Happydoodle Pa it is totally ok. In development casting and converting data structures is normal but doing mathematical calculations might impress the interviewer since all people would think of transforming the integer to string but few would think of using modulo and division
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!!!!
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.
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 ?
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.
Forgive me guys, I was naive. This is how I do it now. Also, Koushik used a 64-bit long, the problem says that you are not supposed to use 64 bit integer at all. class Solution { public int reverse(int x) { int sign = 1; if (x < 0) { x = -x; sign = -1; } int sum = 0; int prev = 0; while (x > 0) { int t = x % 10; x = x / 10; sum = (sum * 10) + t; if ((sum - t) / 10 != prev) { return 0; } prev = sum; } return sum * sign; } }
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
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.
Mayby it will be simplier? StringBuilder reversedNumber = new StringBuilder(Integer.toString(number)); return Integer.parseInt(reversedNumber.reverse().toString());
I think, this could also be achieved by reversing string. String str="54321"; int reveserInt=Integer.parseInt(new StringBuilder(str).reverse().toString());
Thanks for helping the developer community. Keep it up. I think there is a small correction in your code but its not a trivial ...if(reversed > Long.MAX_VALUE || reversed < Long.MIN_VALUE) {....
May be not. We need to be able to convert long reversed into integer right.. But I feel the last return statement is missing.. Since it is just a psudo code.. interviewer will be okay I guess..
It can be used if you cast it to an int while returning. Otherwise you can just start off with an int in the first place, but this requires a bit of recoding inside the while loop.
Can anyone explain me why the reversed is multiplied by 10? Let's say we found the last digit by using modulus (%) of 10 then why do we have to multiply it by 10 again???
The below solution which gives o(n) time complexity : StringBuilder sb=new StringBuilder(); while (n>=10) { int last=n%10; sb.append(last); n/=10; } if(sb!=null){ sb.append(n); } System.out.println(sb.toString()); }
this will not handle negative numbers, -123 will become -123 and this is way more memory intensive than the proposed answer. Usually you trade memory for speed or speed for memory. This doesn't trade at all, it increases memory but gains no speed.
leetcode Message : 1032 / 1032 test cases passed. Runtime: 1 ms, faster than 100.00% of Java online submissions for Reverse Integer. Memory Usage: 33.7 MB, less than 11.66% of Java online submissions for Reverse Integer. class Solution { public int reverse(int input) { long result=0; boolean flag=false; if(input0) { result=input%10+result*10; input=input/10; if(result> Integer.MAX_VALUE || result
Yes, it will work. It will return 1, which is the expected answer. If you thought that is should return 001 then that is not the case, as that is not a valid integer.
if(reversed > Integer.MAX_VALUE || reversed < Integer.MIN_VALUE) Since Integer roll over happens, this did not work. The following code works fine for me. public int reverseInt(int value) { int input = value; int reversed = 0; while(input != 0) { reversed = reversed * 10 + input % 10; input = input / 10; if((reversed < 0 && value > 0) || (reversed > 0 && value < 0) ) { // throw error or return 0 return 0; } } return reversed; }
i will surely work but your will make lotta operations which increment the O(longN) of the algorithms so i think you can find a better solution i have this one from anothe guy and already understood this concept public static int reverseInteger() { int x = 54321; int result = 0; int prev = 0; while (x != 0) { System.out.println("x" + x); int cur = x % 10; System.out.println("cur" + cur); x /= 10; System.out.println("cur / " + x); result = result * 10 + cur; System.out.println("result" + cur); if ((result - cur) / 10 != prev) return 0; prev = result; } return result; } it works like a charm
Let's say our input is a very large number: 2147483647 If we reverse this we get: 7463847412 Guess what will be the problem here? Do you see it? The reverse number far exceeds the Integer.MAX_VALUE. And then the internet will come crashing down, just kidding of course. Hope you see it.
@@authoritycamper class Solution { public int reverse(int x) { long sum=0; int temp=x; int r; while(x!=0){ r=x%10; sum=(sum*10)+r; x=x/10; } if(sum>Integer.MAX_VALUE || sum
this will not work with negative numbers, "-123" will become "321-" and also this is extremely much slower and more memory intensive than the proposed solution in the video. If you would answer that in an interview, you'd probably not get the job, and also this does not return an integer, but a stringbuilder. an integer in java takes up 4 bytes of memory while a String in java takes up at least "the number of chars" * 2 + 32 + rounded off to the nearest number devisable by 8. So the integer 123 takes up 4 bytes. The string 123 takes up 3 * 2 + 32 = 38 and then we round it off to the nearest number that can be divided by 8, and that is 40... so that string will take up at least 40 bytes. That is 10 times as much memory.
@@thomasandolf7365, Yes you're right, but in realtime scenarios I have never seen a need revetse integer in my 4-5 of coding job. Why would people will as such kind of questions in interview?
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I did the same but my code was way longer. Your solution is the best. Just add "
return reversed;"
after while loop ends
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 :)
thats kinda like cheating 😊 ... never convert one data structure to another
Dev Super I am just curious why it is cheating ? And why it is a bad sign during an interview? Could you elaborate a bit more detail please ?
Happydoodle Pa it is totally ok. In development casting and converting data structures is normal but doing mathematical calculations might impress the interviewer since all people would think of transforming the integer to string but few would think of using modulo and division
did you get the job?
@@Your-Average-Gym-Bro I wonder the same... Cuz as I was practicing the same problem, I thought of using the same approach!
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.
.
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I still do!!!
very enjoyable video. Having moments like this reminds me that intellectual satisfaction is highest form of pleasure. thank you sir.
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!!!!
"exceptions are for exceptional situations" - I'm stealing that line. :)
That was awesome!!!
Thank you very much Koushik, today I learned something new, really appreciate!
These videos are easy to follow and understand. Thanks, Koushik!
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.
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 ?
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.
Forgive me guys, I was naive. This is how I do it now. Also, Koushik used a 64-bit long, the problem says that you are not supposed to use 64 bit integer at all.
class Solution {
public int reverse(int x) {
int sign = 1;
if (x < 0) {
x = -x;
sign = -1;
}
int sum = 0;
int prev = 0;
while (x > 0) {
int t = x % 10;
x = x / 10;
sum = (sum * 10) + t;
if ((sum - t) / 10 != prev) {
return 0;
}
prev = sum;
}
return sum * sign;
}
}
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
What about this proposition :
public Integer reverse(Integer myInt){
String s = ""+myInt;
String s2 = "";
for(int i=0;i
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.
I wonder if I can use an array to inverse this array.
Mayby it will be simplier?
StringBuilder reversedNumber = new StringBuilder(Integer.toString(number));
return Integer.parseInt(reversedNumber.reverse().toString());
I think, this could also be achieved by reversing string. String str="54321";
int reveserInt=Integer.parseInt(new StringBuilder(str).reverse().toString());
For String
public static void main(String[] args) {
String a = "aba";
String reverse = "";
for (int i = a.length() - 1 ; i >= 0 ; i--)
reverse = reverse + a.charAt(i);
System.out.println(reverse);
}
Your videos have been so helpful and clear! You rock
We appreciate you ..!!! keep solving more leetcode problems..
great tutorial....pls more of problem coding for interviews
Integer revNum = new Integer (num); String RevString = RevNum.toString ();
String temp = "";
For (int I = RevString.length() -1, I--, I
Subscribed ... very enjoyable
What happened to spring security topics? I was waiting for next video... Please continue on spring security also.
Yes, working on the next Spring Security video right now!
@@Java.Brains Thanks Man
@@Java.Brains nice video
err: lossy conversion from long to int.. did you ëven try it on Leecode or just ran it on your computer and lived happily ever after?
dude use (int)reversed.
@@prag3022 what about negative nos ?
Thank u very much sir for leetcode keep solving more problems
take care of your health kaushik . You don't look Ok to me . Health is very important especially for guys like us who keep sitting for long hours
dont judge a book by its cover
Thanks for helping the developer community. Keep it up. I think there is a small correction in your code but its not a trivial ...if(reversed > Long.MAX_VALUE || reversed < Long.MIN_VALUE) {....
May be not. We need to be able to convert long reversed into integer right.. But I feel the last return statement is missing..
Since it is just a psudo code.. interviewer will be okay I guess..
@@bbvkishore something like this is needed at the end:
return (int)reversed;
public static int revInteger(int quotent) {
int reverse = 0;
while (true) {
if(quotent == 0) break;
reverse = reverse * 10 + quotent % 10;
quotent /= 10;
}
return reverse;
}
Thanks sir , i really got help for your video.
what would be the problem of returning BigInteger? and we dont check if we pass the min/max value
wonderful explanation
what about negative numbers
hello ! i converted the int into a string using the string class and then i reversed it by charAt() method. my question is this way correct ?
I solved it with a long variable and it passed all cases in leetcode. Is it an incorrect or novice approach?
With a 32 bit signed unsigned integer long cannot be used which is a 64 bit data type
It can be used if you cast it to an int while returning. Otherwise you can just start off with an int in the first place, but this requires a bit of recoding inside the while loop.
why are we not returning the int value of the reversed number? The method expects an int return type.
great ! very similar to Armstrong number problem
Can anyone explain me why the reversed is multiplied by 10? Let's say we found the last digit by using modulus (%) of 10 then why do we have to multiply it by 10 again???
Excellent as always 👍
Simple logic it works:
StringBuilder sb=new StringBuilder();
while (n>=10) {
int last=n%10;
sb.append(last);
n/=10;
}
System.out.println(sb.toString());
Thank u so much Sir!!
You shouldn't use long type. It is clearly mentioned that the input is a 32 bit signed integer value. It's just a hack you are applying
Spring security please
Sir, I was thinking, would it be wrong if one just convert the integer to string and reverse the string using string builder
It would be very inefficient. You can mention that in the interview, but the interviewer will likely not accept that as the final solution
@@Java.Brains ok, thanks for pointing out the inefficiency. I was actually thinking about that
Can someone explain once again what happened in reverse *10 line
i didn't understand the code at the line -- input/=10; shouldn't it be input = input/10; ?
why aren't we checking for negative values?
Why do you reverse decimal digits not bits?
Sirrrr❤😍...
The below solution which gives o(n) time complexity :
StringBuilder sb=new StringBuilder();
while (n>=10) {
int last=n%10;
sb.append(last);
n/=10;
}
if(sb!=null){
sb.append(n);
}
System.out.println(sb.toString());
}
this will not handle negative numbers, -123 will become -123 and this is way more memory intensive than the proposed answer. Usually you trade memory for speed or speed for memory. This doesn't trade at all, it increases memory but gains no speed.
What happens if input is 0?
leetcode Message :
1032 / 1032 test cases passed.
Runtime: 1 ms, faster than 100.00% of Java online submissions for Reverse Integer.
Memory Usage: 33.7 MB, less than 11.66% of Java online submissions for Reverse Integer.
class Solution {
public int reverse(int input) {
long result=0;
boolean flag=false;
if(input0) {
result=input%10+result*10;
input=input/10;
if(result> Integer.MAX_VALUE || result
input=~input+1; can you explain the use of ~ please
It does not work for number 10. The result should be 01 but instead it is giving 1
Will this code run for n=100 or any zero digit number?
Because 0*10 is zero got nothing
Yes, it will work. It will return 1, which is the expected answer. If you thought that is should return 001 then that is not the case, as that is not a valid integer.
It is showing error
thank you very much sir for this video
if(reversed > Integer.MAX_VALUE || reversed < Integer.MIN_VALUE)
Since Integer roll over happens, this did not work. The following code works fine for me.
public int reverseInt(int value) {
int input = value;
int reversed = 0;
while(input != 0) {
reversed = reversed * 10 + input % 10;
input = input / 10;
if((reversed < 0 && value > 0) || (reversed > 0 && value < 0) ) {
// throw error or return 0
return 0;
}
}
return reversed;
}
Is it ok if I convert int to stringBuilder by String.valueOf and then reverse that string n convert it back to Integer by Integer.getValue?
i will surely work but your will make lotta operations which increment the O(longN) of the algorithms so i think you can find a better solution
i have this one from anothe guy and already understood this concept
public static int reverseInteger() {
int x = 54321;
int result = 0;
int prev = 0;
while (x != 0) {
System.out.println("x" + x);
int cur = x % 10;
System.out.println("cur" + cur);
x /= 10;
System.out.println("cur / " + x);
result = result * 10 + cur;
System.out.println("result" + cur);
if ((result - cur) / 10 != prev) return 0;
prev = result;
}
return result;
}
it works like a charm
I have much towards you brah..
054321 which is not reversed . Giving result 73722
I didn't understand the integer.max or integer.min part.
Let's say our input is a very large number: 2147483647 If we reverse this we get: 7463847412 Guess what will be the problem here? Do you see it? The reverse number far exceeds the Integer.MAX_VALUE. And then the internet will come crashing down, just kidding of course. Hope you see it.
@@authoritycamper thanks dude! great explanation. god bless Indian people!
@@authoritycamper class Solution {
public int reverse(int x) {
long sum=0;
int temp=x;
int r;
while(x!=0){
r=x%10;
sum=(sum*10)+r;
x=x/10;
}
if(sum>Integer.MAX_VALUE || sum
Why the if-statement is inside the while-loop?
Because you're checking for the limit as you're adding numbers into the reversed variable
Nug
Your code doesn't handle cases as:
100 => 001 which is wrong. Instead, it should be 1
So, please handle such cases too!
The return input is a 'long' number, when 0*10 it's 0 not 00 so it is 1 not 001
StringBuilder sb = new StringBuilder(String.valueOf(number)).reverse();
System.out.println(sb.toString());
this will not work with negative numbers, "-123" will become "321-" and also this is extremely much slower and more memory intensive than the proposed solution in the video. If you would answer that in an interview, you'd probably not get the job, and also this does not return an integer, but a stringbuilder.
an integer in java takes up 4 bytes of memory while a String in java takes up at least "the number of chars" * 2 + 32 + rounded off to the nearest number devisable by 8.
So the integer 123 takes up 4 bytes. The string 123 takes up 3 * 2 + 32 = 38 and then we round it off to the nearest number that can be divided by 8, and that is 40... so that string will take up at least 40 bytes.
That is 10 times as much memory.
@@thomasandolf7365 very clear explanation for how to evaluate the memory allocation. Thanks.
Indeed, it's inefficient and slower. Thanks for the explanation!
When ever interviewer ask without reverse () then it would happen
@@thomasandolf7365, Yes you're right, but in realtime scenarios I have never seen a need revetse integer in my 4-5 of coding job. Why would people will as such kind of questions in interview?
Not reminder it is remainder I guess before being a software engineer you have to learn English well
SKIP this video if you're trying to learn the solution quick. Too much unnecessary jargon
Hey... Slow down a bit... useless explaination
Hey! go to settings and reduce video pace... Useless!
the logic which you gave would fail, when we have this input value 1534236469 which would not return 0