Next Smallest Palindrome | InterviewBit Math | Logic Explained

Next Smallest Palindrome is a tricky InterviewBit question. I explain the solution to this problem in detail using various examples and guiding through the mess of edge cases. The solution we reach to is quite simple, and perhaps even shockingly simple for a "Hard" question. This video covers the logic part, and the next video covers the code part.

Writeup (Summary + Code):

==================================================

Timestamps:
0:00 Introduction
0:25 Problem Discussion
1:09 Odd Even split
2:08 Odd cases
10:00 Even cases

==================================================

Stuff I use:

NOTE: These are affiliate links.
- Doesn't cost you anything
- Gives me a kickback if you buy this

==================================================

Donate:

==================================================

Follow:
#LeetCode #coding #programming #chaudhary1337

==================================================

[sgcreations]

We need more content and people like you. You explained it so simply. Great bro .

[The-masked-girl-songs]

I wish I can someday meet you in person and thank you for this explanation and I adore you for being the best at explaining!!

[anukulsahu]

u explain really good, 😍, u got my subscription

[yujhengfang]

amazing solution and explanation. sadly I just failed my interview on this question today

[sudhiryadav]

Hi Tanishq, I have doubt regarding the approach to deal with Polindrom string itself, having a larger number of 9's suppose my string is
(78 times 9) in this case, adding 1 will going out the memory exception?

[manpatel]

Such a Nice Explanation...😍 You Stole my subscription😇

[chrishemsworth]

Bro sorry to say this, u are wrong:
according to u if "left < right" mid+1;
but it does not satisfy for many cases like
{19263} => {19291}
here we are not adding 1 to mid number

[AMANKUMAR-twjt]

Well, I should say this was a nice explanation- IIT'R-24

[amrithapatil]

Why does the trick of adding 1 to a palindrome work

[tekbssync]

bro please try to provide code in C++ also
More than 70% viewers are expecting c++ code

[Badalkumar-meok]

This is Java version of the problem

import java.util.* ;
import java.io.*;
public class Solution {

public static boolean isPalindrome(String s){
return s.equals(new
}

public static String add1(String s){
StringBuilder num= new StringBuilder();
int carry=1;

for(int i=s.length()-1; i>=0; i--){
int
num.append(sum%10);
carry=sum/10;
}

if(carry!=0){
num.append(carry);
}
return num.reverse().toString();
}
public static int compare(String left, String right){
int i = 0;
while (i < left.length() && i < right.length()) {
int leftValue =
int rightValue =

if (leftValue != rightValue) {
if (leftValue > rightValue) {
return -1;
} else {
return 1;
}
}

i++;
}

// If the loop completes, it means the common prefix is the same.
// Check the lengths to determine the result.
return Integer.compare(left.length(), right.length());
}
public static String handleOdd(String s){
int mid=s.length()/2;
String left= s.substring(0, mid);
String right=s.substring(mid+1);
if(compare(new StringBuilder(left).reverse().toString(), right)== -1){
return left+s.charAt(mid)+ new
}else {
left=left+s.charAt(mid);
left=add1(left);

return left + new
}
}

public static String handleEven(String s){
int mid=s.length()/2;
String left= s.substring(0, mid);
String right=s.substring(mid);
if(compare(new StringBuilder(left).reverse().toString(), right)== -1){
return left+ new
}else {

left=add1(left);
return left+ new
}
}
public static String solve(String number) {
int length=number.length();
// Write your code here.
if (length == 1 && !number.equals("9")) {
int num = Integer.parseInt(number);
return Integer.toString(num + 1);
}


number=add1(number);
}

String ans= new String();
if(number.length()%2==0){
ans= handleEven(number);
}else{
ans=handleOdd(number);
}

return ans;
}
}