Check if a string can be made palindromic by swapping pairs of characters from indices having unequal characters in a Binary String

using System;
using System.Collections.Generic;

class MainClass {
  // Utility function to check if string
  // S can be made palindromic or not
  public static bool CanBecomePalindromeUtil(string S,
                                             string B)
  {
    // Stores the number of distinct
    // characters present in string S
    HashSet<char> set = new HashSet<char>();

    // Traverse the characters of string S
    for (int i = 0; i < S.Length; i++) {
      // Insert current character in S
      set.Add(S[i]);
    }

    // Count frequency of each
    // character of string S
    Dictionary<char, int> map
      = new Dictionary<char, int>();

    // Traverse the characters of string S
    for (int i = 0; i < S.Length; i++) {
      int k = 0;
      if (map.ContainsKey(S[i])) {
        k = map[S[i]];
      }
      map[S[i]] = k + 1;
    }

    bool flag = false;

    // Check for the odd length string
    if (S.Length % 2 == 1) {
      // Stores the count of
      // even and odd frequency
      // characters in the string S
      int count1 = 0, count2 = 0;

      foreach(var e in map)
      {
        if (e.Value % 2 == 1) {
          // Update the count of
          // odd frequent characters
          count2++;
        }
        else {
          // Update the count of
          // even frequent characters
          count1++;
        }
      }

      // If the conditions satisfies
      if (count1 == set.Count - 1 && count2 == 1) {
        flag = true;
      }
    }

    // Check for even length string
    else {
      // Stores the frequency of
      // even and odd characters
      // in the string S
      int count1 = 0, count2 = 0;

      foreach(var e in map)
      {
        if (e.Value % 2 == 1) {
          // Update the count of
          // odd frequent characters
          count2++;
        }
        else {
          // Update the count of
          // even frequent characters
          count1++;
        }
      }

      // If the condition satisfies
      if (count1 == set.Count && count2 == 0) {
        flag = true;
      }
    }

    // If a palindromic string
    // cannot be formed
    if (!flag) {
      return false;
    }

    else {
      // Check if there is
      // atleast one '1' and '0'
      int count1 = 0, count0 = 0;
      for (int i = 0; i < B.Length; i++) {
        // If current character is '1'
        if (B[i] == '1') {
          count1++;
        }
        else {
          count0++;
        }
      }

      // If atleast one '1' and '0' is present
      if (count1 >= 1 && count0 >= 1) {
        return true;
      }

      else {
        return false;
      }
    }
  }

  // Function to determine whether
  // string S can be converted to
  // a palindromic string or not
  static void CanBecomePalindrome(string S, string B)
  {
    if (CanBecomePalindromeUtil(S, B))
      Console.WriteLine("Yes");
    else
      Console.WriteLine("Np");
  }

  // Driver code
  public static void Main(string[] args)
  {
    string S = "ACABB";
    string B = "00010";
    CanBecomePalindrome(S, B);
  }
}

// This code is contributed by phasing17.

using System;

class Program
{
    // Function to check if it's possible to form a palindrome of the given string 'S' using binary string 'B'
    static bool CanFormPalindrome(string S, string B)
    {
        int[] freq = new int[26];  // Create an array to store character frequency for 'S'
        int count1 = 0, count0 = 0, countOdd = 0;  // Initialize counters for '1's, '0's, and odd character frequencies

        // Convert 'S' to lowercase to make the comparison case-insensitive
        S = S.ToLower();

        // Calculate character frequencies for 'S'
        foreach (char c in S)
        {
            freq[c - 'a']++;  // Increment the frequency count for each character
        }

        // Count the number of '1's and '0's in the binary string 'B'
        foreach (char c in B)
        {
            if (c == '1') count1++;  // Count '1's
            if (c == '0') count0++;  // Count '0's
        }

        // Count the number of characters in 'S' with odd frequencies
        foreach (int f in freq)
        {
            if (f % 2 == 1) countOdd++;  // Increment 'countOdd' if a character has an odd frequency
        }

        // Check the conditions for forming a palindrome
        if (S.Length % 2 == 0 && countOdd > 0) return false;  // If 'S' has an even length and there's an odd-frequency character, it's not possible to form a palindrome
        if (S.Length % 2 == 1 && countOdd > 1) return false;  // If 'S' has an odd length and there are more than one odd-frequency characters, it's not possible to form a palindrome
        if (count1 > 0 && count0 > 0) return true;  // If both '1's and '0's are present in 'B', it's possible to form a palindrome

        return false;  // If none of the above conditions are met, it's not possible to form a palindrome
    }

    static void Main()
    {
        string S = "ACABB";
        string B = "00010";

        // Check if it's possible to form a palindrome and print the result
        if (CanFormPalindrome(S, B))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}