Check whether a given number ’n’ is a palindrome number.
Palindrome numbers are the numbers that don't change when reversed.
You don’t need to print anything. Just implement the given function.
Input: 'n' = 51415
Output: true
Explanation: On reversing, 51415 gives 51415.
The first and only line of the input contains the number 'n'.
Return a boolean value True or False.
1032
false
1032, on being reversed, gives 2301, which is a totally different number.
121
true
121, on being reversed, gives 121, which is the same.
The expected time complexity is O(log(n)).
1 <= n <= 10^9
Time Limit: 1 sec
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Interview problems
2 liner C++ Solution in O(n) time complexity
Note:- Since the question didn't mentioned that to solve it without type-casting the value of n from integer to string, I tried this method.
bool palindrome(int n) {
// Write your code here
while(n>=0){
string str= to_string(n); if(str==string(str.rbegin(), str.rend())) return true;
else return false;
}
}
Interview problems
PalindromeNumber Java
public class Solution {
public static boolean palindromeNumber(int n){
// Write your code here.
int duplicate =n;
int reverseNumber=0;
while(n>0){
int rem=n%10;
reverseNumber=(reverseNumber*10)+rem;
n=n/10;
}
if(duplicate==reverseNumber){
return true;
}
else{
return false;
}
}
}
Interview problems
solution
bool palindrome(int n)
{
// Write your code here
int sum=0;
int original=n;
int last=0;
while(n>0){
last=n%10;
n=n/10;
sum=sum*10+last;
}
if(sum==original){
return true;
}
else{
return false;
}
}
Interview problems
Simple java code
public class Solution {
public static boolean palindromeNumber(int n){
// Write your code here.
int dup =n;
int revnum=0;
while(n>0){
int rem = n%10;
revnum = (revnum*10)+rem;
n=n/10;
}
if(dup==revnum){
return true;
}else{
return false;
}
}
}Interview problems
Palindrome cpp easy
bool palindrome(int n)
{
int ori = n;
int rev = 0;
while(n>0) //use while loop as the iteration depends on n not on i
{
int dig = n%10;
rev = (rev * 10) + dig;
n = n/10;
}
if(ori == rev)
{
return true;
}
return false;
}
Interview problems
test case not passing?
n = int(input())
def palindrome(n):
reverse = 0
org = n
while n > 0:
digit = n % 10
reverse = reverse * 10 + digit
n //= 10
return org == reverse
if palindrome(n):
print(f"{n} is a palindrome")
else:
print(f"{n} is not a palindrome")
why this is not working?
Interview problems
java best solution
public class Solution {
public static boolean palindromeNumber(int n){
// Write your code here.
int rev=0;
int orginal=n;
boolean flag=false;
while(n!=0){
int last=n%10;
rev=(rev*10)+last;
n=n/10;
}if(rev==orginal) flag=true;
return flag;
}
}
Interview problems
java code
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int originalNum = n;
int RevNum = 0;
while (n > 0) {
int LastDigit = n % 10;
RevNum = (RevNum * 10) + LastDigit;
n = n / 10;
}
System.out.println(originalNum == RevNum ? "true" : "false");
sc.close();
}
}
Interview problems
palindrome number for beginner it take more time but it is understandable
bool palindrome(int n)
{
int original_n=n;
int reverse_no=0;
while (n > 0) {
int last_digit = n % 10;
reverse_no = reverse_no* 10 + last_digit;
n = n / 10;
}
if(reverse_no==original_n){
return true;
}else{
return false;
}
}
Interview problems
Palindrome number
bool palindrome(int n)
{
int reverse = 0;
int orignal = n;
while(n>0){
int lastdigit = n%10;
reverse = reverse*10+lastdigit;
n = n/10;
}
if(orignal==reverse){
return true;
}
else{
return false;
}
