Find Nth even length palindromic number formed using digits X and Y

Given an integer N, the task is to find the N^th even palindromic number of even length and only comprising of the digits X and Y where X, Y > 0.
Examples: 
 

Input: N = 9, X = 4, Y = 5 
Output: 454454 
Explanation: 
Even length palindromic numbers using 4 & 5 are 
44, 55, 4444, 4554, 5445, 5555, 444444, 445544, 454454, … 
9th term of the above series = 454454
Input: N = 6, X = 1, Y = 2 
Output: 2222 
Explanation: 
Even length palindromic numbers using 1 & 2 are 
11, 22, 1111, 1221, 2112, 2222, 111111, 112211, 121121, … 
6th term of the above series = 2222 
 

Approach: 
 

• Even length palindromic numbers using X & Y are 
   

XX, YY, XXXX, XYYX, YXXY, YYYY, XXXXXX, XXYYXX, ...

• The above sequence can be observed as:
   

XX,       -> Length (L) = 2
YY,       -> Length (L) = 2

XXXX,     -> Length (L) = 4
XYYX,     -> Length (L) = 4
YXXY,     -> Length (L) = 4
YYYY,     -> Length (L) = 4

XXXXXX,   -> Length (L) = 6
XXYYXX,   -> Length (L) = 6
XYXXYX,   -> Length (L) = 6
XYYYYX,   -> Length (L) = 6
YXXXXY,   -> Length (L) = 6
YXYYXY,   -> Length (L) = 6
YYXXYY,   -> Length (L) = 6
YYYYYY,   -> Length (L) = 6

XXXXXXXX, -> Length (L) = 8
...

• If we divide any term into 2 halves, the second half is just the reverse of the first half 
  Example: 
   

Taking the term XXYYXX

Dividing this into 2 halves
XXYYXX = XXY | YXX

So YXX is just the reverse of XXY

• Taking the left half only of the terms and putting X = 0 and Y = 1 to get the Binary String, the numbers of length L can be seen forming a integer sequence from 0 to (2^L/2 – 1), taken as Rank (R). Therefore 0 ≤ R ≤ 2^L/2 – 1 
  Therefore the sequence can be observed as follows:
   

L -> Left Half -> Binary String -> Rank (in Decimal) 

2 -> X    -> 0             -> 0
2 -> Y    -> 1             -> 1

4 -> XX   -> 00            -> 0
4 -> XY   -> 01            -> 1
4 -> YX   -> 10            -> 2
4 -> YY   -> 11            -> 3

6 -> XXX  -> 000           -> 0
6 -> XXY  -> 001           -> 1
6 -> XYX  -> 010           -> 2
6 -> XYY  -> 011           -> 3
6 -> YXX  -> 100           -> 4
6 -> YXY  -> 101           -> 5
6 -> YYX  -> 110           -> 6
6 -> YYY  -> 111           -> 7

8 -> XXXX -> 0000          -> 0
...

• Therefore, For the required term N: 
  • The length (L) of the required Nth term: 
     
    
  • Rank (R) of the required Nth term: 
     
    
  • First Half of the required Nth term = Binary representation of R in L/2 bits by replacing 0 as X and 1 as Y
  • Second Half of the required Nth term = Reverse of the First Half

  Below is the implementation of the above approach: 
   

  C++

  
  

  
  
  

  // C++ program to find nth even // palindromic number of only even // length composing of 4's and 5's.    #include <bits/stdc++.h> using namespace std;    // Utility function to compute // n'th palindrome number string solve(int n, char x, char y) {     // Calculate the length from above     // formula as discussed above     int length = ceil(log2(n + 2)) - 1;        // Calculate rank for length L     int rank = n - (1 << length) + 1;        string left = "", right = "";        for (int i = length - 1; i >= 0; i--) {            // Mask to check if i't bit         // is set or not         int mask = 1 << i;            // If bit is set append '5' else append '4'         bool bit = mask & rank;            if (bit) {             left += y;             right += y;         }         else {             left += x;             right += x;         }     }        reverse(right.begin(), right.end());        return left + right; }    // Driver Code int main() {     int n = 23;     char x = '4', y = '5';     string ans = solve(n, x, y);     cout << ans << '\n';        return 0; }

  Java

  
  

  
  
  

  // Java program to find nth even  // palindromic number of only even  // length composing of 4's and 5's.  import java.util.*;    class GFG {            // Utility function to compute      // n'th palindrome number      static String solve(int n, char x, char y)      {          // Calculate the length from above          // formula as discussed above          int length = (int)Math.ceil(Math.log(n + 2) /                                      Math.log(2)) - 1;                 // Calculate rank for length L          int rank = n - (1 << length) + 1;                 String left = "", right = "";                 for (int i = length -1 ; i >= 0; i--)         {                     // Mask to check if i't bit              // is set or not              int mask = (1 << i);                     // If bit is set append '5' else append '4'              int bit = mask & rank;                             if (bit > 0)             {                  left += y;                  right += y;              }              else              {                  left += x;                  right += x;              }          }                     StringBuilder sb = new StringBuilder(right);          sb.reverse();                     right = sb.toString();                     String res = left + right;         return res;      }             // Driver Code      public static void main (String[] args)     {          int n = 23;          char x = '4', y = '5';          String ans = solve(n, x, y);          System.out.println(ans);      }  }    // This code is contributed by AnkitRai01

  Python3

  
  

  
  
  

  # Python3 program to find nth even  # palindromic number of only even  # length composing of 4's and 5's.  from math import ceil, log2    # Utility function to compute  # n'th palindrome number  def solve(n, x, y) :         # Calculate the length from above      # formula as discussed above      length = ceil(log2(n + 2)) - 1;         # Calculate rank for length L      rank = n - (1 << length) + 1;         left = ""; right = "";         for i in range(length - 1 , -1, -1):            # Mask to check if i't bit          # is set or not          mask = (1 << i);             # If bit is set append '5'          # else append '4'          bit = (mask & rank);             if (bit) :             left += y;              right += y;                         else :             left += x;              right += x;         right = right[::-1];            res = left + right;     return res;    # Driver Code  if __name__ == "__main__" :         n = 23;      x = '4';     y = '5';      ans = solve(n, x, y);      print(ans);         # This code is contributed by kanugargng

  C#

  
  

  
  
  

  // C# program to find nth even  // palindromic number of only even  // length composing of 4's and 5's.  using System;    class GFG {            // Utility function to compute      // n'th palindrome number      static String solve(int n, char x, char y)      {          // Calculate the length from above          // formula as discussed above          int length = (int)Math.Ceiling(Math.Log(n + 2) /                                         Math.Log(2)) - 1;                 // Calculate rank for length L          int rank = n - (1 << length) + 1;                 String left = "", right = "";                 for (int i = length -1; i >= 0; i--)         {                     // Mask to check if i't bit              // is set or not              int mask = (1 << i);                     // If bit is set append '5'             // else append '4'              int bit = mask & rank;                             if (bit > 0)             {                  left += y;                  right += y;              }              else             {                  left += x;                  right += x;              }          }                     right = reverse(right);         String res = left + right;         return res;      }             static String reverse(String input)      {         char[] a = input.ToCharArray();         int l, r = 0;         r = a.Length - 1;            for (l = 0; l < r; l++, r--)          {             // Swap values of l and r              char temp = a[l];             a[l] = a[r];             a[r] = temp;         }         return String.Join("", a);     }             // Driver Code      public static void Main (String[] args)     {          int n = 23;          char x = '4', y = '5';          String ans = solve(n, x, y);          Console.WriteLine(ans);      }  }    // This code is contributed by Rajput-Ji

  Javascript

  
  

  
  
  

  <script>    // Javascript program to find nth even // palindromic number of only even // length composing of 4's and 5's.    // Utility function to compute // n'th palindrome number function solve(n, x, y) {     // Calculate the length from above     // formula as discussed above     var length = Math.ceil(Math.log2(n + 2)) - 1;        // Calculate rank for length L     var rank = n - (1 << length) + 1;        var left = "", right = "";        for (var i = length - 1; i >= 0; i--) {            // Mask to check if i't bit         // is set or not         var mask = 1 << i;            // If bit is set append '5' else append '4'         var bit = mask & rank;            if (bit) {             left += y;             right += y;         }         else {             left += x;             right += x;         }     }        right = right.split('').reverse().join('');        return left + right; }    // Driver Code var n = 23; var x = '4', y = '5'; var ans = solve(n, x, y); document.write( ans + "<br>");    </script>

  Output

  54444445
  

  Time Complexity:where n is the length of string
   

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