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Sum of first K even-length Palindrome numbers
  • Difficulty Level : Easy
  • Last Updated : 28 May, 2021

Given a integer k, find the sum of first k even-length palindrome numbers. 
Even length here refers to the number of digits of a number is even.
Examples: 
 

Input : k = 3
Output : 66
Explanation: 11 + 22 + 33  = 66 (Sum 
of first three even-length palindrome 
numbers)


Input : 10
Output : 1496
Explanation: 11+22+33+44+55+66+77+88+
99+1001 = 1496

 

A naive approach will be to check every even length number, if it is a palindrome number then we sum it up. We repeat the same process for first K even length palindrome numbers and sum them up to get the sum. 
In this case complexity will go high as even length numbers are from 10-99 and then 1000-9999 and then so on… 
10-99, 1000-9999, 100000-999999.. has 9, 90, 900 respectively palindrome numbers in them, so to check k numbers we have to check a lot of numbers which will not be efficient enough.
An efficient approach will be to observe a pattern for even length prime numbers.
 

11, 22, 33, 44, 55, 66, 77, 88, 99, 1001, 1111, 1221, 1331, 1441, 1551, 1661… 
 

1st number is 11, 2nd is 22, third is 33, 16th is 16-rev(16) i.e., 1661
So the Nth number will int(string(n)+rev(string(n)). 
See here for conversion of integer to string and string to integer.
Below is the implementation of the above approach: 
 



C++




#include <bits/stdc++.h>
#include <boost/lexical_cast.hpp>
using namespace std;
 
// function to return the sum of
// first K even length palindrome numbers
int sum(int k)
{
    // loop to get sum of first K even
    // palindrome numbers
    int sum = 0;
    for (int i = 1; i <= k; i++) {
 
        // convert integer to string
        string num = to_string(i);
 
        // Find reverse of num.
        string revNum = num;
        reverse(revNum.begin(), revNum.end());
 
        // string(n)+rev(string(n)
        string strnum = (num + revNum);
 
        // convert string to integer
        int number = boost::lexical_cast<int>(strnum);
 
        sum += number; // summation
    }
    return sum;
}
// driver program to check the above function
int main()
{
    int k = 3;
    cout << sum(k);
    return 0;
}

Java




// Java implementation to find sum of
// first K even-length Palindrome numbers
import java.util.*;
import java.lang.*;
 
public class GfG{
 
public static String reverseString(String str)
{
    StringBuilder sb = new StringBuilder(str);   
    sb.reverse();  
    return sb.toString();
}
 
// function to return the sum of
// first K even length palindrome numbers
static int sum(int k)
{
    // loop to get sum of first K even
    // palindrome numbers
    int sum = 0;
    for (int i = 1; i <= k; i++) {
 
    // convert integer to string
    String num = Integer.toString(i);
 
    // Find reverse of num.
    String revNum = num;
    revNum = reverseString(num);
 
    // string(n)+rev(string(n)
    String strnum = (num + revNum);
 
    // convert string to integer
    int number = Integer.parseInt(strnum);
 
    sum += number; // summation
    }
     
    return sum;
}
 
// driver function
public static void main(String argc[])
{
    int n = 3;
    System.out.println(sum(n));
}
}
 
// This code is contributed by Prerna Saini

Python3




# Python3 implementation of the approach
 
# function to return the sum of
# first K even length palindrome numbers
def summ(k):
 
    # loop to get sum of first K even
    # palindrome numbers
    sum = 0
    for i in range(1, k + 1):
 
        # convert integer to string
        num = str(i)
 
        # Find reverse of num.
        revNum = num
        revNum = ''.join(reversed(revNum))
 
        # string(n)+rev(string(n)
        strnum = num + revNum
 
        # convert string to integer
        number = int(strnum)
 
        sum += number # summation
 
    return sum
 
# Driver Code
if __name__ == "__main__":
    k = 3
    print(summ(k))
 
# This code is contributed by
# sanjeev2552

C#




// C# implementation to find sum of
// first K even-length Palindrome numbers
using System;
 
class GfG
{
     
    // function to return the sum of
    // first K even length palindrome numbers
    static int sum(int k)
    {
         
        // loop to get sum of first K even
        // palindrome numbers
        int sum = 0;
        for (int i = 1; i <= k; i++)
        {
 
            // convert integer to string
            String num = Convert.ToString(i);
 
            // Find reverse of num.
            String revNum = num;
            revNum = reverse(num);
 
            // string(n)+rev(string(n)
            String strnum = (num + revNum);
 
            // convert string to integer
            int number = Convert.ToInt32(strnum);
 
            sum += number; // summation
        }
 
        return sum;
    }
     
    static String reverse(String input)
    {
        char[] temparray = input.ToCharArray();
        int left, right = 0;
        right = temparray.Length - 1;
 
        for (left = 0; left < right; left++, right--)
        {
             
            // Swap values of left and right
            char temp = temparray[left];
            temparray[left] = temparray[right];
            temparray[right] = temp;
        }
        return String.Join("",temparray);
    }
     
    // Driver code
    public static void Main(String []argc)
    {
        int n = 3;
        Console.WriteLine(sum(n));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
 
// function to return the sum of
// first K even length palindrome numbers
function sum(k)
{
    // loop to get sum of first K even
    // palindrome numbers
    var sum = 0;
    for (var i = 1; i <= k; i++) {
 
        // convert integer to string
        var num = (i.toString());
 
        // Find reverse of num.
        var revNum = num;
        revNum = revNum.split('').reverse().join('');
 
        // string(n)+rev(string(n)
        var strnum = (num + revNum);
 
        // convert string to integer
        var number = parseInt(strnum);
 
        sum += number; // summation
    }
    return sum;
}
 
 
// driver program to check the above function
var k = 3;
document.write(sum(k));
 
</script>

Output:  

66

 

 

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