N’th palindrome of K digits

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach

A brute force is to run a loop from smallest kth digit number and check for every number whether it is palindrome or not. If it is palindrome number then decrements the value of k. Therefore the loop runs until k become exhausted.

C++

 // A naive approach of C++ program of finding nth // palindrome of k digit #include 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 palindrom 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 palindrom 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          # palindrom 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 palindrom 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



Output:

6th palindrome of 5 digit = 10501
10th palindrome of 6 digit = 109901

Time complexity: O(10k)
Auxiliary space: O(1)

Efficient approach

An efficient method is to look for a pattern. According to the property of palindrome first half digits is same as the rest half digits in reverse order. Therefore we only need to look for first half digits as rest of them can easily be generated. Let’s take k = 8, smallest palindrome always starts from 1 as leading digit and goes like that for first 4 digit of 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 above steps

C++

 // C++ program of finding nth palindrome // of k digit #include 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

 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 ?>

Output:

6th palindrome of 5 digit = 10501
10th palindrome of 6 digit = 109901

Time complexity: O(k)
Auxiliary space: O(1)

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