Palindrome Number - Leetcode 9 - Python

0:00 - Read the problem
1:10 - Drawing Explanation
8:23 - Coding Explanation

leetcode 9

#palindrome #number #python
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[aspar]

Nice Solution!
We can write the divider value in this way as well 8:37:
if x==0: return True
div = 10**int(math.log10(x))

[nisimkd]

Great solution, thank you!

The solution implemented in c#:
public bool IsPalindrome(int x) {
if (x < 0)
{
return false;
}

long divider = 1;
while (divider * 10 <= x)
{
divider *= 10;
}

while (x > 0)
{
int l = (int) (x / divider);
int r = (int) (x % 10);

if (l != r)
{
return false;
}

// Cut the left digit
x = (int) (x % divider);

// Cut the right digit
x /= 10;

divider /= 100;
}

return true;
}

divider is in long type due to case the given x is one digit less the max int value.


Another solution, write the reserved number and compare it to the given x.
implemented in c#:
public bool IsPalindrome(int x) {
if (x < 0)
{
return false;
}

int reservedNumber = 0;
int originalX = x;

while (x > 0)
{
int lastDigit = x % 10;

reservedNumber = reservedNumber * 10 + lastDigit;
x /= 10;
}

return reservedNumber == originalX;
}

[marioalfredojorge]

A simple solution using JS. The idea is to reverse the number and then compare it to the original one.

var isPalindrome = function(x) {
if (x < 0) return false;

let num = x;
let result = 0;

while (num > 0) {
result = result * 10 + (num % 10);

num = Math.floor(num / 10);
}

return result === x;
};

[anandsharma]

watched this and did it using strings, gave myself a pat on the back.

[babacarkane]

Simple, efficient and straight to the point. Thanks for this answer.

[kaankahveci]

I feel like this is a simpler approach:

public class Solution {
public bool IsPalindrome(int x) {
if(x < 0) return false;
if (x != 0 && x % 10 == 0) return false;

int original = x;
int reversed = 0;
while (x != 0) {
reversed = reversed * 10 + x % 10;
x /= 10;
}

return original == reversed;
}
}

This was my C# solution (which is similar to yours) that I submitted a while ago;

if(x == 0)
return true;
if(x < 0 || x % 10 == 0)
return false;

var reverse = 0;

while(x > reverse)
{
var lastDigit = x % 10;
x /= 10;
reverse = reverse * 10 + lastDigit;
}

return x == reverse || x == reverse / 10;

[iforcollege]

I think this video may need a remake because it will not work when the input is 1410110141 on Java. I tried this and it worked:
if (x < 0 || (x % 10 == 0 && x != 0)) {
return false;
} else {
int original = x, reversed = 0;
while (original > reversed) {
reversed = (reversed * 10) + (original % 10);
original /= 10;
}
return original == reversed || original == reversed / 10;
}

[pavel.pavlov]

you could use decimal log for finding the divider

[zerkinanastya]

Thank you for the solution! But I’m struggling to understand why we should chop off the left and right digits, couldn’t we just compare then and if they are equal return True? Please if someone could explain, I would be much grateful!

[rehatsingh]

Damm. I created a linked list and rebuilt the number in reverse and then compared it. And that too in C. 😭😭

[TheElementFive]

A rather contrived problem. Not a fan. Great explanation though!

[metawrld_dream]

Do you know how to fix this code to work with 0’s on both ends, say you end up with 0110, I think the computer will ignore the first 0 so it’ll try to match 1 and 0 which will return false, idk if this happens on python but i’m doing it in java

[ChangKaiHua]

class Solution:
def isPalindrome(self, x: int) -> bool:
list1=[]
import math
y = [a for a in str(x)]
for i in range(0, math.floor(len(y)/2)+1):
if y[i]!=y[-i-1]:
list1.append(1)
else:
list1.append(0)
if 1 in list1:
return False
return True

[bennychristiyan]

s = str(x)
return s == s[::-1]

This code is more effective and very simple to understand. Hope it was useful

[gumogumo]

class Solution:
def isPalindrome(self, x: int) -> bool:
return str(x) == str(x)[::-1]

[KumarMukand-dmiw]

Easiest Way Ever :
def isPalindrome(self, x):

num = x
reverse = 0
while(x>0):
lastdigit = x%10
reverse = reverse*10+lastdigit
x = x//10

return True if reverse == num else False

[Xuze]

This solution will fail if the number contains "0" in the middle of a number right? Like 1000021

[prabinlamsal]

Will this solution work for this testcase?

1000021

Because, once we remove the first and the last letter and proceed to the next step,
we have
000021 which is evaluated as 21

[sangpark]

why do you assume 21/10 rounds down to 2?