N’th palindrome of K digits

Given two integers n and k, Find the lexicographical n^th 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 k^th 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. 

C++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

<?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
?>

Javascript

<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(10^k) 
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 

1. Determine the number of first half digits = floor(5/2) = 2
2. Use formula: num = (6-1) + 10² = 105
3. 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++

// 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

// 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

# 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#

// 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
// 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
?>

Javascript

<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
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