N’th palindrome of K digits

N’th palindrome of K digits

Given two integers n and k, Find the lexicographical nth palindrome of k digits.

Input  : n = 5, k = 4
Output : 1441
Explanation:
4 digit lexicographical palindromes are:
1001, 1111, 1221, 1331, 1441
5th palindrome = 1441

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 

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");

}

A brute force is to run a loop from smallest k^th 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.

Time complexity: O(10k)

// 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;

}

N’th palindrome of K digits