Java Coding Test Course, 'Finding Good Numbers'

Problem Description

Given a number N, write a computer program that can transform N such that the sum of its digits is less than N.
Here, a number where the sum of its digits is less than N is called a “good number”.
In other words, this is an algorithm problem to find the largest number among all numbers less than N, where the sum of its digits is less than N.

Example Problem

For example, if N is 23:

22 (2 + 2 = 4) → ‘good number’
21 (2 + 1 = 3) → ‘good number’
20 (2 + 0 = 2) → ‘good number’
19 (1 + 9 = 10) → ‘good number’
18 (1 + 8 = 9) → ‘good number’
17 (1 + 7 = 8) → ‘good number’
16 (1 + 6 = 7) → ‘good number’
15 (1 + 5 = 6) → ‘good number’
14 (1 + 4 = 5) → ‘good number’
13 (1 + 3 = 4) → ‘good number’
12 (1 + 2 = 3) → ‘good number’
11 (1 + 1 = 2) → ‘good number’
10 (1 + 0 = 1) → ‘good number’
9 (9 = 9) → ‘good number’

The largest ‘good number’ is 22.

Approach to Solve the Problem

To solve this problem, we first calculate the sum of the digits for each number starting from N-1 going downwards.
If the sum of the digits is less than N, we return that number as the result and terminate the algorithm.

Algorithm Implementation

        
        public class GoodNumber {
            public static void main(String[] args) {
                int N = 23; // Set the value of N to test here.
                int result = findGoodNumber(N);
                System.out.println("The 'good number' is: " + result);
            }

            public static int findGoodNumber(int N) {
                for (int i = N - 1; i > 0; i--) {
                    if (digitSum(i) < N) {
                        return i; // Return the largest 'good number'
                    }
                }
                return -1; // Return -1 if not found
            }

            public static int digitSum(int num) {
                int sum = 0;
                while (num > 0) {
                    sum += num % 10; // Add each digit
                    num /= 10; // Divide by 10 to reduce the number of digits
                }
                return sum;
            }
        }

Code Explanation

In the Java code above, the main method sets the value of N, then calls the findGoodNumber method to find the ‘good number’.
The findGoodNumber method starts from N-1 and calculates the sum of the digits, returning the number if that value is less than N.

digitSum method contains the logic to calculate the sum of each digit of the given number.
This method uses a loop to separate the digits one by one and returns the summed result.

Time Complexity Analysis

The time complexity of this algorithm is O(N * D), where N is the input value and D is the number of digits in N.
It requires O(D) operations to sum the digits for each number, resulting in a worst-case time complexity of O(N * D).

Space Complexity

The space complexity is O(1). It does not use additional space, and only the input/output variables and constant space within methods are used,
thus minimizing the memory used.

Conclusion and Tips

The problem of finding a ‘good number’ can be easily solved through a simple loop and digit sum process.
However, if N is very large, performance degradation may be a concern, so algorithm optimization may be necessary.
Testing this problem with various input values and attempting to solve it with different methods or structures can be a good learning experience.

Finally, it is important to practice these types of problems thoroughly while preparing for coding tests.
Facing new problems every time and cultivating diverse thinking skills will greatly assist in job preparation.

Java Coding Test Course, ‘Finding Good Numbers’