# 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

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.

```// 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 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;
}
```
```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.

Below is C++ implementation of 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;
}
```

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

```Output:
6th palindrome of 5 digit = 10501
10th palindrome of 6 digit = 109901
```

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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