How to resolve the algorithm Find the missing permutation step by step in the C# programming language

Published on 12 May 2024 09:40 PM
#C_Sharp

How to resolve the algorithm Find the missing permutation step by step in the C# programming language

Table of Contents

Problem Statement

Listed above are   all-but-one   of the permutations of the symbols   A,   B,   C,   and   D,   except   for one permutation that's   not   listed.

Find that missing permutation.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Find the missing permutation step by step in the C# programming language

First code:

The first code finds the missing permutation of a given set of strings. It uses a recursive function to generate all possible permutations of a given string, and then checks if the permutation is present in the given set of strings. If the permutation is not present, it is added to a list of missing permutations. The function permuteString takes three parameters:

  • result: a reference to a list of strings, where the missing permutations will be stored
  • beginningString: the beginning of the permutation
  • endingString: the remaining string to be permuted

The function works by recursively calling itself for each possible character in the endingString. For each character, the function creates a new string by removing the character from the endingString, and then calls itself again with the new string and the beginningString plus the removed character. The function stops when the endingString is empty, and adds the beginningString to the result list.

Second code:

The second code is a more efficient way to find the missing permutation of a given set of strings. It uses bitwise XOR to find the missing character. The function works by iterating over the given strings and XORing each character with the corresponding character in the missing permutation. The result of the XOR operation is stored in an array of integers. After iterating over all the strings, the missing permutation is found by XORing the values in the array. The function then converts the integer values to characters and prints the missing permutation.

Source code in the csharp programming language

using System;
using System.Collections.Generic;

namespace MissingPermutation
{
    class Program
    {
        static void Main()
        {
            string[] given = new string[] { "ABCD", "CABD", "ACDB", "DACB", 
                                            "BCDA", "ACBD", "ADCB", "CDAB", 
                                            "DABC", "BCAD", "CADB", "CDBA", 
                                            "CBAD", "ABDC", "ADBC", "BDCA", 
                                            "DCBA", "BACD", "BADC", "BDAC", 
                                            "CBDA", "DBCA", "DCAB" };
            
            List<string> result = new List<string>();
            permuteString(ref result, "", "ABCD");
            
            foreach (string a in result)            
                if (Array.IndexOf(given, a) == -1)
                    Console.WriteLine(a + " is a missing Permutation");
        }

        public static void permuteString(ref List<string> result, string beginningString, string endingString)
        {
            if (endingString.Length <= 1)
            {                 
                result.Add(beginningString + endingString);
            }
            else
            {
                for (int i = 0; i < endingString.Length; i++)
                {                     
                    string newString = endingString.Substring(0, i) + endingString.Substring(i + 1);
                    permuteString(ref result, beginningString + (endingString.ToCharArray())[i], newString);                    
                }
            }
        }
    }
}


using System;
using System.Linq;

public class Test
{
    public static void Main()
    {
        var input = new [] {"ABCD","CABD","ACDB","DACB","BCDA",
            "ACBD","ADCB","CDAB","DABC","BCAD","CADB",
            "CDBA","CBAD","ABDC","ADBC","BDCA","DCBA",
            "BACD","BADC","BDAC","CBDA","DBCA","DCAB"};
        
        int[] values = {0,0,0,0};
        foreach (string s in input)
            for (int i = 0; i < 4; i++)
                values[i] ^= s[i];
        Console.WriteLine(string.Join("", values.Select(i => (char)i)));
    }
}

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