N’th palindrome of K digits

Difficulty Level : Hard
Last Updated : 06 Sep, 2022

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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
5^th 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
1^st: 1000
2^nd: 1001
3^rd: 1002
...
...
100^th: 1099

So we can easily write the above sequence for n^th
palindrome as: (n-1) + 1000
For k digit number, we can generalize above formula as:

If k is odd
=> num = (n-1) + 10^k/2
else 
=> num = (n-1) + 10^{k/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++

// 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
This article is contributed by Shubham Bansal. 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 the GeeksforGeeks 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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Data Structures & Algorithms in Python - Self Paced

Competitive Programming - Live

Complete Interview Preparation - Self Paced

Python Programming Foundation -Self Paced

Data Structures and Algorithms - Self Paced

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Complete Machine Learning & Data Science Program

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