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Write a program that takes number and gives the resulting palindrome (if one exists). If it took more than 1, 000 iterations (additions) or yield a
palindrome that is greater than 4, 294, 967, 295, assume that no palindrome exist for the given number.
Examples:
Input: N = 195 Output: 9339 Input: N = 265 Output: 45254 Input: N = 196 Output: No palindrome exist
Approach: Create a reverse and add function to start with a number, reverses its digits, and adds the
reverse to the original. If the sum is not a palindrome, repeat this procedure until it does.
C++
#include <bits/stdc++.h>
using namespace std;
long long reverse Digits(long long num)
{
long long rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
bool isPalindrome(long long num)
{
return (reverse Digits(num) == num);
}
void ReverseandAdd(long long num)
{
long long rev_num = 0;
while (num <= 4294967295) {
rev_num = reverse Digits(num);
num
= num + rev_num;
if (isPalindrome(num)) {
printf("%lld\n", num);
break;
}
else if (num > 4294967295) {
printf("No palindrome exist");
}
}
}
int main()
{
ReverseandAdd(195);
ReverseandAdd(265);
return 0;
}
Java
public class ReverseAdd {
long revers eDigits(long num)
{
long rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
boolean isPalindrome(long num)
{
return (reversDigits(num) == num);
}
void ReverseandAdd(long num)
{
long rev_num = 0;
while (num <= 4294967295l) {
rev_num = reversDigits(num);
num
= num + rev_num;
if (isPalindrome(num)) {
System.out.println(num);
break;
}
else if (num > 4294967295l) {
System.out.println("No palindrome exist");
}
}
}
public static void main(String[] args)
{
ReverseAdd ob = new ReverseAdd();
ob.ReverseandAdd(195l);
ob.ReverseandAdd(265l);
}
}
Python3
def reverse Digits(num):
rev_num = 0
while
(num > 0):
rev_num = rev_num * 10 + num % 10
num = num//10
return rev_num
def isPalindrome(num):
return (reverse Digits(num) == num)
def ReverseandAdd(num):
rev_num = 0
while (num <= 4294967295):
rev_num = reverse Digits(num)
num =
num + rev_num
if(isPalindrome(num)):
print (num)
break
else:
if (num > 4294967295):
print ("No palindrome exist")
ReverseandAdd(195)
ReverseandAdd(265)
C#
using System;
class GFG {
static long reverse Digits(long num)
{
long
rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = num / 10;
}
return rev_num;
}
static bool isPalindrome(long num)
{
return (reverse Digits(num) == num);
}
static void ReverseandAdd(long num)
{
long rev_num = 0;
while (num <= 4294967295) {
rev_num = reverse Digits(num);
num = num + rev_num;
if
(isPalindrome(num)) {
Console.WriteLine(num);
break;
}
else if (num > 4294967295) {
Console.WriteLine("No palindrome exist");
}
}
}
public static void Main()
{
ReverseandAdd(195);
ReverseandAdd(265);
}
}
PHP
<?php
function reverse Digits($num)
{
$rev_num = 0;
while ($num > 0)
{
$rev_num = $rev_num * 10 + $num % 10;
$num = (int)($num / 10);
}
return $rev_num;
}
function isPalindrome($num)
{
return (reverse Digits($num) == $num);
}
function ReverseandAdd($num)
{
$rev_num
= 0;
while ($num <= 4294967295)
{
$rev_num = reverse Digits($num);
$num = $num + $rev_num;
if (isPalindrome($num))
{
print($num . "\n");
break;
}
else if ($num > 4294967295)
{
print("No palindrome exist");
}
}
}
ReverseandAdd(195);
ReverseandAdd(265);
?>
Javascript
<script>
function reverse Digits(num)
{
let rev_num = 0;
while (num > 0)
{
rev_num = rev_num * 10 + num % 10;
num = parseInt(num / 10, 10);
}
return rev_num;
}
function isPalindrome(num)
{
return(reverse Digits(num) == num);
}
function
ReverseandAdd(num)
{
let rev_num = 0;
while (num <= 4294967295)
{
rev_num = reverse Digits(num);
num = num + rev_num;
if (isPalindrome(num))
{
document.write(num + "</br>");
break;
}
else if (num > 4294967295)
{
document.write("No palindrome exist" +
"</br>");
}
}
}
ReverseandAdd(195);
ReverseandAdd(265);
</script>
Time complexity: O(log N) for a given input
Auxiliary space: O(1) because constant variables have been used
References: https://app.assembla.com/spaces/AASU_Fall2008_ProgrammingTeam/wiki
This article is contributed by Rahul Agrawal.
If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to . See your article appearing on the GeeksforGeeks main page and help other Geeks.
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