TFind 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
A naive approach would be, to traverse through all the integers from 1 to 2^31 – 1 and increment the palindrome count, if the number is a palindrome. When the palindrome count reaches the required n, break the loop and return the current integer.
C++
// 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// representationint isKthBitSet(int x, int k){ return (x & (1 << (k - 1))) ? 1 : 0;}// Returns the position of leftmost set bit// in the binary representationint leftmostSetBit(int x){ int count = 0; while (x) { count++; x = x >> 1; } return count;}// Finds whether the integer in binary// representation is palindrome or notint 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 codeint main(){ int n = 9; // Function Call cout << findNthPalindrome(n);}// This code is contributed// by Akanksha Rai |
C
// 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// representationint isKthBitSet(int x, int k){ return (x & (1 << (k - 1))) ? 1 : 0;}// Returns the position of leftmost set bit// in the binary representationint leftmostSetBit(int x){ int count = 0; while (x) { count++; x = x >> 1; } return count;}// Finds whether the integer in binary// representation is palindrome or notint 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 codeint main(){ int n = 9; // Function Call printf("%d", findNthPalindrome(n));} |
Java
// 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. |
Python3
# 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 representationdef isKthBitSet(x, k): return 1 if (x & (1 << (k - 1))) else 0# Returns the position of leftmost# set bit in the binary representationdef leftmostSetBit(x): count = 0 while (x): count += 1 x = x >> 1 return count# Finds whether the integer in binary# representation is palindrome or notdef 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 1def 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 codeif __name__ == "__main__": n = 9 # Function Call print(findNthPalindrome(n))# This code is contributed# by ChitraNayal |
C#
// 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 |
PHP
<?php// PHP program to find n-th number whose// binary representation is palindrome.// Finds if the kth bit is set in// the binary representationfunction isKthBitSet($x, $k){ return ($x & (1 << ($k - 1))) ? 1 : 0;}// Returns the position of leftmost set// bit in the binary representationfunction leftmostSetBit($x){ $count = 0; while ($x) { $count++; $x = $x >> 1; } return $count;}// Finds whether the integer in binary// representation is palindrome or notfunction isBinPalindrome($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 Callecho (findNthPalindrome($n));// This code is contributed by jit_t?> |
Javascript
<script>// Javascript program to find n-th// number whose binary// representation is palindrome.let INT_MAX = 2147483647;// Finds if the kth bit// is set in the binary// representationfunction isKthBitSet(x, k){ return ((x & (1 << (k - 1))) > 0) ? 1 : 0;}// Returns the position of// leftmost set bit in the// binary representationfunction leftmostSetBit(x){ let count = 0; while (x > 0) { count++; x = x >> 1; } return count;}// Finds whether the integer// in binary representation is// palindrome or notfunction isBinPalindrome(x){ let l = leftmostSetBit(x); let 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){ let pal_count = 0; // Start from 1, traverse // through all the integers let 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 codelet n = 9; // Function Calldocument.write(findNthPalindrome(n));// This code is contributed by suresh07</script> |
Output
27
Time complexity: O(x) where x is a resultant number.
Auxiliary Space: O(1), the space complexity is O(1) since we are only using a constant amount of memory to store the values.
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
- if the curr string of even length then add “0” and “1” at the mid of curr string and add it into the queue.
- 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:
C++
// C++ program to find n-th palindrome#include <bits/stdc++.h>using namespace std;// utility function which is used to// convert binary string into integerint 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 numberint 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 Codeint main(){ int n = 9; // Function Call cout << getNthNumber(n);}// This code is contributed by Sagar Jangra and Naresh// Saharan |
Java
// Java program to find n-th palindromeimport 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 |
Python3
# Python program to find n-th palindrome# utility function which is used to# convert binary string into integerdef convertStringToInt(s): ans = 0 # convert binary string into integer for i in range(len(s)): ans = ans * 2 + (ord(s[i]) - ord('0')) return ans# function to find nth binary palindrome numberdef getNthNumber(n): # stores the binary palindrome string q = [] # base case if(n == 1): return 1 n = n - 1 # add 2nd binary palindrome string q.append("11") # runs till the nth binary palindrome number while(len(q) != 0): # remove curr binary palindrome string from # queue curr = q.pop(0) n -= 1 # 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 lenn = len(curr) # if length is even .so it is our first case # we have two choices if (len(curr) % 2 == 0): q.append(curr[0:int(lenn/2)]+"0"+curr[int(lenn/2):]) q.append(curr[0:int(lenn/2)]+"1"+curr[int(lenn/2):]) # if length is odd .so it is our second case # we have only one choice else: midChar = curr[int(lenn/2)] q.append(curr[0:int(lenn/2)]+midChar+curr[int(lenn/2):]) return 0 # Driver coden = 9# Function Callprint(getNthNumber(n))# This code is contributed by avanitrachhadiya2155 |
C#
// C# program to find n-th palindromeusing System;using System.Collections.Generic;class GFG{// Utility function which is used to// convert binary string into integerpublic static int convertStringToInt(String s){ int ans = 0; // Convert binary string into integer for(int i = 0; i < s.Length; i++) { if (s[i] == '1') ans += 1 << i; } return ans;}// Function to find nth binary palindrome numberpublic static int getNthNumber(int n){ // Stores the binary palindrome string Queue<String> q = new Queue<String>(); // Base case if (n == 1) return 1; n = n - 1; // Add 2nd binary palindrome string q.Enqueue("11"); // Runs till the nth binary palindrome number while (n-- > 0) { // Remove curr binary palindrome // string from queue String curr = q.Dequeue(); // 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.Enqueue(curr.Substring(0, len / 2) + "0" + curr.Substring(len / 2)); q.Enqueue(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[len / 2]; q.Enqueue(curr.Substring(0, len / 2) + midChar + curr.Substring(len / 2)); } } return -1;}// Driver codepublic static void Main(String[] args){ int n = 9; // Function Call Console.WriteLine(getNthNumber(n));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to find n-th palindrome// utility function which is used to // convert binary string into integerfunction convertStringToInt(s){ let ans = 0; // convert binary string into integer for (let i = 0; i < s.length; i++) { if (s[i] == '1') ans += 1 << i; } return ans;}// function to find nth binary palindrome numberfunction getNthNumber(n){ // stores the binary palindrome string let q = []; // base case if (n == 1) return 1; n = n - 1; // add 2nd binary palindrome string q.push("11"); // runs till the nth binary palindrome number while (n-- > 0) { // remove curr binary palindrome string from // queue let curr = q.shift(); // 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 let len = curr.length; let len2 = Math.floor(len / 2); // if length is even .so it is our first case // we have two choices if (len % 2 == 0) { q.push(curr.substring(0, len2) + "0" + curr.substring(len2)); q.push(curr.substring(0, len2) + "1" + curr.substring(len2)); } // if length is odd .so it is our second case // we have only one choice else { let midChar = curr[len2]; q.push(curr.substring(0, len2) + midChar + curr.substring(len2)); } } return -1;}// Driver codelet n = 9; // Function Calldocument.write(getNthNumber(n));// This code is contributed by ab2127</script> |
Output
27
Time complexity: O(N)
Auxiliary Space: 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 the 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 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 the 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 the 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++
// Efficient C++ program to find n-th palindrome#include <iostream>#include <bits/stdc++.h>using namespace std;// Construct the nth binary palindrome with the// given group number, aux_number and operation// typeint constructNthNumber(int group_no, int aux_num, int op){ int INT_SIZE = 32 ; 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++; } } // Insert bit 1 in the middle else { // 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 numberint 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))) { // No need to put extra bit in middle op = 2; // 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) // Need to Insert 0 at middle op = 0; else // Need to Insert 1 at middle op = 1; aux_num = ((group_offset) - (1 << (group_no - 1))) / 2; } return constructNthNumber(group_no, aux_num, op);}// Driver codeint main(){ int n = 9; // Function Call cout << getNthNumber(n) ; return 0;}// This code is contributed by Khushboogoyal499 |
C
// 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// typeint 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 numberint 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 codeint main(){ int n = 9; // Function Call printf("%d", getNthNumber(n)); return 0;} |
Java
// Efficient Java program to find n-th palindromeclass 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 */ |
Python3
# Efficient Python3 program to find n-th palindromeINT_SIZE = 32# Construct the nth binary palindrome with the# given group number, aux_number and operation typedef 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 numberdef 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 codeif __name__ == "__main__": n = 9 # Function Call print(getNthNumber(n))# This code is contributed by Rituraj Jain |
C#
// Efficient C# program to find n-th palindromeusing 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 |
PHP
<?php// Efficient PHP program to find n-th palindrome$INT_SIZE=32;/* Construct the nth binary palindrome with thegiven group number, aux_number and operationtype */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?> |
Javascript
<script>// Efficient javascript program to find n-th palindrome var INT_SIZE = 32;// Construct the nth binary palindrome with the// given group number, aux_number and operation// typefunction constructNthNumber(group_no , aux_num,op){ var a = Array.from({length: INT_SIZE}, (_, i) => 0); var num = 0, len_f=0; var 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++; } } // Insert bit 1 in the middle else { // 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 numberfunction getNthNumber(n){ var group_no = 0, group_offset; var count_upto_group = 0, count_temp = 1; var 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))) { // No need to put extra bit in middle op = 2; // 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) // Need to Insert 0 at middle op = 0; else // Need to Insert 1 at middle op = 1; aux_num = ((group_offset) - (1 << (group_no - 1))) / 2; } return constructNthNumber(group_no, aux_num, op);}// Driver codevar n = 9;// Function Calldocument.write(getNthNumber(n));// This code is contributed by Princi Singh</script> |
Output
27
Time Complexity: O(1).
Auxiliary Space: 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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on GeeksforGeek’s main page and help other Geeks.
Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.
