conclusion

In this lesson, we wrote a program to calculate the simple sum of numbers using Java.
It is very important to learn string processing methods to handle various input formats
as part of your coding test preparation.
Based on this foundation, we encourage you to challenge more complex problems.

This article aims to address the commonly asked problem of “Finding the Order of a Permutation” in coding tests using Java. We will detail the necessary theories, approaches, and code examples required to solve this problem.

Problem Definition

This problem involves determining the position of a specific number’s permutation in a given number array. For example, when the array is [1, 2, 3], all permutations are listed as follows:

• 1, 2, 3
• 1, 3, 2
• 2, 1, 3
• 2, 3, 1
• 3, 1, 2
• 3, 2, 1

For example, for the number 2, when the permutation is 2, 1, 3, this permutation will be in the 3rd position. We will deduce from this information to find out the position of the given number’s permutation.

Problem Approach

To solve this problem, you can follow these steps:

1. Calculate the total number of permutations based on the length of the given array.
2. Calculate the position based on the target permutation using the necessary formulas at each step.
3. Write a recursive function to determine the position of the permutation at each step.

Calculating the Number of Permutations

When the length of the given number array is n, there are n! (n factorial) permutations. This is the product of all integers from 1 to n.

For example, if the length of the array is 3, the number of permutations is 3! = 6.

Negative Calculation and Recursion for Order Calculation

To calculate the order, we need to consider the cases after a specific number has been selected. For instance, if we assume we choose 1, we recursively call the remaining numbers’ permutations to compute the order.

This approach helps us find out how many permutations are possible for each number and perform the calculations.

Java Code Implementation

Below is the Java code to solve the “Finding the Order of a Permutation” problem:

public class PermutationOrder {
    
    public static void main(String[] args) {
        int[] numbers = {1, 2, 3}; // Given array
        int[] target = {2, 1, 3};   // Target permutation to calculate the order
        int rank = findRank(numbers, target);
        System.out.println("Order of the target permutation: " + rank);
    }
    
    public static int findRank(int[] numbers, int[] target) {
        int n = numbers.length;
        boolean[] used = new boolean[n];
        return getRank(numbers, target, used, 0);
    }
    
    private static int getRank(int[] numbers, int[] target, boolean[] used, int index) {
        int rank = 1; // Default order starts from 1
        int n = numbers.length;
        
        for (int i = 0; i < n; i++) {
            // If the number is smaller than the target number
            if (!used[i]) {
                for (int j = 0; j < target[index]; j++) {
                    if (j == numbers[i]) {
                        break;
                    }
                    rank += factorial(n - index - 1); // Add (n-1)! for each smaller number
                }
            }
            if (target[index] == numbers[i]) {
                used[i] = true; // Use that number
                break;
            }
        }
        
        if (index < n - 1) {
            rank += getRank(numbers, target, used, index + 1); // Recursive call for the next index
        }
        
        return rank;
    }

    private static int factorial(int n) {
        if (n == 0) return 1;
        int result = 1;
        for (int i = 1; i <= n; i++) {
            result *= i;
        }
        return result;
    }
}
            

This code uses a recursive approach to calculate the order based on the given number array and target permutation. The main function getRank incrementally checks each number in the target permutation and determines the order based on combinations with other numbers.

Example Test Cases

After writing the code, you can validate the results through several examples as follows:

• Input: [1, 2, 3], Target: [2, 1, 3] → Output: 3
• Input: [1, 2, 3], Target: [1, 3, 2] → Output: 2
• Input: [1, 2, 3], Target: [3, 2, 1] → Output: 6

You can iterate as much as needed using various arrays and target permutations with the above code to obtain the correct output. Since various test cases can be generated just from the combinations of permutations, it is essential to evaluate the accuracy and efficiency of the algorithm.

Conclusion

In this article, we addressed the problem of finding the order of permutations using Java. We emphasized how the combination of recursive approaches, the calculation of the number of permutations, and the algorithm for checking the exact order effectively solved the problem. This should enhance the understanding of coding tests and greatly aid in solving a wider range of problems.