Finding Reverse of a Number | Java | Detailed Explanation

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[@wheresisgod2295]

beautiful explanation...better than other KZbinrs! thanks much

[@mithank2666]

Thank you so much, RESPECT from karnataka

[@Anikets_Art_Gallery]

Extremely good explanation it became like crystal clear for me......

[@BTechComputerScience]

Glad to hear!

[@payalbhuyan9409]

thank god i watched ur video i understood the logic so well thank u so much

[@sulethagopi2898]

Thanks for clear explanation...

[@aarulmozhinandhini1540]

After a Long Gap u are back....thanks for the video bro🤩

[@BTechComputerScience]

Thanks for being supportive😊

[@sandhiyamanju7470]

The way you are teaching is awesome.🎉

[@puneethn9104]

One of the best explanation 🙏

[@techbaap999]

Awesome explanation brother. We need more videos from you.

[@shivu2012]

Awesome bro.. how did you learn this deeper....Keep up this work....

[@aisumolda6794]

Great explanation!

[@shanthinij8328]

Sir upload Java full concept video. It is very much useful for us.

[@SindhuVenkatachalam]

You are just amazing 👍✨

[@lillytheresa551]

Ur videos is very helpful

[@Wasim2294]

Nice explanation 👍

[@rajeshkanna3853]

But if user enter 100 , we get output 1 and not as 001....how can we solve this?

[@animeshkumar7950]

superb bro

[@ashwiniashwaniperka8482]

Nice explanation

[@gujjerajesh2492]

nice explanation

[@l.d.vgamer9345]

After very long time

[@rekhareddy8432]

Please teach from reverse a string

[@rekalit8192]

Thanks

[@abhiram9849]

Bro please explain string programs also

[@HeavendeepMunjal]

love you bhai

[@ravi54217]

public class Playground { public static void main(String[] args) { final var number = -102; System.out.println(reverse0(number)); System.out.println(reverse1(number)); System.out.println(reverse2(number)); System.out.println(reverse3(number)); } // Will not work for negative numbers private static int reverse0(final int number) { if (number >= 0 && number 0) { int lastDigit = input % 10; reverse = reverse * power + lastDigit; input /= 10; } return reverse; } private static int reverse1(final int number) { if (number >= 0 && number 0) { int lastDigit = input % 10; reverse = reverse * power + lastDigit; input /= 10; } return isNegative ? reverse * -1 : reverse; } private static int reverse2(final int number) { if (number >= 0 && number = 0 && number

[@luckybhaiyt-3095]

Anna nuvu thop anna thurrum anna

[@zhawu1361]

Class reverse { Public static void main (string[]args){ Int n=1234,r; While(n>0) { r=n%10; System.out.print (r); r=n/10; } } }

[@BTechComputerScience]

Here we are not generating the reverse number, we r just printing it. In coding contest, usually we are asked to write a function that returns the reverse number, here we cannot return since we didn't generate a reverse number. So learn accordingly 😊

[@coctamizhan19]

Soo... Palindrome number and reverse a given number are same?

[@sukanshigarg6051]

if on reversed number is same as the original number then it is indeed a palindrome number

[@payalbhuyan9409]

for palindrome number the logic is same , but the in the 2nd part u need to compare the input number with the rev number and write if and else condition.

[@rm.121]

The above method of reversing a number won't work if the last digit os number is 0 . e.g: 5430 Can you please tell me how to achieve that

[@payalbhuyan9409]

actually when u reverse the number then the 0 comes in front which is why it dont show up but it is working

[@manojkumar-if9xu]

please make a series in pattern printing program and one solution with explaination of this pattern 1 2 3 4 5 10 9 8 7 6 11 12 13 14 15 20 19 18 17 16

[@zhawu1361]

For my prospective you can also use looping In your code ...

[@amanmahato3897]

Good But it would be better if you've done it neatly.

[@BTechComputerScience]

Yes I do create a mess 😅

[@priya_world]

@@BTechComputerScience yes that's why little bit confusing it but great explanation sir

[@venkatvenkat6357]

Bro simple ah reverse for loop laye potrulame why this much explanation. For(int i=array.length-1;i>=0;i--){ }

[@BTechComputerScience]

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