Reverse an Integer value - LeetCode Interview Coding Challenge [Java Brains]

Interview Question: Reverse an integer in Java

Difficulty level: Easy
Language: Java

Welcome to the Java Brains Java Interview challenge series of videos where we tackle various coding challenges ranging from the beginner level to advanced level - especially in the context of coding interviews.

Any coding problems you would like me to explore? Please let me know in the comments what you think!

#java #interview #javabrains

[DK-scvn]

I did the same but my code was way longer. Your solution is the best. Just add "
return reversed;"
after while loop ends

[nabilelhaouari]

I got this challenge in an interview, I have just converted the integer to string, I reverse the string and convert back to a number :)

[RameenFallschirmjager]

very enjoyable video. Having moments like this reminds me that intellectual satisfaction is highest form of pleasure. thank you sir.

[sirjonaz]

"exceptions are for exceptional situations" - I'm stealing that line. :)

[fufuisgood]

Wow this is the first video I have seen from your channel and it is amazing! You go in deepth and make it so understandable, and then later implement the code! love it!!!!

[dkryeziu]

That was awesome!!!

Thank you very much Koushik, today I learned something new, really appreciate!

[HammamMounir]

What about this proposition :
public Integer reverse(Integer myInt){
String s = ""+myInt;
String s2 = "";

for(int i=0;i<s.length();i++){

}

return Integer.valueOf(s2);
}

[sharathchandrareddy]

Two things to mention here
1. The code did handle the exception cases of integer range boundaries BUT didn't return the "reversed" value outside the loop to handle the happy case
2. I understand that A smaller number in the range of integer might possible cross the integer range while reversing ( Reverse of 12345 is 54321 ) which is bigger than original . Is the min range meant to handle the negative integer numbers ?

[meeradad]

I think the sign of the integer being reversed needs to be handled outside the reversal. So one has to take the absolute value of the number, reverse that absolute value, and then multiply the reversed number with 1 or -1 depending on the sign of the original number. At least in Python, the remainders of -ve numbers do not work in a sign-agnostic way.

[thanga]

We appreciate you ..!!! keep solving more leetcode problems..

[diegoguzman]

These videos are easy to follow and understand. Thanks, Koushik!

[sagarmeena]

great tutorial....pls more of problem coding for interviews

[ShinAkuma]

int a = 54321;
StringBuffer str = new StringBuffer( ""+a );
System.out.println( str.reverse().toString() );

In this solution, Need to check if it's +ve or -ve and add the sign later on accordingly.
This is shorter but requires parsing int to string which is less efficient, but it does the job for lazy people like me. I'm sure the client won't mind waiting 1 extra second for his reversed number.

[abhijitmadchetti]

I think while condition should be replaced with while (input > 9) and after while loop 1 more statement reversed = reversed * 10 + input; for last digit. It will work for input < 10.

[Aaron-yuzo]

I wonder if I can use an array to inverse this array.

[rushikeshr]

I think, this could also be achieved by reversing string. String str="54321";
int

[antonslonkin]

Mayby it will be simplier?
StringBuilder reversedNumber = new
return

[msk]

This problem is also a good candidate for recursion as we are decreasing the input at every iteration

private static long reverse(int input) {
if (input <= 0) {
return reversed;
}
reversed = reversed * 10 + input % 10;
reverse(input / 10);

return reversed;
}

[parshuramrv]

Thank u very much sir for leetcode keep solving more problems

[ivanviveros]

Your videos have been so helpful and clear! You rock