Given a positive integer K, the task is to find Kth smallest palindromic number odd length.
Examples:
Input: K = 5
Output: 5
Explanation:
The palindromic numbers of odd lengths is {1, 2, 3, 4, 5, 6, 7, …, }. The 5th smallest palindromic numbers is 5.
Input: K = 10
Output: 101
Approach: The given problem can be solved based on the following observations:
- The first Palindromic Numbers of length 1 are 1, 2, 3, 4, 5, 6, 7, 8, and 9.
- The first Palindromic Numbers of length 3 is 101, which is the 10th smallest odd length palindrome number. Similarly, 11th, 12th, 13th, …, 99th smallest palindromic numbers are 111, 121, 131 …, 999 respectively.
- Therefore, the Kth smallest odd length palindrome number can be formed by joining K and the reverse of K except the last digit.
From the above observations, the Kth smallest odd length palindromic number is given by appending the reverse of all the digits of K except the last one at the end of K.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int oddLengthPalindrome( int k)
{
int palin = k;
k = k / 10;
while (k > 0)
{
int rev = k % 10;
palin = (palin * 10) + rev;
k = k / 10;
}
return palin;
}
int main()
{
int k = 504;
cout << oddLengthPalindrome(k);
}
|
Java
import java.util.*;
import java.lang.*;
class GFG{
static int oddLengthPalindrome( int k)
{
int palin = k;
k = k / 10 ;
while (k > 0 )
{
int rev = k % 10 ;
palin = (palin * 10 ) + rev;
k = k / 10 ;
}
return palin;
}
public static void main(String[] args)
{
int k = 504 ;
System.out.println(oddLengthPalindrome(k));
}
}
|
Python3
def oddLengthPalindrome(K):
palin = K
K = K / / 10
while (K > 0 ):
rev = K % 10
palin = palin * 10 + rev
K = K / / 10
return palin
if __name__ = = '__main__' :
K = 504
print (oddLengthPalindrome(K))
|
C#
using System;
class GFG{
static int oddLengthPalindrome( int k)
{
int palin = k;
k = k / 10;
while (k > 0)
{
int rev = k % 10;
palin = (palin * 10) + rev;
k = k / 10;
}
return palin;
}
static void Main( string [] args)
{
int k = 504;
Console.WriteLine(oddLengthPalindrome(k));
}
}
|
Javascript
<script>
function oddLengthPalindrome(k)
{
let palin = k;
k = Math.floor(k / 10);
while (k > 0)
{
let rev = k % 10;
palin = (palin * 10) + rev;
k = Math.floor(k / 10);
}
return palin;
}
let k = 504;
document.write(oddLengthPalindrome(k));
</script>
|
Time Complexity: O(log10K)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!
Last Updated :
22 Jul, 2021
Like Article
Save Article