Category: Java Coding Test

It is very important to solve various algorithm problems in preparing for coding tests. In this course, we will develop basic problem-solving skills through the problem of ‘Making an Integer Equal to 1’ and look at the process of designing efficient algorithms. This problem asks for the minimum number of operations needed to turn a given integer into 1.

Problem Definition

Given an integer N, you can transform the number by choosing one of the following three operations:

• Decrement: N - 1
• Divide by 2: N / 2 (only possible if N is even)
• Divide by 3: N / 3 (only possible if N is divisible by 3)

The goal is to find the minimum number of operations to make N equal to 1.

Example Input and Output

• Input: N = 10
• Output: 3 (10 → 9 → 3 → 1)

Approach to Solve the Problem

There are several approaches to solving this problem. Here we will introduce two methods: DFS (Depth-First Search) and DP (Dynamic Programming).

1. DFS (Depth-First Search)

First, we can consider using DFS to explore all possible paths. However, this approach can lead to a very high time complexity. For example, the number of possible paths can be quite large for N=10. Nevertheless, let’s approach it with DFS.

DFS Implementation Code

import java.util.HashMap;

public class Main {
    static int minSteps = Integer.MAX_VALUE;
    
    public static void main(String[] args) {
        int N = 10;
        findSteps(N, 0);
        System.out.println("Minimum steps to reach 1: " + minSteps);
    }
    
    private static void findSteps(int N, int steps) {
        if (N == 1) {
            minSteps = Math.min(minSteps, steps);
            return;
        }
        
        findSteps(N - 1, steps + 1); // Decrement
        if (N % 2 == 0) {
            findSteps(N / 2, steps + 1); // Divide by 2
        }
        if (N % 3 == 0) {
            findSteps(N / 3, steps + 1); // Divide by 3
        }
    }
}

The above code explores all possible paths from the given N. However, this method has a time complexity of O(3^N) due to duplicate calls and inefficient path exploration.

2. DP (Dynamic Programming)

Thus, a more efficient method is to use DP. By using DP, we can store previously computed results and reuse them when needed, reducing unnecessary calculations.

DP Implementation Code

public class Main {
    public static void main(String[] args) {
        int N = 10;
        System.out.println("Minimum steps to reach 1: " + minStepsDP(N));
    }

    private static int minStepsDP(int N) {
        int[] dp = new int[N + 1];
        dp[1] = 0; // Minimum operations to reach 1 is 0
        
        for (int i = 2; i <= N; i++) {
            dp[i] = dp[i - 1] + 1; // Decrement

            if (i % 2 == 0) {
                dp[i] = Math.min(dp[i], dp[i / 2] + 1); // Divide by 2
            }
            if (i % 3 == 0) {
                dp[i] = Math.min(dp[i], dp[i / 3] + 1); // Divide by 3
            }
        }
        
        return dp[N];
    }
}

The above DP implementation code uses an array dp to store the minimum steps needed to reach each number. The algorithm has a time complexity of O(N). It calculates the minimum steps by referencing the values of the previous numbers for each number.

Time Complexity Analysis

The DFS approach has a time complexity of O(3^N), making it very inefficient. In contrast, the DP approach has a time complexity of O(N), which allows for deriving the minimum steps for all numbers with at most one calculation.

Conclusion

In this course, we examined various approaches to making an integer equal to 1 and learned efficient problem-solving methods through DFS and DP. Understanding actual algorithms and developing the ability to solve problems based on them are important in preparing for coding tests. Practice learning various approaches to complex problems like this and choose optimized methods.

Practice Problems

Try to solve the additional problems below for practice:

• For integer N = 15, find the minimum number of operations to make it 1.
• For integer N = 25, find the minimum number of operations to make it 1.

As you solve the problems, try to output results for various input values. In coding tests, it is essential not just to solve the problem but to find the optimal solution.

One of the common problems encountered in coding tests is the implementation of data structures that efficiently store and manage data. In this article, we will explain a data structure called absolute value heap (absmin heap) and explore how to implement it using Java. The absolute value heap prioritizes based on the absolute value of the input data.

Problem Description

This problem involves implementing an absolute value heap for a given list of integers. The absolute value heap should support the following functionalities:

• Remove and return the smallest absolute value.
• Add the given integer to the absolute value heap.

The functioning of the implemented absolute value heap is as follows:

• Elements are sorted in ascending order of their absolute values.
• If absolute values are the same, the original smaller value is prioritized.

Input/Output Format

The input is provided in the following format:

    N
    operation[0]
    operation[1]
    ...
    operation[N-1]
    

Here, N is the number of operations, and each operation[i] is as follows:

• 0: Remove and print the smallest value from the absolute value heap.
• Other integer x: Add x to the absolute value heap.

Example

Input

    7
    5
    3
    6
    0
    -2
    4
    0
    

Output

    -2
    3
    

In the above example, the following process occurs:

• Adding 5, 3, and 6 stores [5, 3, 6] in the heap.
• When 0 is input, the smallest absolute value, -2, is returned.
• Additionally, the smallest absolute value of 3 is printed.

Implementation of Absolute Value Heap

Now, let’s implement the absolute value heap in Java. Essentially, Java allows us to implement heaps using PriorityQueue. However, in this case, we need to set the priority based on absolute values, so we will create a Comparator as follows.

Step 1: Define Heap Node

We create a data structure to store in the heap, where we will save each number’s original value and absolute value.

    class AbsHeapNode {
        int value;

        public AbsHeapNode(int value) {
            this.value = value;
        }

        public int getAbs() {
            return Math.abs(value);
        }
    }
    

Step 2: Define Comparator

We define a Comparator that can compare based on absolute values.

    class AbsMinComparator implements Comparator<AbsHeapNode> {
        @Override
        public int compare(AbsHeapNode a, AbsHeapNode b) {
            if (a.getAbs() != b.getAbs()) {
                return Integer.compare(a.getAbs(), b.getAbs());
            }
            return Integer.compare(a.value, b.value);
        }
    }
    

Step 3: Implement Absolute Value Heap

Now, we create a class that implements the actual absolute value heap.

    import java.util.PriorityQueue;

    public class AbsHeap {
        private PriorityQueue<AbsHeapNode> heap;

        public AbsHeap() {
            heap = new PriorityQueue<>(new AbsMinComparator());
        }

        public void insert(int value) {
            heap.offer(new AbsHeapNode(value));
        }

        public Integer removeMin() {
            AbsHeapNode node = heap.poll();
            return (node != null) ? node.value : null;
        }
    }
    

Step 4: Implement Main Method

Using the previously implemented classes, we write the main method.

    import java.util.Scanner;

    public class Main {
        public static void main(String[] args) {
            Scanner scanner = new Scanner(System.in);
            AbsHeap absHeap = new AbsHeap();
            int N = scanner.nextInt();

            for (int i = 0; i < N; i++) {
                int operation = scanner.nextInt();
                if (operation == 0) {
                    Integer minValue = absHeap.removeMin();
                    System.out.println(minValue != null ? minValue : 0);
                } else {
                    absHeap.insert(operation);
                }
            }

            scanner.close();
        }
    }
    

Conclusion

In the process of implementing the absolute value heap, it is crucial to define a suitable Comparator. This allows us to build an efficient data structure capable of solving the given problem. Problems like this often appear in coding tests as similar algorithm questions, so it is important to practice sufficiently to master them. Keep practicing various algorithms and data structures to become a better programmer. Thank you!

In this lecture, we will take a closer look at the Binomial Coefficient and write efficient code to calculate it in Java. The binomial coefficient is used to count the number of combinations and is denoted as “nCr”. Through this problem, you will understand the importance of algorithms and have the opportunity to enhance the skills needed for coding tests.

What is a Binomial Coefficient?

The binomial coefficient refers to the number of ways to choose r elements from n elements given n and r. Mathematically, it is expressed as follows:

C(n, r) = n! / (r! * (n - r)!)

Here, n! denotes the factorial of n, which is n! = n × (n-1) × … × 1. Although the formula for calculating the binomial coefficient may seem simple, caution is needed for large numbers since the factorial value increases rapidly as n gets larger.

Problem Definition

Let’s solve the following problem:

Problem: Given integers in the range of 0 ≤ n ≤ 30 and 0 ≤ r ≤ n, write a Java program to efficiently calculate the binomial coefficient C(n, r).

Approach to the Problem

To solve this problem, we can start with a basic method. However, since the maximum value of n is 30, using simple factorial calculations may be inefficient. Therefore, we will utilize the Dynamic Programming technique to solve this problem.

Calculating Binomial Coefficient using Dynamic Programming

By using dynamic programming, we can avoid redundant calculations. We utilize the following property to calculate the binomial coefficient:

C(n, r) = C(n-1, r-1) + C(n-1, r)

Using this property, we can initialize the base cases such as C(n, 0) = C(n, n) = 1 and fill in the remaining values to calculate the binomial coefficient.

Java Code Implementation

Below is the Java code to calculate the binomial coefficient using dynamic programming:

public class BinomialCoefficient {
    public static int binomialCoefficient(int n, int r) {
        int[][] dp = new int[n + 1][r + 1];

        // C(n, 0) = 1
        for (int i = 0; i <= n; i++) {
            dp[i][0] = 1;
        }

        // C(n, n) = 1
        for (int i = 0; i <= n; i++) {
            dp[i][i] = 1;
        }

        // Fill the dp table
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= Math.min(i, r); j++) {
                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
            }
        }

        return dp[n][r];
    }

    public static void main(String[] args) {
        int n = 5; // Sample value
        int r = 2; // Sample value
        System.out.println("C(" + n + ", " + r + ") = " + binomialCoefficient(n, r));
    }
}

Code Explanation

The above code is a dynamic programming approach to calculate the binomial coefficient. The explanation for each step is as follows:

• Creating a 2D array: A 2D array of size (n+1) x (r+1) is created to store the binomial coefficients.
• Initializing base cases: C(n, 0) and C(n, n) are initialized to 1.
• Filling the DP table: Each binomial coefficient is calculated through a nested loop. Values are filled according to the formula C(n, r) = C(n-1, r-1) + C(n-1, r).
• Returning the result: The final result is stored in dp[n][r].

Testing and Results

Let's run the code to check the binomial coefficients for various n and r values. For instance, let's see what happens when n=5 and r=2.

C(5, 2) = 10

This means that there are 10 ways to choose 2 elements from a set of 5 elements. By running the code for various cases, we can verify the number of combinations.

Complexity Analysis

The time complexity of this code is O(n * r). The space complexity is also O(n * r), making it sufficiently efficient when considering the space needed to store the DP table. Since n can go up to 30, it will work effectively even for larger problems.

Conclusion and Additional Learning Material

In this lecture, we explored how to calculate the binomial coefficient and the process of writing Java code for it. The algorithm presented here is a useful technique for coding tests. To gain a deeper understanding of the binomial coefficients, it is also beneficial to study additional materials on combinatorial theory and probability.

I encourage you to solve various algorithm problems to prepare for coding tests and build your skills.

Problem Description

The binomial coefficient is a concept used in combinatorics when selecting two items, represented in the form of
C(n, k). Here,
C(n, k) refers to the number of ways to choose k items from n items.
In this problem, you are required to write a program to calculate the binomial coefficient.

Problem

Given two integers n and k, write a program to output the number of ways to choose k items from n items.

Input

• On the first line, two integers n (0 ≤ n ≤ 30) and k (0 ≤ k ≤ n) are provided.

Output

• Output the binomial coefficient for the given n and k.

Definition of Binomial Coefficient

The binomial coefficient is defined by the following formula:

    C(n, k) = n! / (k! * (n - k)!)
    

Here, n! (n factorial) is the product of all integers from 1 to n.
For example,
5! = 5 × 4 × 3 × 2 × 1 = 120.

Problem Solving Process

Step 1: Input and Output Handling

Process the user input data, which is the starting point of the problem.
Since both n and k are integers,
you can use the Scanner class to read the input.

    import java.util.Scanner;

    public class BinomialCoefficient {
        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            int n = sc.nextInt();
            int k = sc.nextInt();
            // Use n and k to calculate the binomial coefficient in the following code
        }
    }
    

Step 2: Implementing Factorial Calculation Function

Implement a function to calculate the factorial.
Here, we will compute n factorial using a recursive function.

    public static int factorial(int num) {
        if (num == 0) return 1; // 0! = 1
        return num * factorial(num - 1);
    }
    

Step 3: Implementing Binomial Coefficient Calculation Function

Write a dedicated function to calculate the binomial coefficient.
Utilizing the factorial function implemented in the previous step.

    public static int binomialCoefficient(int n, int k) {
        return factorial(n) / (factorial(k) * factorial(n - k));
    }
    

Step 4: Writing the Final Code

Now, we combine all functionalities to complete the final code.

    import java.util.Scanner;

    public class BinomialCoefficient {
        public static void main(String[] args) {
            Scanner sc = new Scanner(System.in);
            int n = sc.nextInt();
            int k = sc.nextInt();
            System.out.println(binomialCoefficient(n, k));
        }

        public static int factorial(int num) {
            if (num == 0) return 1;
            return num * factorial(num - 1);
        }

        public static int binomialCoefficient(int n, int k) {
            return factorial(n) / (factorial(k) * factorial(n - k));
        }
    }
    

Step 5: Time Complexity Analysis

The above algorithm uses recursion to calculate the factorial.
The time complexity of the factorial function is O(n).
Therefore, the overall time complexity of the algorithm can be analyzed as O(n) * 3 (since factorial is called three times), which is O(n).

Conclusion

We have examined the method of calculating the binomial coefficient.
This problem helps in understanding the fundamental concepts of combinatorics and is a type of problem often encountered in coding tests.
Additionally, it can enhance understanding of Java programming through the use of recursion and factorial concepts.

Next, we will learn about dynamic programming for binomial coefficients and discuss ways to improve performance.
We will also take time to explore various applications of the binomial coefficient.

Additional References

• Wikipedia: Probability by Position
• CMU 112: Factorials