Given two integers n and k, Find the lexicographical nth palindrome of k digits.
Examples:
Input : n = 5, k = 4 Output : 1441 Explanation: 4 digit lexicographical palindromes are: 1001, 1111, 1221, 1331, 1441 5th palindrome = 1441 Input : n = 4, k = 6 Output : 103301
Naive Approach
A brute force is to run a loop from the smallest kth digit number and check for every number whether it is palindrome or not. If it is a palindrome number then decrements the value of k. Therefore the loop runs until k becomes exhausted.
// A naive approach of C++ program of finding nth // palindrome of k digit #include<bits/stdc++.h> using namespace std;
// Utility function to reverse the number n int reverseNum( int n)
{ int rem, rev=0;
while (n)
{
rem = n % 10;
rev = rev * 10 + rem;
n /= 10;
}
return rev;
} // Boolean Function to check for palindromic // number bool isPalindrom( int num)
{ return num == reverseNum(num);
} // Function for finding nth palindrome of k digits int nthPalindrome( int n, int k)
{ // Get the smallest k digit number
int num = ( int ) pow (10, k-1);
while ( true )
{
// check the number is palindrome or not
if (isPalindrom(num))
--n;
// if n'th palindrome found break the loop
if (!n)
break ;
// Increment number for checking next palindrome
++num;
}
return num;
} // Driver code int main()
{ int n = 6, k = 5;
printf ( "%dth palindrome of %d digit = %d\n" ,
n, k, nthPalindrome(n, k));
n = 10, k = 6;
printf ( "%dth palindrome of %d digit = %d" ,
n, k, nthPalindrome(n, k));
return 0;
} |
// A naive approach of Java program of finding nth // palindrome of k digit import java.util.*;
class GFG
{ // Utility function to reverse the number n static int reverseNum( int n)
{ int rem, rev = 0 ;
while (n > 0 )
{
rem = n % 10 ;
rev = rev * 10 + rem;
n /= 10 ;
}
return rev;
} // Boolean Function to check for palindromic // number static boolean isPalindrom( int num)
{ return num == reverseNum(num);
} // Function for finding nth palindrome of k digits static int nthPalindrome( int n, int k)
{ // Get the smallest k digit number
int num = ( int )Math.pow( 10 , k- 1 );
while ( true )
{
// check the number is palindrome or not
if (isPalindrom(num))
--n;
// if n'th palindrome found break the loop
if (n == 0 )
break ;
// Increment number for checking next palindrome
++num;
}
return num;
} // Driver code public static void main(String[] args)
{ int n = 6 , k = 5 ;
System.out.println(n + "th palindrome of " + k + " digit = " + nthPalindrome(n, k));
n = 10 ; k = 6 ;
System.out.println(n + "th palindrome of " + k + " digit = " + nthPalindrome(n, k));
} } // This code is contributed by mits |
# A naive approach of Python3 program # of finding nth palindrome of k digit import math;
# Utility function to # reverse the number n def reverseNum(n):
rev = 0 ;
while (n):
rem = n % 10 ;
rev = (rev * 10 ) + rem;
n = int (n / 10 );
return rev;
# Boolean Function to check for # palindromic number def isPalindrom(num):
return num = = reverseNum(num);
# Function for finding nth # palindrome of k digits def nthPalindrome(n, k):
# Get the smallest k digit number
num = math. pow ( 10 , k - 1 );
while ( True ):
# check the number is
# palindrome or not
if (isPalindrom(num)):
n - = 1 ;
# if n'th palindrome found
# break the loop
if ( not n):
break ;
# Increment number for checking
# next palindrome
num + = 1 ;
return int (num);
# Driver code n = 6 ;
k = 5 ;
print (n, "th palindrome of" ,k, "digit =" ,nthPalindrome(n, k));
n = 10 ;
k = 6 ;
print (n, "th palindrome of" ,k, "digit =" ,nthPalindrome(n, k));
# This code is contributed by mits |
// A naive approach of C# program of finding nth // palindrome of k digit using System;
class GFG
{ // Utility function to reverse the number n static int reverseNum( int n)
{ int rem, rev = 0;
while (n > 0)
{
rem = n % 10;
rev = rev * 10 + rem;
n /= 10;
}
return rev;
} // Boolean Function to check for palindromic // number static bool isPalindrom( int num)
{ return num == reverseNum(num);
} // Function for finding nth palindrome of k digits static int nthPalindrome( int n, int k)
{ // Get the smallest k digit number
int num = ( int )Math.Pow(10, k-1);
while ( true )
{
// check the number is palindrome or not
if (isPalindrom(num))
--n;
// if n'th palindrome found break the loop
if (n == 0)
break ;
// Increment number for checking next palindrome
++num;
}
return num;
} // Driver code public static void Main()
{ int n = 6, k = 5;
Console.WriteLine(n + "th palindrome of " + k + " digit = " + nthPalindrome(n, k));
n = 10; k = 6;
Console.WriteLine(n + "th palindrome of " + k + " digit = " + nthPalindrome(n, k));
} } // This code is contributed // by Akanksha Rai |
<?php // A naive approach of PHP program // of finding nth palindrome of k digit // Utility function to // reverse the number n function reverseNum( $n )
{ $rem ;
$rev = 0;
while ( $n )
{
$rem = $n % 10;
$rev = ( $rev * 10) + $rem ;
$n = (int)( $n / 10);
}
return $rev ;
} // Boolean Function to check for // palindromic number function isPalindrom( $num )
{ return $num == reverseNum( $num );
} // Function for finding nth // palindrome of k digits function nthPalindrome( $n , $k )
{ // Get the smallest k digit number
$num = pow(10, $k - 1);
while (true)
{
// check the number is
// palindrome or not
if (isPalindrom( $num ))
-- $n ;
// if n'th palindrome found
// break the loop
if (! $n )
break ;
// Increment number for checking
// next palindrome
++ $num ;
}
return $num ;
} // Driver code $n = 6;
$k = 5;
echo $n , "th palindrome of " , $k , " digit = " ,
nthPalindrome( $n , $k ), "\n" ;
$n = 10;
$k = 6;
echo $n , "th palindrome of " , $k , " digit = " ,
nthPalindrome( $n , $k ), "\n" ;
// This code is contributed by ajit ?> |
<script> // A naive approach of Javascript
// program of finding nth
// palindrome of k digit
// Utility function to
// reverse the number n
function reverseNum(n)
{
let rem, rev = 0;
while (n > 0)
{
rem = n % 10;
rev = rev * 10 + rem;
n = parseInt(n / 10);
}
return rev;
}
// Boolean Function to
// check for palindromic
// number
function isPalindrom(num)
{
return num == reverseNum(num);
}
// Function for finding nth
// palindrome of k digits
function nthPalindrome(n, k)
{
// Get the smallest k digit number
let num = Math.pow(10, k-1);
while ( true )
{
// check the number is
// palindrome or not
if (isPalindrom(num))
--n;
// if n'th palindrome found
// break the loop
if (n == 0)
break ;
// Increment number for checking
// next palindrome
++num;
}
return num;
}
let n = 6, k = 5;
document.write(n + "th palindrome of " + k +
" digit = " + nthPalindrome(n, k) + "</br>" );
n = 10; k = 6;
document.write(n + "th palindrome of " + k +
" digit = " + nthPalindrome(n, k));
</script> |
Output:
6th palindrome of 5 digit = 10501 10th palindrome of 6 digit = 109901
Time complexity: O(10k)
Auxiliary space: O(1), since no extra space has been taken.
Efficient approach
An efficient method is to look for a pattern. According to the property of palindrome first, half digits are the same as the rest half digits in reverse order. Therefore, we only need to look for the first half digits as the rest of them can easily be generated. Let’s take k = 8, the smallest palindrome always starts from 1 as the leading digit and goes like that for the first 4 digits of the number.
First half values for k = 8 1st: 1000 2nd: 1001 3rd: 1002 ... ... 100th: 1099 So we can easily write the above sequence for nth palindrome as: (n-1) + 1000 For k digit number, we can generalize above formula as: If k is odd => num = (n-1) + 10k/2 else => num = (n-1) + 10k/2 - 1 Now rest half digits can be expanded by just printing the value of num in reverse order. But before this if k is odd then we have to truncate the last digit of a value num
Illustration:
n = 6 k = 5
- Determine the number of first half digits = floor(5/2) = 2
- Use formula: num = (6-1) + 102 = 105
- Expand the rest half digits by reversing the value of num.
Final answer will be 10501
Below is the implementation of the above steps
// C++ program of finding nth palindrome // of k digit #include<bits/stdc++.h> using namespace std;
void nthPalindrome( int n, int k)
{ // Determine the first half digits
int temp = (k & 1) ? (k / 2) : (k/2 - 1);
int palindrome = ( int ) pow (10, temp);
palindrome += n - 1;
// Print the first half digits of palindrome
printf ( "%d" , palindrome);
// If k is odd, truncate the last digit
if (k & 1)
palindrome /= 10;
// print the last half digits of palindrome
while (palindrome)
{
printf ( "%d" , palindrome % 10);
palindrome /= 10;
}
printf ( "\n" );
} // Driver code int main()
{ int n = 6, k = 5;
printf ( "%dth palindrome of %d digit = " ,n ,k);
nthPalindrome(n ,k);
n = 10, k = 6;
printf ( "%dth palindrome of %d digit = " ,n ,k);
nthPalindrome(n, k);
return 0;
} |
// Java program of finding nth palindrome // of k digit class GFG{
static void nthPalindrome( int n, int k)
{ // Determine the first half digits
int temp = (k & 1 )!= 0 ? (k / 2 ) : (k/ 2 - 1 );
int palindrome = ( int )Math.pow( 10 , temp);
palindrome += n - 1 ;
// Print the first half digits of palindrome
System.out.print(palindrome);
// If k is odd, truncate the last digit
if ((k & 1 )> 0 )
palindrome /= 10 ;
// print the last half digits of palindrome
while (palindrome> 0 )
{
System.out.print(palindrome % 10 );
palindrome /= 10 ;
}
System.out.println( "" );
} // Driver code public static void main(String[] args)
{ int n = 6 , k = 5 ;
System.out.print(n+ "th palindrome of " +k+ " digit = " );
nthPalindrome(n ,k);
n = 10 ;
k = 6 ;
System.out.print(n+ "th palindrome of " +k+ " digit = " );
nthPalindrome(n, k);
} } // This code is contributed by mits |
# Python3 program of finding nth palindrome # of k digit def nthPalindrome(n, k):
# Determine the first half digits
if (k & 1 ):
temp = k / / 2
else :
temp = k / / 2 - 1
palindrome = 10 * * temp
palindrome = palindrome + n - 1
# Print the first half digits of palindrome
print (palindrome, end = "")
# If k is odd, truncate the last digit
if (k & 1 ):
palindrome = palindrome / / 10
# print the last half digits of palindrome
while (palindrome):
print (palindrome % 10 , end = "")
palindrome = palindrome / / 10
# Driver code if __name__ = = '__main__' :
n = 6
k = 5
print (n, "th palindrome of" , k, " digit = " , end = " " )
nthPalindrome(n, k)
print ()
n = 10
k = 6
print (n, "th palindrome of" , k, "digit = " ,end = " " )
nthPalindrome(n, k)
# This code is contributed by # Sanjit_Prasad |
// C# program of finding nth palindrome // of k digit using System;
class GFG
{ static void nthPalindrome( int n, int k)
{ // Determine the first half digits
int temp = (k & 1) != 0 ? (k / 2) : (k / 2 - 1);
int palindrome = ( int )Math.Pow(10, temp);
palindrome += n - 1;
// Print the first half digits
// of palindrome
Console.Write(palindrome);
// If k is odd, truncate the last digit
if ((k & 1) > 0)
palindrome /= 10;
// print the last half digits
// of palindrome
while (palindrome>0)
{
Console.Write(palindrome % 10);
palindrome /= 10;
}
Console.WriteLine( "" );
} // Driver code static public void Main ()
{ int n = 6, k = 5;
Console.Write(n+ "th palindrome of " + k +
" digit = " );
nthPalindrome(n, k);
n = 10;
k = 6;
Console.Write(n+ "th palindrome of " + k +
" digit = " );
nthPalindrome(n, k);
} } // This code is contributed by ajit |
<?php // PHP program of finding nth palindrome // of k digit function nthPalindrome( $n , $k )
{ // Determine the first half digits
$temp = ( $k & 1) ?
(int)( $k / 2) : (int)( $k / 2 - 1);
$palindrome = (int)pow(10, $temp );
$palindrome += $n - 1;
// Print the first half digits of palindrome
print ( $palindrome );
// If k is odd, truncate the last digit
if ( $k & 1)
$palindrome = (int)( $palindrome / 10);
// print the last half digits of palindrome
while ( $palindrome > 0)
{
print ( $palindrome % 10);
$palindrome = (int)( $palindrome / 10);
}
print ( "\n" );
} // Driver code $n = 6;
$k = 5;
print ( $n . "th palindrome of $k digit = " );
nthPalindrome( $n , $k );
$n = 10;
$k = 6;
print ( $n . "th palindrome of $k digit = " );
nthPalindrome( $n , $k );
// This code is contributed by mits ?> |
<script> // Javascript program of finding nth palindrome of k digit
function nthPalindrome(n, k)
{
// Determine the first half digits
let temp = (k & 1) != 0 ? parseInt(k / 2, 10) : (parseInt(k / 2, 10) - 1);
let palindrome = parseInt(Math.pow(10, temp), 10);
palindrome += n - 1;
// Print the first half digits
// of palindrome
document.write(palindrome);
// If k is odd, truncate the last digit
if ((k & 1) > 0)
palindrome = parseInt(palindrome / 10, 10);
// print the last half digits
// of palindrome
while (palindrome>0)
{
document.write(palindrome % 10);
palindrome = parseInt(palindrome / 10, 10);
}
document.write( "" + "</br>" );
}
let n = 6, k = 5;
document.write(n+ "th palindrome of " + k + " digit = " );
nthPalindrome(n, k);
n = 10;
k = 6;
document.write(n+ "th palindrome of " + k + " digit = " );
nthPalindrome(n, k);
</script> |
Output:
6th palindrome of 5 digit = 10501 10th palindrome of 6 digit = 109901
Time complexity: O(k)
Auxiliary space: O(1), since no extra space has been taken.
Reference:
http://stackoverflow.com/questions/11925840/how-to-calculate-nth-n-digit-palindrome-efficiently