LeetCode #9: Palindrome Number

A step-by-step visualization of #Leetcode question 9: Palindrome Number.

0:00 Problem Description
0:28 Strategy Overview
1:35 Code Walkthrough
4:30 2nd Example

#coding #programming #algorithms #tech #faang #softwareengineer

[SimranMallar]

Add more videos !!! Great explaination

[nihalbhamrah]

great explanation i knew what way to think but didnt know how to implement it

[LDanya]

eee good, спасибо большое = thank you very

[FRAMEDSKATEKREW]

JESUS, dude this video was amazing
I have adhd so its hard to focus on videos or books but I saw your solution on leetcode and saw you attached the video.
I swear that about 6 minutes ago I didn't even know where to begin and now I feel like I could explain this someone else.
Believe me, that is an achievement lol Definitely subbing and I wish you a lot of success. You are definitely special and are going to help a lot of people.

[librahuang]

As a beginner and not an English speaker this explanation is easy for me to understand👍

[husseinhadliye]

your explanation is so good, keep it up🤙

[codecompass]

It wasn't easy you made it sound easy!1! brilliant!!!

[KookyPiranha]

math coding if mod or floor/ceiling function never existed

[jacksonlow]

The explanation was good, even though I approached it using Java.

[alexito-pvrlx]

how to do it into the first way? :)
i cant find info, what is "x = str(x), x == x[::-1]"

[bhaktipatekar]

what is the time and space complexity of second solution?

[Unkonw]

IDK why but when taking input as 121 it shows false even though it is true?

[Richard-yzgy]

why you need to convert x into a string??

[alijohnnaqvi]

#include<bits/stdc++.h>
class Solution {
public:

bool isPalindrome(int x)

{
vector <int> hash;
if (x<0)
return false;
else
{
if (x>0 && x <=9) // single digits are palindrome
return true;
else
{
while (x>9)
{
int y = x%10;
x = (x-y)/10;
hash.push_back(y); // increment count
}
hash.push_back(x);
bool decision = true;
auto itr1 = hash.begin();
auto itr2 = hash.rbegin();
for ( int i=0; i<hash.size()/2; ++i, ++itr1, ++itr2)
{
if (*itr1!= *itr2)
decision = false;
}
return decision;


}

}

}
};
I maintained two pointers, one from start and one from end after splitting separate digits. I know it's not near to optimal. I just started CP. How do you come up with such elegant solutions at first sight?