Category: C# Coding Test Tutorials

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.

This article explains in detail how to solve the card sorting algorithm problem using C#. We will start from the problem definition and proceed step by step through the approach, code implementation, and optimization methods. Through this, you can simultaneously improve your algorithmic thinking and C# programming skills needed for coding tests.

Problem Definition

The definition of the problem is as follows.

There are N cards given. Each card is represented by a unique integer from 1 to N. The task is to implement an algorithm that sorts these cards in ascending order. However, instead of a fixed sorting algorithm, we need to find the optimal method during the sorting process.

Problem Solving Process

Step 1: Problem Analysis

The first step in solving the problem is to analyze it closely. Given the number of cards N, we need to think about how to sort the data. If the number of cards is small, we can use simple direct methods (like bubble sort, selection sort, etc.), but when the numbers become larger, we should choose a more efficient algorithm.

Step 2: Algorithm Selection

Efficient sorting algorithms such as Quick Sort or Merge Sort can be considered. These sorting algorithms typically have an average time complexity of O(N log N), allowing them to handle large numbers of cards quickly.

Step 3: Implementation

Now, let’s implement a sorting algorithm in C#. Below is an example code implementing Merge Sort.

using System;
using System.Linq;

class Program
{
    static void Main(string[] args)
    {
        int[] cards = { 5, 3, 8, 1, 2, 7 };

        Console.WriteLine("Cards before sorting: " + string.Join(", ", cards));
        cards = MergeSort(cards);
        Console.WriteLine("Cards after sorting: " + string.Join(", ", cards));
    }

    static int[] MergeSort(int[] array)
    {
        if (array.Length <= 1)
            return array;

        int midPoint = array.Length / 2;
        int[] left = MergeSort(array.Take(midPoint).ToArray());
        int[] right = MergeSort(array.Skip(midPoint).ToArray());

        return Merge(left, right);
    }

    static int[] Merge(int[] left, int[] right)
    {
        int[] result = new int[left.Length + right.Length];
        int leftIndex = 0, rightIndex = 0, resultIndex = 0;

        while (leftIndex < left.Length && rightIndex < right.Length)
        {
            if (left[leftIndex] <= right[rightIndex])
            {
                result[resultIndex++] = left[leftIndex++];
            }
            else
            {
                result[resultIndex++] = right[rightIndex++];
            }
        }

        while (leftIndex < left.Length)
        {
            result[resultIndex++] = left[leftIndex++];
        }

        while (rightIndex < right.Length)
        {
            result[resultIndex++] = right[rightIndex++];
        }

        return result;
    }
}

The above code implements a function that sorts cards using Merge Sort. It is designed to display the state of the cards before and after sorting.

Step 4: Testing

After writing the code, we need to check the accuracy and performance of the algorithm through various test cases. For example, one can consider scenarios such as when there is only one card, when all cards have the same number, and when the cards are sorted in reverse order.

static void TestSort()
{
    int[] testCards1 = { 4, 3, 2, 1 };
    int[] sortedCards1 = MergeSort(testCards1);
    Console.WriteLine("Cards after sorting 1: " + string.Join(", ", sortedCards1));  // 1, 2, 3, 4

    int[] testCards2 = { 1, 1, 1, 1 };
    int[] sortedCards2 = MergeSort(testCards2);
    Console.WriteLine("Cards after sorting 2: " + string.Join(", ", sortedCards2));  // 1, 1, 1, 1

    int[] testCards3 = { 3, 5, 2, 8, 7 };
    int[] sortedCards3 = MergeSort(testCards3);
    Console.WriteLine("Cards after sorting 3: " + string.Join(", ", sortedCards3));  // 2, 3, 5, 7, 8
}

Through this process, we can confirm that the given cards are sorted correctly.

Step 5: Optimization and Improvement

Once all implementations are completed, it’s necessary to optimize the performance of the code. In C#, using LINQ can make the code more concise, but it’s often implemented manually to consider performance issues. Additionally, utilizing multidimensional arrays or other data structures can reduce unnecessary resource consumption.