Author: root

Hello, blog readers! Today, as part of the C# coding test course, we will address the line-up problem. This problem may seem simple, but without the right algorithm, it can become quite complex. Additionally, it is one of the types of problems frequently encountered in actual job coding tests.

Problem Description

This problem involves lining up students based on their height information. Students should be sorted in ascending order by height, and those with the same height must maintain their original order. In other words, stability must be maintained during sorting.

Problem Definition

    Given an array of students' heights, the task is to sort the array in ascending order.

    Input
    - Integer N: The number of students (1 ≤ N ≤ 100,000)
    - Integer array heights: Heights of the students (1 ≤ heights[i] ≤ 200 cm)

    Output
    - The sorted heights of the students should be returned.
    

Sample Input and Output

    Input: [160, 155, 170, 160, 175]
    Output: [155, 160, 160, 170, 175]
    

Problem-Solving Strategy

It’s important to devise a strategy for solving the problem. Since we need to sort the students’ heights, we should apply an efficient sorting algorithm. Among various sorting algorithms, we can use Merge Sort or Quick Sort, which both have a time complexity of O(N log N).

1. Merge Sort Explanation

Merge sort is a divide-and-conquer algorithm that splits an array into two sub-arrays, sorts each, and then merges the sorted sub-arrays to produce a final sorted array. This method is stable and useful when a sorted array is required.

2. Quick Sort Explanation

Quick sort is also a divide-and-conquer algorithm. It splits the array into two parts based on a pivot and sorts them recursively. The average time complexity is O(N log N), but in the worst case, it can be O(N^2). However, performance can be improved with an appropriate pivot selection.

C# Code Implementation

In this section, we will write code to solve the problem using the C# language. Below is the code that sorts students’ heights using merge sort.

using System;

class Sort
{
    public static void MergeSort(int[] heights, int left, int right)
    {
        if (left < right)
        {
            int mid = (left + right) / 2;
            MergeSort(heights, left, mid);
            MergeSort(heights, mid + 1, right);
            Merge(heights, left, mid, right);
        }
    }

    public static void Merge(int[] heights, int left, int mid, int right)
    {
        int n1 = mid - left + 1;
        int n2 = right - mid;

        int[] leftArray = new int[n1];
        int[] rightArray = new int[n2];

        for (int i = 0; i < n1; i++)
            leftArray[i] = heights[left + i];
        for (int j = 0; j < n2; j++)
            rightArray[j] = heights[mid + 1 + j];

        int k = left;
        int iIndex = 0, jIndex = 0;

        while (iIndex < n1 && jIndex < n2)
        {
            if (leftArray[iIndex] <= rightArray[jIndex])
            {
                heights[k] = leftArray[iIndex];
                iIndex++;
            }
            else
            {
                heights[k] = rightArray[jIndex];
                jIndex++;
            }
            k++;
        }

        while (iIndex < n1)
        {
            heights[k] = leftArray[iIndex];
            iIndex++;
            k++;
        }

        while (jIndex < n2)
        {
            heights[k] = rightArray[jIndex];
            jIndex++;
            k++;
        }
    }

    static void Main(string[] args)
    {
        int[] heights = { 160, 155, 170, 160, 175 };
        MergeSort(heights, 0, heights.Length - 1);

        Console.WriteLine("Sorted heights: " + string.Join(", ", heights));
    }
}
    

Code Explanation

The code above is an example of sorting a given array using the merge sort algorithm. The MergeSort method recursively divides the array, and the Merge method merges the two sub-arrays to achieve sorted order.

Code Flow

• MergeSort Method: Recursively divides the array into halves to sort it.
• Merge Method: Merges two sorted sub-arrays into one sorted array.
• Main Method: The starting point of the program, defining a sample array, sorting it, and outputting the result.

Performance Analysis

The time complexity of merge sort is O(N log N) and remains the same in the worst case. The space complexity is O(N). This is advantageous when processing large datasets. Because sorting is stable, for example, in cases where heights are the same, the original order can be guaranteed.

Applications of This Problem

The line-up problem can be applied in many fields. For example, it can be useful in queue management, student performance analysis, and various other scenarios. While solving this problem, it's good to consider the potential for expanding to other data sorting problems.

Conclusion

In this post, we explored how to solve the line-up problem using C#. Through defining the problem and implementing algorithms and code to solve it, we can develop algorithmic thinking. It is important to enhance algorithmic thinking through various problems.

Next time, we will address another type of algorithm problem. Please leave any questions or feedback in the comments. Thank you!

1. Problem Description

This problem is to find the largest single square in a given binary matrix and return its size.
A binary matrix consists of elements that are either 0 or 1.
We need to find the largest square composed of 1s, where the size is defined as the length of the square’s side.

Example

• Input: matrix = [[0,1,1,1,0],[0,1,1,1,1],[0,1,1,1,1],[0,0,0,0,0]]
• Output: 3 (length of the side of the largest square)

2. Approach to Solve the Problem

To solve this problem, we can use the Dynamic Programming technique.
The basic idea is to use the information from the left, above, and diagonally above positions (minimum size) to determine
the size of the square that can be formed at the current position.

2.1 State Definition

dp[i][j] is defined as the maximum size of the square that can be formed at row i and column j.
That is, the value of dp[i][j] is set to 1 plus the minimum of dp[i-1][j], dp[i][j-1], dp[i-1][j-1] when matrix[i][j] is 1.
If matrix[i][j] is 0, then dp[i][j] becomes 0.

2.2 Recurrence Relation

dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1 (if matrix[i][j] == 1)

2.3 Initial Conditions

For the first row and the first column, if matrix[i][j] is 1, then dp[i][j] is also set to 1,
and if it is 0, it is initialized to 0.

3. Algorithm Implementation

Below is the C# code based on the above approach.

C#
public class Solution {
    public int MaximalSquare(char[][] matrix) {
        if (matrix.Length == 0) return 0;
        
        int maxSide = 0;
        int rows = matrix.Length;
        int cols = matrix[0].Length;
        int[][] dp = new int[rows][];
        
        for (int i = 0; i < rows; i++) {
            dp[i] = new int[cols];
        }

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (matrix[i][j] == '1') {
                    if (i == 0 || j == 0) {
                        dp[i][j] = 1; // Set directly to 1 for edge cases
                    } else {
                        dp[i][j] = Math.Min(Math.Min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1;
                    }
                    maxSide = Math.Max(maxSide, dp[i][j]); // Update maximum size
                }
            }
        }
        
        return maxSide * maxSide; // Return area of the square
    }
}

4. Complexity Analysis

The time complexity of this algorithm is O(m * n), where m is the number of rows in the matrix and n is the number of columns.
This complexity arises because each element is checked only once.
The space complexity is also O(m * n) due to the use of a dynamic array.

5. Conclusion

This problem showcases a powerful application of dynamic programming.
Through this problem of finding squares in a binary matrix, we learned how to think in terms of dynamic programming and how to design algorithms effectively.
It is important to gain a thorough understanding of how to solve such problems in C#.

6. Additional Learning Resources

• LeetCode Problem Link
• GeeksforGeeks Resource
• Recommended books on C# Data Structures and Algorithms

Problem Description

A “good number” refers to the odd divisors among all the divisors of a given number N. Your goal is to find all odd divisors of N when a given number N is provided.
For example, if N is 12, the divisors of this number are 1, 2, 3, 4, 6, and 12,
and the odd divisors among them are 1 and 3. Write a program to find and output the odd divisors of the given number in ascending order.

Input Format

The input is an integer N (1 ≤ N ≤ 10^6).

Output Format

Print the odd divisors of N in ascending order on one line.
If there are no odd divisors, print “There are no odd divisors.”

Example Input and Output

    Input:
    12
    Output:
    1 3
    

    Input:
    7
    Output:
    1 7
    

Solution Approach

Step 1: Understand the Problem

To solve the problem, it is necessary to define the divisors of the given number N accurately.
A divisor means a number that divides another number with a remainder of 0.
For example, when N is 12, the divisors are all numbers that can divide 12.

Step 2: Finding Divisors

To find all divisors of N, you need to iterate from 1 to N, checking if N is divisible by the current number
and if the remainder is 0.

Method:
– Iterate through all integers from 1 to N.
– If N divided by the current number i has a remainder of 0, then i is a divisor of N.
– Additionally, check if i is odd, and only save it in the list if it is odd.

Step 3: Storing and Outputting Odd Divisors

Create a list to store the odd divisors, then sort this list in ascending order and output it.
If the list is empty, print a message saying “There are no odd divisors.”

Step 4: Implementing in C#

    
    using System;
    using System.Collections.Generic;

    class Program
    {
        static void Main()
        {
            int N = int.Parse(Console.ReadLine());
            List oddDivisors = new List();

            for (int i = 1; i <= N; i++)
            {
                if (N % i == 0 && i % 2 != 0)
                {
                    oddDivisors.Add(i);
                }
            }

            if (oddDivisors.Count == 0)
            {
                Console.WriteLine("There are no odd divisors.");
            }
            else
            {
                oddDivisors.Sort();
                Console.WriteLine(string.Join(" ", oddDivisors));
            }
        }
    }
    
    

Code Explanation

1. Getting Input

<code>int N = int.Parse(Console.ReadLine());</code>
part is where the program takes an integer N as input from the user. The Console.ReadLine() method reads
the user’s input as a string, and then it is converted to an integer using the int.Parse() method.

2. Finding Odd Divisors

<code>for (int i = 1; i <= N; i++)</code> starts the loop, executing it from 1 to N. Then the <code>if (N % i == 0 && i % 2 != 0)</code> statement checks if N is divisible by i and whether i is odd. If both conditions are satisfied, i is added to the odd divisor list.

3. Outputting Results

Finally, the length of the odd divisor list is checked. If there are no divisors, it prints “There are no odd divisors.”
If there are divisors, they are sorted in ascending order and printed as a space-separated string.
Here, <code>string.Join(” “, oddDivisors)</code> converts the list values to a string for output.

Performance Analysis

The time complexity of the above algorithm is O(N). Since N can be at most 10^6, this can feel slow.
However, considering the range of the input values, it is generally fast enough for calculations.
Additionally, to improve performance, the range for finding divisors can be reduced to the square root of N.
Doing so can decrease the time complexity to O(√N).

Going Further: Optimizing Odd Divisor Calculation of N

    
    using System;
    using System.Collections.Generic;

    class Program
    {
        static void Main()
        {
            int N = int.Parse(Console.ReadLine());
            List oddDivisors = new List();

            for (int i = 1; i * i <= N; i++)
            {
                if (N % i == 0)
                {
                    if (i % 2 != 0)
                    {
                        oddDivisors.Add(i);
                    }

                    int pairedDivisor = N / i;
                    if (pairedDivisor != i && pairedDivisor % 2 != 0)
                    {
                        oddDivisors.Add(pairedDivisor);
                    }
                }
            }

            if (oddDivisors.Count == 0)
            {
                Console.WriteLine("There are no odd divisors.");
            }
            else
            {
                oddDivisors.Sort();
                Console.WriteLine(string.Join(" ", oddDivisors));
            }
        }
    }
    
    

In this code, it checks both i and N/i pairs of divisors at the same time by looping from 1 to the square root of N.
This improves efficiency.

Conclusion

In this article, we solved an algorithm problem to find the odd divisors of a given integer N using C#.
The solution process was detailed in steps, and methods for code improvement through optimization were suggested.
This process will be beneficial for improving coding test or algorithm skills.