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Find the n-th number whose binary representation is a palindrome
  • Difficulty Level : Expert
  • Last Updated : 14 Oct, 2020

Find the nth number whose binary representation is a palindrome. Do not consider the leading zeros, while considering the binary representation. Consider the 1st number whose binary representation is a palindrome as 1, instead of 0 

Examples: 

Input : 1
Output : 1
1st Number whose binary representation 
is palindrome is 1 (1)

Input : 9
Output : 27
9th Number whose binary representation
is palindrome is 27 (11011)

Method 1: Naive
Naive approach would be, traverse through all the integers from 1 to 2^31 – 1 and increment palindrome count, if the number is palindrome. When the palindrome count reaches the required n, break the loop and return the current integer. 
 

C++

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// C++ program to find n-th number whose binary
// representation is palindrome.
#include <bits/stdc++.h>
using namespace std;
 
// Finds if the kth bit is set in the binary
// representation
int isKthBitSet(int x, int k)
{
    return (x & (1 << (k - 1))) ? 1 : 0;
}
 
// Returns the position of leftmost set bit
// in the binary representation
int leftmostSetBit(int x)
{
    int count = 0;
    while (x) {
        count++;
        x = x >> 1;
    }
    return count;
}
 
// Finds whether the integer in binary
// representation is palindrome or not
int isBinPalindrome(int x)
{
    int l = leftmostSetBit(x);
    int r = 1;
 
    // One by one compare bits
    while (l > r) {
 
        // Compare left and right bits and converge
        if (isKthBitSet(x, l) != isKthBitSet(x, r))
            return 0;
        l--;
        r++;
    }
    return 1;
}
 
int findNthPalindrome(int n)
{
    int pal_count = 0;
 
    // Start from 1, traverse through
    // all the integers
    int i = 0;
    for (i = 1; i <= INT_MAX; i++) {
        if (isBinPalindrome(i)) {
            pal_count++;
        }
        // If we reach n, break the loop
        if (pal_count == n)
            break;
    }
 
    return i;
}
 
// Driver code
int main()
{
    int n = 9;
   
    // Function Call
    cout << findNthPalindrome(n);
}
 
// This code is contributed
// by Akanksha Rai

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C

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// C program to find n-th number whose binary
// representation is palindrome.
#include <stdio.h>
#define INT_MAX 2147483647
 
// Finds if the kth bit is set in the binary
// representation
int isKthBitSet(int x, int k)
{
    return (x & (1 << (k - 1))) ? 1 : 0;
}
 
// Returns the position of leftmost set bit
// in the binary representation
int leftmostSetBit(int x)
{
    int count = 0;
    while (x) {
        count++;
        x = x >> 1;
    }
    return count;
}
 
// Finds whether the integer in binary
// representation is palindrome or not
int isBinPalindrome(int x)
{
    int l = leftmostSetBit(x);
    int r = 1;
 
    // One by one compare bits
    while (l > r) {
 
        // Compare left and right bits and converge
        if (isKthBitSet(x, l) != isKthBitSet(x, r))
            return 0;
        l--;
        r++;
    }
    return 1;
}
 
int findNthPalindrome(int n)
{
    int pal_count = 0;
 
    // Start from 1, traverse through
    // all the integers
    int i = 0;
    for (i = 1; i <= INT_MAX; i++) {
        if (isBinPalindrome(i)) {
            pal_count++;
        }
        // If we reach n, break the loop
        if (pal_count == n)
            break;
    }
 
    return i;
}
 
// Driver code
int main()
{
    int n = 9;
   
    // Function Call
    printf("%d", findNthPalindrome(n));
}

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Java

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// Java program to find n-th
// number whose binary
// representation is palindrome.
import java.io.*;
 
class GFG {
    static int INT_MAX = 2147483647;
 
    // Finds if the kth bit
    // is set in the binary
    // representation
    static int isKthBitSet(int x, int k)
    {
        return ((x & (1 << (k - 1))) > 0) ? 1 : 0;
    }
 
    // Returns the position of
    // leftmost set bit in the
    // binary representation
    static int leftmostSetBit(int x)
    {
        int count = 0;
        while (x > 0) {
            count++;
            x = x >> 1;
        }
        return count;
    }
 
    // Finds whether the integer
    // in binary representation is
    // palindrome or not
    static int isBinPalindrome(int x)
    {
        int l = leftmostSetBit(x);
        int r = 1;
 
        // One by one compare bits
        while (l > r) {
 
            // Compare left and right
            // bits and converge
            if (isKthBitSet(x, l) != isKthBitSet(x, r))
                return 0;
            l--;
            r++;
        }
        return 1;
    }
 
    static int findNthPalindrome(int n)
    {
        int pal_count = 0;
 
        // Start from 1, traverse
        // through all the integers
        int i = 0;
        for (i = 1; i <= INT_MAX; i++) {
            if (isBinPalindrome(i) > 0) {
                pal_count++;
            }
 
            // If we reach n,
            // break the loop
            if (pal_count == n)
                break;
        }
 
        return i;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 9;
       
        // Function Call
        System.out.println(findNthPalindrome(n));
    }
}
 
// This code is contributed
// by anuj_67.

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Python 3

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# Python 3 program to find n-th number
# whose binary representation is palindrome.
INT_MAX = 2147483647
 
# Finds if the kth bit is set in
# the binary representation
 
 
def isKthBitSet(x, k):
 
    return 1 if (x & (1 << (k - 1))) else 0
 
# Returns the position of leftmost
# set bit in the binary representation
 
 
def leftmostSetBit(x):
 
    count = 0
    while (x):
        count += 1
        x = x >> 1
 
    return count
 
# Finds whether the integer in binary
# representation is palindrome or not
 
 
def isBinPalindrome(x):
 
    l = leftmostSetBit(x)
    r = 1
 
    # One by one compare bits
    while (l > r):
 
        # Compare left and right bits
        # and converge
        if (isKthBitSet(x, l) != isKthBitSet(x, r)):
            return 0
        l -= 1
        r += 1
    return 1
 
 
def findNthPalindrome(n):
    pal_count = 0
 
    # Start from 1, traverse
    # through all the integers
    i = 0
    for i in range(1, INT_MAX + 1):
        if (isBinPalindrome(i)):
            pal_count += 1
 
        # If we reach n, break the loop
        if (pal_count == n):
            break
 
    return i
 
 
# Driver code
if __name__ == "__main__":
    n = 9
     
    # Function Call
    print(findNthPalindrome(n))
 
# This code is contributed
# by ChitraNayal

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C#

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// C# program to find n-th
// number whose binary
// representation is palindrome.
using System;
 
class GFG {
 
    static int INT_MAX = 2147483647;
 
    // Finds if the kth bit
    //  is set in the binary
    // representation
    static int isKthBitSet(int x, int k)
    {
        return ((x & (1 << (k - 1))) > 0) ? 1 : 0;
    }
 
    // Returns the position of
    // leftmost set bit in the
    // binary representation
    static int leftmostSetBit(int x)
    {
        int count = 0;
        while (x > 0) {
            count++;
            x = x >> 1;
        }
        return count;
    }
 
    // Finds whether the integer
    // in binary representation is
    // palindrome or not
    static int isBinPalindrome(int x)
    {
        int l = leftmostSetBit(x);
        int r = 1;
 
        // One by one compare bits
        while (l > r) {
 
            // Compare left and right
            // bits and converge
            if (isKthBitSet(x, l) != isKthBitSet(x, r))
                return 0;
            l--;
            r++;
        }
        return 1;
    }
 
    static int findNthPalindrome(int n)
    {
        int pal_count = 0;
 
        // Start from 1, traverse
        // through all the integers
        int i = 0;
        for (i = 1; i <= INT_MAX; i++) {
            if (isBinPalindrome(i) > 0) {
                pal_count++;
            }
 
            // If we reach n,
            // break the loop
            if (pal_count == n)
                break;
        }
 
        return i;
    }
 
    // Driver code
    static public void Main()
    {
        int n = 9;
       
        // Function Call
        Console.WriteLine(findNthPalindrome(n));
    }
}
 
// This code is contributed ajit

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PHP

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<?php
// PHP program to find n-th number whose
// binary representation is palindrome.
 
// Finds if the kth bit is set in
// the binary representation
function isKthBitSet($x, $k)
{
    return ($x & (1 << ($k - 1))) ? 1 : 0;
}
 
// Returns the position of leftmost set
// bit in the binary representation
function leftmostSetBit($x)
{
    $count = 0;
    while ($x)
    {
        $count++;
        $x = $x >> 1;
    }
    return $count;
}
 
// Finds whether the integer in binary
// representation is palindrome or not
function isBinPalindrome($x)
{
    $l = leftmostSetBit($x);
    $r = 1;
 
    // One by one compare bits
    while ($l > $r)
    {
 
        // Compare left and right bits
        // and converge
        if (isKthBitSet($x, $l) !=
            isKthBitSet($x, $r))
            return 0;
        $l--;
        $r++;
    }
    return 1;
}
 
function findNthPalindrome($n)
{
    $pal_count = 0;
 
    // Start from 1, traverse through
    // all the integers
    $i = 0;
    for ($i = 1; $i <= PHP_INT_MAX; $i++)
    {
        if (isBinPalindrome($i))
        {
            $pal_count++;
        }
         
        // If we reach n, break the loop
        if ($pal_count == $n)
            break;
    }
    return $i;
}
 
// Driver code
$n = 9;
 
// Function Call
echo (findNthPalindrome($n));
 
// This code is contributed by jit_t
?>

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Output

27

The time complexity of this solution is O(x) where x is a resultant number. 
Note that the value of x is generally much larger than n.



Method 2: Using BFS

In this approach first, we simply add   “11” this string into the queue. And then for every string, we have two cases. i.e 

  1. if the curr string of even length then add “0” and “1” at the mid of curr string and add it into the queue.
  2. if the curr string is of odd length then add mid char of the curr string into the resultant string and then add it into the queue.

if curr string is of even length

if curr string is of odd length

Below is the implementation of the above approach:

Java

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// Java program to find n-th palindrome
import java.io.*;
import java.util.*;
class GFG {
 
    // utility function which is used to
    // convert binary string into integer
    public static int convertStringToInt(String s)
    {
        int ans = 0;
 
        // convert binary string into integer
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '1')
                ans += 1 << i;
        }
        return ans;
    }
 
    // function to find nth binary palindrome number
    public static int getNthNumber(int n)
    {
        // stores the binary palindrome string
        Queue<String> q = new LinkedList<>();
 
        // base case
        if (n == 1)
            return 1;
        n = n - 1;
 
        // add 2nd binary palindrome string
        q.add("11");
 
        // runs till the nth binary palindrome number
        while (n-- > 0) {
 
            // remove curr binary palindrome string from
            // queue
            String curr = q.remove();
 
            // if n==0 then we find the n'th binary
            // palindrome so we return our answer
            if (n == 0)
                return convertStringToInt(curr);
 
            // calculate length of curr binary palindrome
            // string
            int len = curr.length();
 
            // if length is even .so it is our first case
            // we have two choices
            if (len % 2 == 0) {
                q.add(curr.substring(0, len / 2) + "0"
                      + curr.substring(len / 2));
                q.add(curr.substring(0, len / 2) + "1"
                      + curr.substring(len / 2));
            }
 
            // if length is odd .so it is our second case
            // we have only one choice
            else {
                char midChar = curr.charAt(len / 2);
                q.add(curr.substring(0, len / 2) + midChar
                      + curr.substring(len / 2));
            }
        }
        return -1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 9;
       
        // Function Call
        System.out.println(getNthNumber(n));
    }
}
// This code is contributed by Naresh Saharan and Sagar
// Jangra

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C++

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// C++ program to find n-th palindrome
#include <bits/stdc++.h>
using namespace std;
 
// utility function which is used to
// convert binary string into integer
int convertStringToInt(string s)
{
    int num = 0;
 
    // convert binary string into integer
    for (int i = 0; i < s.size(); i++) {
        num = num * 2 + (s[i] - '0');
    }
    return num;
}
 
// function to find nth binary palindrome number
int getNthNumber(int n)
{
 
    // base case
    if (n == 1)
        return 1;
    n--;
 
    // stores the binary palindrome string
    queue<string> q;
 
    // add 2nd binary palindrome string
    q.push("11");
 
    // runs till the nth binary palindrome number
    while (!q.empty()) {
        // remove curr binary palindrome string from queue
        string curr = q.front();
        q.pop();
        n--;
 
        // if n==0 then we find the n'th binary palindrome
        // so we return our answer
        if (!n) {
            return convertStringToInt(curr);
        }
 
        int mid = curr.size() / 2;
 
        // if length is even .so it is our first case
        // we have two choices
        if (curr.size() % 2 == 0) {
            string s0 = curr, s1 = curr;
            s0.insert(mid, "0");
            s1.insert(mid, "1");
            q.push(s0);
            q.push(s1);
        }
         
        // if length is odd .so it is our second case
        // we have only one choice
        else {
            string ch(1, curr[mid]);
            string temp = curr;
            temp.insert(mid, ch);
            q.push(temp);
        }
    }
 
    return 0;
}
 
// Driver Code
int main()
{
    int n = 9;
     
    // Function Call
    cout << getNthNumber(n);
}
 
// This code is contributed by Sagar Jangra and Naresh
// Saharan

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Output

27

Time complexity: O(N)
Space complexity:O(N)

Method 3: Constructing the nth palindrome

We can construct the nth binary palindrome in its binary representation directly using the below approach. 
If we observe first few binary palindromes 



 *         | nth Binary  |
       n   | Palindrome  |     Group
           |             |
--------------------------- Group 0
    1 --->  1 (1)
    

Group 1 (Will have binary representation of length 2*(1)
and 2*(1) + 1)

    Fix the first and last bit as 1 and insert nothing
    (|) in between. Length is 2*(1)
    2 --->  1|1 (3)     

    Fix the first and last bit as 1 and insert bit 0
    in between. Length is 2*(1) + 1
    3 --->  101 (5)    

    Fix the first and last bit as 1 and insert bit 1 
    in between. Length is 2*(1) + 1 
    4 --->  111 (7)    
    F

Group 2 (Will have binary representation of length 
    2*(2) and 2*(2) + 1).  Fix the first and last 
    bit as 1 and insert nothing (|) at middle. 
    And put 0 in binary format in both directions 
    from middle. Length is 2*(2)
    5 --->  10|01       
    Fix the first and last bit as 1 and insert 
    nothing (|) at middle. And put 1 in binary 
    format in both directions from middle. 
    Length is 2*(2)
    6 --->  11|11      
    
    7 --->  10001      
    8 --->  10101     
    9 --->  11011     
    10 --->  11111      
    
Group 3 (Will have binary representation of 
        length 2*(3) and 2*(3) + 1)
    11 ---> 100|001    
    12 ---> 101|101    
    13 ---> 110|011    
    14 ---> 111|111  
    
    15 ---> 1000001  
    16 ---> 1001001  
    17 ---> 1010101  
    18 ---> 1011101  
    19 ---> 1100011  
    20 ---> 1101011  
    21 ---> 1110111    
    22 ---> 1111111 
-------------------- 

Algorithm: 
1) We can divide the set of palindrome numbers into some groups. 
2) n-th group will have (2^(n-1) + 2^n = 3 * 2 ^(n-1) ) number of binary palindromes 
3) With the given number, we can find the group to which it belongs to and the offset in that group. 
4) As the leading zeros are not to be considered, we should use bit 1 as the starting bit and ending bit of the number in binary representation 
5) And we will fill other bits based on the groupno and groupoffset 
6) Based on the offset, we can find which bit should be inserted at the middle (|(nothing) or 0 or 1) and 
which number(in binary form) (1 or 2 or 3 or 4 or ..) should be placed in both directions from middle

Consider Below Example 

Let us construct the 8th binary palindrome number
The group number will be 2, and no.of elements 
before that group are 1 + 3 * 2^1 which is 4

So the offset for the 8th element will be 8 - 4
 - 1 = 3

And first 2^(groupno - 1) = 2^1, elements will 
have even length(in binary representation) of
2*groupno, next 2^groupno elements will have odd 
length(in binary representation) of 2*groupno + 1

Place bit 1 as the first bit and as the last bit 
(firstbit: 0, last bit: 2*groupno or 2*groupno - 1)

As the offset is 3, 4th(3 + 1) element in the
group, will have odd length and have 1 in the 
middle

Below is the table of middle bit to be used for 
the given offset for the group 2
offset    middle bit
  0            |
  1            |
  2            0
  3            1
  4            0
  5            1
And we should be filling the binary representation 
of number 0(((groupoffset) - 2^(groupno-1)) /2)
from middle n both directions
1 0 1 0 1
FirstElement Number MiddleElement Number LastElement
    1           0         1         0         1

The 8th number will be 21

Below is the implementation of the above idea : 

C

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// Efficient C program to find n-th palindrome
#include <stdio.h>
#define INT_SIZE 32
 
// Construct the nth binary palindrome with the
// given group number, aux_number and operation
// type
int constructNthNumber(int group_no, int aux_num, int op)
{
    int a[INT_SIZE] = { 0 };
 
    int num = 0, len_f;
    int i = 0;
 
    // No need to insert any bit in the middle
    if (op == 2) {
         
        // Length of the final binary representation
        len_f = 2 * group_no;
 
        // Fill first and last bit as 1
        a[len_f - 1] = a[0] = 1;
 
        // Start filling the a[] from middle,
        // with the aux_num binary representation
        while (aux_num) {
             
            // Get the auxiliary number's ith bit and
            // fill around middle
            a[group_no + i]
                = a[group_no - 1 - i]
                = aux_num & 1;
            aux_num = aux_num >> 1;
            i++;
        }
    }
 
    // Insert bit 0 in the middle
    else if (op == 0) {
         
        // Length of the final binary representation
        len_f = 2 * group_no + 1;
 
        // Fill first and last bit as 1
        a[len_f - 1] = a[0] = 1;
        a[group_no] = 0;
 
        // Start filling the a[] from middle, with
        // the aux_num binary representation
        while (aux_num) {
            // Get the auxiliary number's ith bit and fill
            // around middle
            a[group_no + 1 + i]
                = a[group_no - 1 - i]
                = aux_num & 1;
            aux_num = aux_num >> 1;
            i++;
        }
    }
    else // Insert bit 1 in the middle
    {
        // Length of the final binary representation
        len_f = 2 * group_no + 1;
 
        // Fill first and last bit as 1
        a[len_f - 1] = a[0] = 1;
        a[group_no] = 1;
 
        // Start filling the a[] from middle, with
        // the aux_num binary representation
        while (aux_num) {
             
            // Get the auxiliary number's ith bit and fill
            // around middle
            a[group_no + 1 + i]
                = a[group_no - 1 - i]
                = aux_num & 1;
            aux_num = aux_num >> 1;
            i++;
        }
    }
 
    // Convert the number to decimal from binary
    for (i = 0; i < len_f; i++)
        num += (1 << i) * a[i];
    return num;
}
 
// Will return the nth binary palindrome number
int getNthNumber(int n)
{
    int group_no = 0, group_offset;
    int count_upto_group = 0, count_temp = 1;
    int op, aux_num;
 
    // Add number of elements in all the groups,
    // until the group of the nth number is found
    while (count_temp < n) {
        group_no++;
 
        // Total number of elements until this group
        count_upto_group = count_temp;
        count_temp += 3 * (1 << (group_no - 1));
    }
 
    // Element's offset position in the group
    group_offset = n - count_upto_group - 1;
 
    // Finding which bit to be placed in the
    // middle and finding the number, which we
    // will fill from the middle in both
    // directions
    if ((group_offset + 1) <= (1 << (group_no - 1))) {
        op = 2; // No need to put extra bit in middle
 
        // We need to fill this auxiliary number
        // in binary form the middle in both directions
        aux_num = group_offset;
    }
    else {
        if (((group_offset + 1)
             - (1 << (group_no - 1))) % 2)
            op = 0; // Need to Insert 0 at middle
        else
            op = 1; // Need to Insert 1 at middle
        aux_num
            = ((group_offset) - (1 << (group_no - 1))) / 2;
    }
 
    return constructNthNumber(group_no, aux_num, op);
}
 
// Driver code
int main()
{
    int n = 9;
   
    // Function Call
    printf("%d", getNthNumber(n));
    return 0;
}

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Java

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// Efficient Java program to find n-th palindrome
class GFG {
 
    static int INT_SIZE = 32;
 
    // Construct the nth binary palindrome with the
    // given group number, aux_number and operation
    // type
    static int constructNthNumber(int group_no, int aux_num,
                                  int op)
    {
        int a[] = new int[INT_SIZE];
 
        int num = 0, len_f;
        int i = 0;
 
        // No need to insert any bit in the middle
        if (op == 2) {
             
            // Length of the final binary representation
            len_f = 2 * group_no;
 
            // Fill first and last bit as 1
            a[len_f - 1] = a[0] = 1;
 
            // Start filling the a[] from middle,
            // with the aux_num binary representation
            while (aux_num > 0) {
                // Get the auxiliary number's ith bit and
                // fill around middle
                a[group_no + i]
                    = a[group_no - 1 - i]
                    = aux_num & 1;
                aux_num = aux_num >> 1;
                i++;
            }
        }
 
        // Insert bit 0 in the middle
        else if (op == 0) {
            // Length of the final binary representation
            len_f = 2 * group_no + 1;
 
            // Fill first and last bit as 1
            a[len_f - 1] = a[0] = 1;
            a[group_no] = 0;
 
            // Start filling the a[] from middle, with
            // the aux_num binary representation
            while (aux_num > 0) {
                 
                // Get the auxiliary number's ith bit and
                // fill around middle
                a[group_no + 1 + i]
                    = a[group_no - 1 - i]
                    = aux_num & 1;
                aux_num = aux_num >> 1;
                i++;
            }
        }
        else // Insert bit 1 in the middle
        {
            // Length of the final binary representation
            len_f = 2 * group_no + 1;
 
            // Fill first and last bit as 1
            a[len_f - 1] = a[0] = 1;
            a[group_no] = 1;
 
            // Start filling the a[] from middle, with
            // the aux_num binary representation
            while (aux_num > 0) {
                // Get the auxiliary number's ith bit and
                // fill around middle
                a[group_no + 1 + i]
                    = a[group_no - 1 - i]
                    = aux_num & 1;
                aux_num = aux_num >> 1;
                i++;
            }
        }
 
        // Convert the number to decimal from binary
        for (i = 0; i < len_f; i++)
            num += (1 << i) * a[i];
        return num;
    }
 
    // Will return the nth binary palindrome number
    static int getNthNumber(int n)
    {
        int group_no = 0, group_offset;
        int count_upto_group = 0, count_temp = 1;
        int op, aux_num;
 
        // Add number of elements in all the groups,
        // until the group of the nth number is found
        while (count_temp < n) {
            group_no++;
 
            // Total number of elements until this group
            count_upto_group = count_temp;
            count_temp += 3 * (1 << (group_no - 1));
        }
 
        // Element's offset position in the group
        group_offset = n - count_upto_group - 1;
 
        // Finding which bit to be placed in the
        // middle and finding the number, which we
        // will fill from the middle in both
        // directions
        if ((group_offset + 1) <= (1 << (group_no - 1))) {
            op = 2; // No need to put extra bit in middle
 
            // We need to fill this auxiliary number
            // in binary form the middle in both directions
            aux_num = group_offset;
        }
        else {
            if (((group_offset + 1)
                 - (1 << (group_no - 1))) % 2 == 1)
                op = 0; // Need to Insert 0 at middle
            else
                op = 1; // Need to Insert 1 at middle
            aux_num
                = ((group_offset)
                - (1 << (group_no - 1))) / 2;
        }
 
        return constructNthNumber(group_no, aux_num, op);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 9;
       
        // Function Call
        System.out.printf("%d", getNthNumber(n));
    }
}
 
/* This code contributed by PrinciRaj1992 */

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Python3

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# Efficient Python3 program to find n-th palindrome
INT_SIZE = 32
 
# Construct the nth binary palindrome with the
# given group number, aux_number and operation type
 
 
def constructNthNumber(group_no, aux_num, op):
 
    a = [0] * INT_SIZE
    num, i = 0, 0
 
    # No need to insert any bit in the middle
    if op == 2:
 
        # Length of the final binary representation
        len_f = 2 * group_no
 
        # Fill first and last bit as 1
        a[len_f - 1] = a[0] = 1
 
        # Start filling the a[] from middle,
        # with the aux_num binary representation
        while aux_num:
 
            # Get the auxiliary number's ith
            # bit and fill around middle
            a[group_no + i] = a[group_no - 1 - i] = \
                aux_num & 1
            aux_num = aux_num >> 1
            i += 1
 
    # Insert bit 0 in the middle
    elif op == 0:
 
        # Length of the final binary representation
        len_f = 2 * group_no + 1
 
        # Fill first and last bit as 1
        a[len_f - 1] = a[0] = 1
        a[group_no] = 0
 
        # Start filling the a[] from middle, with
        # the aux_num binary representation
        while aux_num:
 
            # Get the auxiliary number's ith
            # bit and fill around middle
            a[group_no + 1 + i] = a[group_no - 1 - i] = \
                aux_num & 1
            aux_num = aux_num >> 1
            i += 1
 
    else# Insert bit 1 in the middle
 
        # Length of the final binary representation
        len_f = 2 * group_no + 1
 
        # Fill first and last bit as 1
        a[len_f - 1] = a[0] = 1
        a[group_no] = 1
 
        # Start filling the a[] from middle, with
        # the aux_num binary representation
        while aux_num:
 
            # Get the auxiliary number's ith
            # bit and fill around middle
            a[group_no + 1 + i] = a[group_no - 1 - i] = \
                aux_num & 1
            aux_num = aux_num >> 1
            i += 1
 
    # Convert the number to decimal from binary
    for i in range(0, len_f):
        num += (1 << i) * a[i]
    return num
 
# Will return the nth binary palindrome number
 
 
def getNthNumber(n):
 
    group_no = 0
    count_upto_group, count_temp = 0, 1
 
    # Add number of elements in all the groups,
    # until the group of the nth number is found
    while count_temp < n:
 
        group_no += 1
 
        # Total number of elements until this group
        count_upto_group = count_temp
        count_temp += 3 * (1 << (group_no - 1))
 
    # Element's offset position in the group
    group_offset = n - count_upto_group - 1
 
    # Finding which bit to be placed in the
    # middle and finding the number, which we
    # will fill from the middle in both directions
    if (group_offset + 1) <= (1 << (group_no - 1)):
 
        op = 2  # No need to put extra bit in middle
 
        # We need to fill this auxiliary number
        # in binary form the middle in both directions
        aux_num = group_offset
 
    else:
 
        if (((group_offset + 1) -
             (1 << (group_no - 1))) % 2):
            op = 0  # Need to Insert 0 at middle
        else:
            op = 1  # Need to Insert 1 at middle
        aux_num = (((group_offset) -
                    (1 << (group_no - 1))) // 2)
 
    return constructNthNumber(group_no, aux_num, op)
 
 
# Driver code
if __name__ == "__main__":
 
    n = 9
     
    # Function Call
    print(getNthNumber(n))
 
# This code is contributed by Rituraj Jain

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C#

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// Efficient C# program to find n-th palindrome
using System;
 
class GFG {
 
    static int INT_SIZE = 32;
 
    // Construct the nth binary palindrome with the
    // given group number, aux_number and operation
    // type
    static int constructNthNumber(int group_no, int aux_num,
                                  int op)
    {
        int[] a = new int[INT_SIZE];
 
        int num = 0, len_f;
        int i = 0;
 
        // No need to insert any bit in the middle
        if (op == 2) {
            
            // Length of the final binary representation
            len_f = 2 * group_no;
 
            // Fill first and last bit as 1
            a[len_f - 1] = a[0] = 1;
 
            // Start filling the a[] from middle,
            // with the aux_num binary representation
            while (aux_num > 0) {
                 
                // Get the auxiliary number's ith bit and
                // fill around middle
                a[group_no + i] = a[group_no - 1 - i]
                    = aux_num & 1;
                aux_num = aux_num >> 1;
                i++;
            }
        }
 
        // Insert bit 0 in the middle
        else if (op == 0) {
            
            // Length of the final binary representation
            len_f = 2 * group_no + 1;
 
            // Fill first and last bit as 1
            a[len_f - 1] = a[0] = 1;
            a[group_no] = 0;
 
            // Start filling the a[] from middle, with
            // the aux_num binary representation
            while (aux_num > 0) {
                 
                // Get the auxiliary number's ith bit and
                // fill around middle
                a[group_no + 1 + i] = a[group_no - 1 - i]
                    = aux_num & 1;
                aux_num = aux_num >> 1;
                i++;
            }
        }
        else // Insert bit 1 in the middle
        {
            // Length of the final binary representation
            len_f = 2 * group_no + 1;
 
            // Fill first and last bit as 1
            a[len_f - 1] = a[0] = 1;
            a[group_no] = 1;
 
            // Start filling the a[] from middle, with
            // the aux_num binary representation
            while (aux_num > 0) {
                // Get the auxiliary number's ith bit and
                // fill around middle
                a[group_no + 1 + i] = a[group_no - 1 - i]
                    = aux_num & 1;
                aux_num = aux_num >> 1;
                i++;
            }
        }
 
        // Convert the number to decimal from binary
        for (i = 0; i < len_f; i++)
            num += (1 << i) * a[i];
        return num;
    }
 
    // Will return the nth binary palindrome number
    static int getNthNumber(int n)
    {
        int group_no = 0, group_offset;
        int count_upto_group = 0, count_temp = 1;
        int op, aux_num;
 
        // Add number of elements in all the groups,
        // until the group of the nth number is found
        while (count_temp < n) {
            group_no++;
 
            // Total number of elements until this group
            count_upto_group = count_temp;
            count_temp += 3 * (1 << (group_no - 1));
        }
 
        // Element's offset position in the group
        group_offset = n - count_upto_group - 1;
 
        // Finding which bit to be placed in the
        // middle and finding the number, which we
        // will fill from the middle in both
        // directions
        if ((group_offset + 1) <= (1 << (group_no - 1))) {
            op = 2; // No need to put extra bit in middle
 
            // We need to fill this auxiliary number
            // in binary form the middle in both directions
            aux_num = group_offset;
        }
        else {
            if (((group_offset + 1) - (1 << (group_no - 1)))
                    % 2
                == 1)
                op = 0; // Need to Insert 0 at middle
            else
                op = 1; // Need to Insert 1 at middle
            aux_num
                = ((group_offset) - (1 << (group_no - 1)))
                  / 2;
        }
 
        return constructNthNumber(group_no, aux_num, op);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n = 9;
       
        // Function Call
        Console.Write("{0}", getNthNumber(n));
    }
}
 
// This code contributed by Rajput-Ji

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<?php
// Efficient PHP program to find n-th palindrome
$INT_SIZE=32;
 
/* Construct the nth binary palindrome with the
given group number, aux_number and operation
type */
function constructNthNumber($group_no, $aux_num,$op)
{
    global $INT_SIZE;
    $a=array_fill(0,$INT_SIZE,0);
 
    $num = 0;
    $i = 0;
    $len_f=0;
 
    // No need to insert any bit in the middle
    if ($op == 2)
    {
        // Length of the final binary representation
        $len_f = 2 * $group_no;
 
        // Fill first and last bit as 1
        $a[$len_f - 1] = $a[0] = 1;
 
        /* Start filling the a[] from middle,
        with the aux_num binary representation */
        while ($aux_num)
        {
            // Get the auxiliary number's ith bit and
            // fill around middle
            $a[$group_no + i] = $a[$group_no - 1 - $i]
                            = $aux_num & 1;
            $aux_num = $aux_num >> 1;
            $i++;
        }
    }
 
    // Insert bit 0 in the middle
    else if ($op == 0)
    {
        // Length of the final binary representation
        $len_f = 2 * $group_no + 1;
 
        // Fill first and last bit as 1
        $a[$len_f - 1] = $a[0] = 1;
        $a[$group_no] = 0;
 
        /* Start filling the a[] from middle, with
        the aux_num binary representation */
        while ($aux_num)
        {
            // Get the auxiliary number's ith bit and fill
            // around middle
            $a[$group_no + 1 + $i] = $a[$group_no - 1 - $i]
                                = $aux_num & 1;
            $aux_num = $aux_num >> 1;
            $i++;
        }
    }
    else     // Insert bit 1 in the middle
    {
        // Length of the final binary representation
        $len_f = 2 * $group_no + 1;
 
        // Fill first and last bit as 1
        $a[$len_f - 1] = $a[0] = 1;
        $a[$group_no] = 1;
 
        /* Start filling the a[] from middle, with
        the aux_num binary representation */
        while ($aux_num)
        {
            // Get the auxiliary number's ith bit and fill
            // around middle
            $a[$group_no + 1 + $i] = $a[$group_no - 1 - $i]
                                = $aux_num & 1;
            $aux_num = $aux_num >> 1;
            $i++;
        }
    }
 
    /* Convert the number to decimal from binary */
    for ($i = 0; $i < $len_f; $i++)
        $num += (1 << $i) * $a[$i];
    return $num;
}
 
/* Will return the nth binary palindrome number */
function getNthNumber($n)
{
    $group_no = 0;
    $count_upto_group = 0;
    $count_temp = 1;
    $op=$aux_num=0;
 
    /* Add number of elements in all the groups,
    until the group of the nth number is found */
    while ($count_temp < $n)
    {
        $group_no++;
 
        // Total number of elements until this group
        $count_upto_group = $count_temp;
        $count_temp += 3 * (1 << ($group_no - 1));
    }
 
    // Element's offset position in the group
    $group_offset = $n - $count_upto_group - 1;
 
    /* Finding which bit to be placed in the
    middle and finding the number, which we
    will fill from the middle in both
    directions */
    if (($group_offset + 1) <= (1 << ($group_no - 1)))
    {
        $op = 2; // No need to put extra bit in middle
 
        // We need to fill this auxiliary number
        // in binary form the middle in both directions
        $aux_num = $group_offset;
    }
    else
    {
        if ((($group_offset + 1) - (1 << ($group_no - 1))) % 2)
            $op = 0; // Need to Insert 0 at middle
        else
            $op = 1; // Need to Insert 1 at middle
        $aux_num = (int)((($group_offset) - (1 << ($group_no - 1))) / 2);
    }
 
    return constructNthNumber($group_no, $aux_num, $op);
}
 
    // Driver code
    $n = 9;
 
    // Function Call
    print(getNthNumber($n));
 
// This code is contributed by mits
?>

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Output

27

Time Complexity: O(1).

Reference: 
https://www.codeproject.com/Articles/1162038/Finding-nth-Binary-Palindrome-in-Csharp
This article is contributed by Kamesh Relangi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeek’s main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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