Given two integers **n** and **k**, Find the lexicographical n^{th} palindrome of k digits.

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 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.

`// 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(10^{k})

**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 = 81^{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) + 1000For 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 ofnumin reverse order. But before this if k is odd then we have to truncate the last digit of a valuenum

**Illustration:**

n = 6 k = 5

- Determine the number of first half digits = floor(5/2) = 2
- Use formula: num = (6-1) + 10
^{2}= 105 - Expand the rest half digits by reversing the value of num.

Final answer will be 10501

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

**Reference:**

http://stackoverflow.com/questions/11925840/how-to-calculate-nth-n-digit-palindrome-efficiently

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