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Each time I read an element of the array from the input, I directly assign it to the array at the speific index after shifting. d left rotations equal (n-d) right rotations. The array has a size of n, so when the index (i+n-d) is equal to n, the index actually goes back to the starting index. I use mod to get the correct index where I want to assign the element when the index is larger than n. I'm not an english native speaker. Hope you will understand what I explain. :)

I see a problem here. You've assumed that we can sort the array at the time of initialization itself. Though it's good, but IMHO, the real test is to sort an already existing array with space constraints.

I'm failing to see where it actually does a left rotate of an existing array.

What I'm seeing is (from assumption of ...)
array[0..n] is empty initially.
getArray[0..n] stores input from stdin.
store rotated result into array[0..n].

The existing array is getArray. That array is still in an unrotated order. The array[0..n] isn't considered the exiting array as it is empty.

Challenge to you: get rid of getArray array and store input into array[] array. Rotate that array and then print the end result of that array.

This code is actually quite efficient. So for the following values, n=5 (number of elements in the array), d=2 (number of rotations)and assuming i goes from 0-4.
(i + n - d)%n with the iteration of i from 0-4:
(0+5-2)%5=3
(1+5-2)%5=4
.
.
.
(4+5-2)%5=2
So essentially, you are changing the locations of the elements in the array. How the "%" works is that it takes the remainder. So for example, 1%7=1 since 1 cannot be divided by 7 which means 1 is the remainder.

That's not the aim of the exercise... You have to fill the array and after shift!!! In this way is easy and you are not resolving the left shift problem.

For me, it's just an alternative way. There's not only one way to think for the same problem whatever the solution is good or bad. I shared my solution because I hadn't seen anyone think the same way and I thought it was good to learn from people's comments on my solution.

I like your solution, elegant and efficient. It took my two days for mine and even now has timeout issues when testing. I have good envy now. Well done!

This solution will not work when rotation is greater than array size, meaning d > n. Java returns negative result if you perform % operation on a negative number. And that negative number will throw an

java.lang.ArrayIndexOutOfBoundsException

Eg. n = 4, d = 7

Starting with i=0,
a[(i+n-d)%n] will give
a[(0 + 4 - 7)%4] => a[(0 + (-3))%4] => a[(-3)%4] => a[-3]

It's the far superior solution. Why shift all the elements in the array individually, when you can just shift the index to achieve the same result? It's a total waste of processing to shift the entire array. What if the array was 1,000,000,000 elements long, and you had to shift multiple times? In an interview they would probably expect you to do it this way, with an explanation of your logic.

In my case, they were expecting the function that returns a vector with all the elements rotated any number of times. So, the best way is to take the number of shiftings and move the index that number of times, and do the shift only one time.

Have you tried running this? You actually run into an issue where you just propagate the first value in the array across all other indices in the array using your method.

The second for loop moves the value at index 0, the 1, over to index 1, overwriting the value 2 at index 1, and then this 1 gets written into all the other indices as a result.

Solid answer.
I would like to point out to people that this does require double the memory of the intial answer as this requires a second array to store all the properly ordered array.
If you're on a memory tight situation, this would not be a favorable solution.

To solve this, I've shifted the elements in the array in O(n) time and O(n) memory (vs O(2n)). Not as easy on the eyes, though :(

functionprocessData(input){// get `n` and `d` from inputconstlines=input.split('\n');constfirstLine=lines[0].split(' ').map(Number);constn=firstLine[0];constd=firstLine[1];// process each linelines.slice(1,lines.length).forEach(line=>{// no need to shift in these casesif(n===1||d===n){console.info(line);}else{// shift digitsconsta=line.split(' ').map(Number);letlastLastItem=null;letcount=0;leti=0;while(count<n){i++;conststart=i;letj=start;do{count++;letlastItem=lastLastItem;lastLastItem=a[j];a[j]=lastItem;j=shiftLeft(n,d,j);}while(j!==start);a[start]=lastLastItem;}console.info(a.reduce((acc,value)=>{returnacc+' '+value;}));}});}/** * @param {Number} n total number of elements * @param {Number} d number of shifts to left * @param {Number} i index to begin shifting from * @returns {Number} new index after shifting to left */functionshiftLeft(n,d,i){return(n-d+i)%n;}

I'm ignoring the processing time on input string manipulation and such. This examples assumes they gave us an existing array to manipulate.

Hey! Could you include comments in your code please? I use C and dont kow other language as of now but would like to learn from your solution O(n) is preety nice

Revisited this with an easier-on-the-eyes solution that finishes in O(n) time with O(n) memory.

functionmain(){// ...loopUntilAllSwapped(a,d);console.log(a.join(' ')+'\n');}/** * If a.length can be divided by d evenly, swapSeries will end * where it started without swapping all numbers. * This function ensures that all numbers are swapped. * * Variables: * a: array of numbers * d: number of left-rotations to perform * c: count of numbers swapped * i: index to start swapSeries() from */functionloopUntilAllSwapped(a,d){letc=0;leti=0;while(c<a.length){c+=swapSeries(a,i,d);i++;}}/** * Swaps elements in an array in-place. * * Variables: * a: array of numbers * i: index to start with * d: number of left-rotations to perform * c: count of numbers swapped, returned to loopUntilAllSwapped() * iS: index source (index of number to be moved) * iD: index destination (index iS will be moved to) * q: a queue that acts as a temporary buffer for numbers as they * move from source to destination * * Algorithm: * 1. Find index destination (iD) of a[iS] after d rotations to * left * 2. Place destination in temporary buffer (q). * 3. Replace destination with index source (iS) value (a[iS]). * 4. Repeat until we end where we started (iS === i) */functionswapSeries(a,i,d){letc=0;letiS=i;letq=[a[iS]];do{letiD=h(a.length,d,iS);q.push(a[iD]);a[iD]=q.shift();iS=iD;c++;}while(iS!==i);returnc;}/** * A constant-time formula to find index destination of i * after d rotations to left. */functionh(n,d,i){return((n-d%n)%n+i)%n;}

YOU don't need second vector for this. Just store the 0th index of a vector in some temp variable and when the whole array gets shifted by one left rotation ,store it at the n-1th index. Repeat the process till d becomes zero.

One drawback i can see here is that you are actually holding the IO operation while your code is getting executed. This will not affect for an input of 100, but think about an input of millions

Given an array of n integers and a number, d, perform d left rotations on the array.
So this means that the array is already given. Then we have to perform left rotation on that. Your solution may produce the required results, but as others have pointed out, I think that is not what has been asked in the solution. We have been given a complete array, and on that we have to perform the rotation.

your code is very small and still work fine,
will you explain me how [(i+n-d)%n] is working here. please
actually why you are using % here thats making me confuse

% (modulus) gives you the remainder of division.so 4%5=4,5%5=0 likewise 6%5=1, it helps you to stay inside the limit i.e lesser than the value n (i.e till n-1) eg:if n=5 it goes from 0 to 4.try this simple mod in calc or put it in a loop you will understand.we are getting the input value to the rotated index thats why (-d).if right rotation +d.

exactly...it's better to follow up with "Reversal algorithm". Follows O(n) time complexity without the usage of any temp storage. Hence saves storage and time both

At the very least, you would have to allocate memory to store one integer. The question then becomes what you consider to be important: memory, or run-time, or something in between. If your primary focus is memory, then your solution is just to move EVERYTHING in the array one unit over, multiplied by the number of iterations. (You can mitigate this somewhat by right-shifting if the number of iterations is more than 1/2 the size of the array.) If your focus is time, then just transfer everything to another array, and back again, giving ~ O(2n). A decent in-between (which is what I went with) is to move only the number of elements out that you need to move (max 1/2 n) to another array, shift everything else, then re-insert the moved elements.

"A left rotation operation on an array of size n shifts each of the array's elements d unit to the left."

This isn't a correct solution. It's merely masking itself to be a solution. It's taking input and stuffing it pre-ordered into a new array. You should be doing a left rotation on an existing array.

Not sure what the n-d is to move outside the loop. Thanks for the feedback.

I'm trying to figure the site out to do a test for employment. I keep getting time out so I have been experimenting to see what is causing the times outs. I don't want to fail an employment related challenge b/c of a site timeout.

I find I have to write code that is less and less readable to try and get around the timeout issues.

your time complexity : O(n + n)
your space complexity : n

you can optimize it a little bit with :

time complexity : n + d
space complexity : d

public static void main(String[] args) {
Scanner c = new Scanner(System.in);
int n = c.nextInt();
int d = c.nextInt();
int skipPart[] = new int[d];
int j =0;
for(int i = 0; i < n; i++) {
if(i >= d) {
System.out.print(c.nextInt() + " ");
} else {
skipPart[j] = c.nextInt();
j++;
}
}
for(int i = 0; i < skipPart.length; i++) {
System.out.print(skipPart[i] + " ");
}
}

Got it. I'm guessing being judged in a challenge your way of thinking is better -- do the challenge int he fewest steps possible. I really do not think of optimzing the stdin/stdout as they are just mechanism for giving you the data needed and demonstrating the algorithm is correct. I'm guessing if company is viewing my code, they will think with your mindset.

There are a lot of people solving the problem this way. Reread the problem statement and you'll see that this doesn't do what was asked. There only prints out the input in the order expected after a left rotation d times. The real problem is left rotating an existing array and then print the final content of the array. So what you need to do is read in your input and then store it into an unsorted, unrotated, etc array. Left shift that array, and then print it out from array[0] to array [n].

That's what I was thinking, too. The easiest solution to the expected output isn't necessarily solving the stated problem. But, that's also the nature of the problems. Perhaps a better problem would be one that simply passed in an array and you had to modify said array in place. That would at least ensure the right problem is being solved.

it is just awesome. I want to ask how you think this way, while everybody else is concentrating on different point, your solution is really different and cool :) My understanding is you approach the problem right at the beginning, the only issue is the coming up with this formula "(i+n-d)%n" good thinking akueisara39

intmain(){/* Enter your code here. Read input from STDIN. Print output to STDOUT */inti,d,k,n,temp,j;cin>>n;inta[n];for(i=0;i<n;i++)cin>>a[i];cin>>d;for(i=0;i<n;i++)cout<<" "<<a[(i+(n-d))%n];return0;}

It can be optimized in terms of space if you don't store the whole array, but only the rotated elements. If the elements are 1 2 3 ... 100000, and the number of left rotations is 2, you can save the first two elements (1 and 2) and directly print the rest as soon as you read them. Finally, print the saved ones. The difference in space is from 100000 ints to 2 ints.

no no, if you have 1000 rotations in an array of 30, you have 1000/30 complete rotations (rotations of the whole array) and 1000%30 effective left rotations. So you store only the first 1000%30 elements. Sorry for the initial incomplete explanation, my fault.

How come you reached to this logic array[(i+n-d)%n], i understood what you did here but, having hard time to understand how did you get to this formula, does it comes with expreince or some mathematical practice i am lacking.
Thanks

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int d = in.nextInt();
int[] array = new int[n];
int c=0,i,temp;
for(i=0; i

I was thining on the same lines after I got the timeout error. But didn't reach anywhere, I was still using the same number of For loops some how! Thanks a lot!

suppose if i enter value of n=5;d=2;
then it will accept five numbers not more than that but if i wil not hit the Enter key then it will have been accepting the many values i will hav to take only 5 values not more than that

¿This works for you? In my analysis, this does a right rotation instead of a left rotation. Isn't it? You should implement the logic for translate right to left rotations or to invert the direction.

Hi akueisara39 : i appreciate your code as it appears to be out of box solution . but i feel it throws an exception when the number of shift operations is greater than the size(d>n) of the array so just added few lines of code to it .

int[] array = new int[n];
if(d>n) d=d%n;
for(int i=0; i
array[(i+n-d)%n] = scan.nextInt();
}

Here's my solution in Kotlin. I did it using circular LinkedList.
data class Node(val value: Int, var nextNode: Node?)

fun main(args: Array) {
val (n, d) = readLine()!!.split(" ").map(String::toInt)
val list = readLine()!!.split(" ").map(String::toInt)
var head = Node(list[0], null)
var current = head
var rotatedHead = head
for (i in 1 until n) {
val node = Node(list[i], null)
current.nextNode = node
current = node
if (i == d % n)
rotatedHead = node
}
current.nextNode = head

current = rotatedHead
repeat(n) {
print(current.value.toString() + " ")
current = current.nextNode!!
}

I liked your approach.But I would like to pin point one thing here as you have been already given an array and N number of rotation.ON that array you have to perform operations.You are changing it while reading from console.
please have a look at this and let me know if this can be optimized.

static int[] leftRotation(int[] a, int d) {
// Complete this function

int[] b= new int[a.length];
for(int iLoop=0;iLoop<a.length;iLoop++){
b[iLoop]=a[iLoop];
}
for(int iLoop=0;iLoop<a.length;iLoop++){
a[iLoop]=b[(iLoop+d)%a.length];
}
return a;
}
But Really good solution though.

hey, your logic is cool . but i dont understand it . like i am not able to figure out how it works (in the logical sense ) . why do you add n in(i+n-d)? It would be really nice if you explain . thank you

Just calculate the effective number of rotations d % n and then simply change the list accordingly. Simple Pythonic solution with O(1) complexity:

def rotate(a, r):
l = a[r:] + a[:r]
return l
if __name__ == '__main__':
nd = input().split()
n = int(nd[0])
d = int(nd[1])
a = list(map(int, input().rstrip().split()))
r = d % n
print(*rotate(a,r))

## Left Rotation

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This is my solution in Java

can i get an explanation,pls?

Each time I read an element of the array from the input, I directly assign it to the array at the speific index after shifting. d left rotations equal (n-d) right rotations. The array has a size of n, so when the index (i+n-d) is equal to n, the index actually goes back to the starting index. I use mod to get the correct index where I want to assign the element when the index is larger than n. I'm not an english native speaker. Hope you will understand what I explain. :)

thank you so much.very helpful.

This is Great

I see a problem here. You've assumed that we can sort the array at the time of initialization itself. Though it's good, but IMHO, the real test is to sort an already existing array with space constraints.

Good thinking! I guess it could be handled with another array logically. My two cents.

I like you code some much.PLEASE can you add some explainatioin.I will be very glad

It just generates an array first from Stdin, then illustrates the way to left-rotate an EXISTING array.

I'm failing to see where it actually does a left rotate of an existing array.

What I'm seeing is (from assumption of ...) array[0..n] is empty initially. getArray[0..n] stores input from stdin. store rotated result into array[0..n].

The existing array is getArray. That array is still in an unrotated order. The array[0..n] isn't considered the exiting array as it is empty.

Challenge to you: get rid of getArray array and store input into array[] array. Rotate that array and then print the end result of that array.

This code is actually quite efficient. So for the following values, n=5 (number of elements in the array), d=2 (number of rotations)and assuming i goes from 0-4. (i + n - d)%n with the iteration of i from 0-4: (0+5-2)%5=3 (1+5-2)%5=4 . . . (4+5-2)%5=2 So essentially, you are changing the locations of the elements in the array. How the "%" works is that it takes the remainder. So for example, 1%7=1 since 1 cannot be divided by 7 which means 1 is the remainder.

that's great, can you provide any mathematical illustration to arrive (i+n-d)%n ?

That's not the aim of the exercise... You have to fill the array and after shift!!! In this way is easy and you are not resolving the left shift problem.

For me, it's just an alternative way. There's not only one way to think for the same problem whatever the solution is good or bad. I shared my solution because I hadn't seen anyone think the same way and I thought it was good to learn from people's comments on my solution.

A good solution by you!

I like your solution, elegant and efficient. It took my two days for mine and even now has timeout issues when testing. I have good envy now. Well done!

I really like your solution as you have reduce the time complexity to O(n^2) while is very efficeint compare to my program which has a T(n) = O(n^3)

How the complexity is quadratic? It is O(n), right?

This solution will not work when rotation is greater than array size, meaning d > n. Java returns negative result if you perform % operation on a negative number. And that negative number will throw an

Eg. n = 4, d = 7

Starting with i=0, a[(i+n-d)%n] will give a[(0 + 4 - 7)%4] => a[(0 + (-3))%4] => a[(-3)%4] => a[-3]

Range of a is from 0 to n. Hence the error

for this case you can check for d>n and d = d%n then d will be in range of 0 to n and the above solution will work perfectly :)

this is the simplest and most efficient solution. I was struggling Time out error with mine.

It's the far superior solution. Why shift all the elements in the array individually, when you can just shift the index to achieve the same result? It's a total waste of processing to shift the entire array. What if the array was 1,000,000,000 elements long, and you had to shift multiple times? In an interview they would probably expect you to do it this way, with an explanation of your logic.

I agree. If they wanted something to shift elements of an array, they should have asked for a function to do so.

In my case, they were expecting the function that returns a vector with all the elements rotated any number of times. So, the best way is to take the number of shiftings and move the index that number of times, and do the shift only one time.

Have you tried running this? You actually run into an issue where you just propagate the first value in the array across all other indices in the array using your method.

The second for loop moves the value at index 0, the 1, over to index 1, overwriting the value 2 at index 1, and then this 1 gets written into all the other indices as a result.

Solid answer. I would like to point out to people that this does require double the memory of the intial answer as this requires a second array to store all the properly ordered array. If you're on a memory tight situation, this would not be a favorable solution.

To solve this, I've shifted the elements in the array in O(n) time and O(n) memory (vs O(2n)). Not as easy on the eyes, though :(

I'm ignoring the processing time on input string manipulation and such. This examples assumes they gave us an existing array to manipulate.

Hey! Could you include comments in your code please? I use C and dont kow other language as of now but would like to learn from your solution O(n) is preety nice

i think your solutions is O(n^2) for the worst case.

It looks like it, because of the nested loops. But it'll break out of all loops after

`n`

loops.Revisited this with an easier-on-the-eyes solution that finishes in O(n) time with O(n) memory.

YOU don't need second vector for this. Just store the 0th index of a vector in some temp variable and when the whole array gets shifted by one left rotation ,store it at the n-1th index. Repeat the process till d becomes zero.

In my opinion, the 3rd for-loop is unnecessary.. Instead it should be merged with the 2nd one.

This should be called circle array,haha

Doesn't this use extra space?

for me is: (d + i) % n

what does it mean...????

%.....?

One drawback i can see here is that you are actually holding the IO operation while your code is getting executed. This will not affect for an input of 100, but think about an input of millions

nice!! different way of thinking! you cahnged the way of thinging..

getArray[i] = scan.next(); this is an error.It cannot be converted into string.

getArray[i] = scan.nextInt(); this is correct statement.

Very nice. I didn't even think to assign the integers to their "new" positions in the array as they are read in.

Concise code and Wonderful explanation.This should go to Editorial.

why can't we do the left rotation. like you said d left rotation is equal to n-d right rotation.what is problem in that...

Given an array of n integers and a number, d, perform d left rotations on the array. So this means that the array is already given. Then we have to perform left rotation on that. Your solution may produce the required results, but as others have pointed out, I think that is not what has been asked in the solution. We have been given a complete array, and on that we have to perform the rotation.

thanks

very convenient solution

pretty rad :),loved it.

your code is very small and still work fine, will you explain me how [(i+n-d)%n] is working here. please actually why you are using % here thats making me confuse

% (modulus) gives you the remainder of division.so 4%5=4,5%5=0 likewise 6%5=1, it helps you to stay inside the limit i.e lesser than the value n (i.e till n-1) eg:if n=5 it goes from 0 to 4.try this simple mod in calc or put it in a loop you will understand.we are getting the input value to the rotated index thats why (-d).if right rotation +d.

It's a great solution but how would someone come up with the idea to use a modulus in this case?

This is my C# sharp code.

THANK YOU SO MUCH FOR THIS SOLUTION!!!!!!!!!!!!!!!!!!!! :')

I've been searching for both right & left shift formulae!!!

This code will fail the test. This is the fix:

for (int i = 0; i < n; i++) { Console.Write(array[i]+" "); }

Console.WriteLine();

Very slow. & it will not work out 1,000,000,000 records

exactly...it's better to follow up with "Reversal algorithm". Follows O(n) time complexity without the usage of any temp storage. Hence saves storage and time both

Not the desired approach but it gets the job done though

static void Main(String[] args) { string[] token=Console.ReadLine().Split(' '); int n=Convert.ToInt32(token[0]); int k=Convert.ToInt32(token[1]); string[] arr1=Console.ReadLine().Split(' '); int [] arr=Array.ConvertAll(arr1,int.Parse);

Great solution ajaycarnet. I compared yours with my own and yours is much faster.

My C# code:

Did anyone consider if we cant alloc a new array to store sorted array, only operate the old array to left rotation?

At the very least, you would have to allocate memory to store one integer. The question then becomes what you consider to be important: memory, or run-time, or something in between. If your primary focus is memory, then your solution is just to move EVERYTHING in the array one unit over, multiplied by the number of iterations. (You can mitigate this somewhat by right-shifting if the number of iterations is more than 1/2 the size of the array.) If your focus is time, then just transfer everything to another array, and back again, giving ~ O(2n). A decent in-between (which is what I went with) is to move only the number of elements out that you need to move (max 1/2 n) to another array, shift everything else, then re-insert the moved elements.

It is not the best solution but it requires only one array:

Yes. But I think it will slower than

Correct me if I am wrong.

My in-place O(n) solution. No allocation required

Why has everyone ignored this?

Is it possible to do the same in C# using only one array (as one array must hold the splitted values and other to place them accordingly) ?

It is not the best solution but it requires only one array:

厉害，佩服

Thumbs up, for sharp thinking. Complicated my solution before stumbling on the same short solution myself :D

"A left rotation operation on an array of size n shifts each of the array's elements d unit to the left."

This isn't a correct solution. It's merely masking itself to be a solution. It's taking input and stuffing it pre-ordered into a new array. You should be doing a left rotation on an existing array.

move that n-d outside the loop its always same. other then that your solution is best.

Not sure what the n-d is to move outside the loop. Thanks for the feedback.

I'm trying to figure the site out to do a test for employment. I keep getting time out so I have been experimenting to see what is causing the times outs. I don't want to fail an employment related challenge b/c of a site timeout.

I find I have to write code that is less and less readable to try and get around the timeout issues.

your time complexity : O(n + n) your space complexity : n

you can optimize it a little bit with :

time complexity : n + d space complexity : d

public static void main(String[] args) { Scanner c = new Scanner(System.in); int n = c.nextInt(); int d = c.nextInt(); int skipPart[] = new int[d]; int j =0; for(int i = 0; i < n; i++) { if(i >= d) { System.out.print(c.nextInt() + " "); } else { skipPart[j] = c.nextInt(); j++; } } for(int i = 0; i < skipPart.length; i++) { System.out.print(skipPart[i] + " "); } }

Got it. I'm guessing being judged in a challenge your way of thinking is better -- do the challenge int he fewest steps possible. I really do not think of optimzing the stdin/stdout as they are just mechanism for giving you the data needed and demonstrating the algorithm is correct. I'm guessing if company is viewing my code, they will think with your mindset.

There are a lot of people solving the problem this way. Reread the problem statement and you'll see that this doesn't do what was asked. There only prints out the input in the order expected after a left rotation d times. The real problem is left rotating an existing array and then print the final content of the array. So what you need to do is read in your input and then store it into an unsorted, unrotated, etc array. Left shift that array, and then print it out from array[0] to array [n].

thanks. Got it.

That's what I was thinking, too. The easiest solution to the

expected outputisn't necessarily solving the stated problem. But, that's also the nature of the problems. Perhaps a better problem would be one that simply passed in an array and you had to modify said array in place. That would at least ensure the right problem is being solved.I like this I wish I'd came to this algorithm myself ...

Similarly:

Great Solution!

b=a[d : ]

b.append(a[:d])

print(*b)

A really good solution. I was unable to figure it out mathematically XD.

very clever, thanks!

What a great hack, akueisara39! Given that the defaulted code that read the elements into an array, I saw myself forced to use additional space.

Can someone explain how you came about the formula (i+n-d)? I am not able to understand how that formula is derived?

good logic akueisara. i used similar logic like "(i+n-d)%n" in circular linked list.

it is just awesome. I want to ask how you think this way, while everybody else is concentrating on different point, your solution is really different and cool :) My understanding is you approach the problem right at the beginning, the only issue is the coming up with this formula "(i+n-d)%n" good thinking akueisara39

It is a greate solution.

I think something similar, but your solution is better.

pls let me knw wats wrng in this?

cin >> d; should be after cin >> n;

you need to use vector instead of array for memory allocation at runtime else if you are using array they you have to use dynamic array.

vector leftRotation(vector a, int d) { int temp; for(int j=0;j

//whats wrong one test case not working(timeout error)

write a[(i+d)%n] in place of a[(i + (n - d)) % n]

My C solution was similar:

Hand down! Best solution for me!

Nailed it!

It can be optimized in terms of space if you don't store the whole array, but only the rotated elements. If the elements are 1 2 3 ... 100000, and the number of left rotations is 2, you can save the first two elements (1 and 2) and directly print the rest as soon as you read them. Finally, print the saved ones. The difference in space is from 100000 ints to 2 ints.

what if I have thousand rotation??then I have to store thousands of int?

no no, if you have 1000 rotations in an array of 30, you have 1000/30 complete rotations (rotations of the whole array) and 1000%30 effective left rotations. So you store only the first 1000%30 elements. Sorry for the initial incomplete explanation, my fault.

appreciated!!bt still don't you think it will require alot memory??even if u will use 1000%30

I think on the place to store those elements it will b more efficient to jst show those elements.

How come you reached to this logic array[(i+n-d)%n], i understood what you did here but, having hard time to understand how did you get to this formula, does it comes with expreince or some mathematical practice i am lacking. Thanks

Pure Magic !!!

The logic does it in one loop. Brilliant.

Very creative solution!

3 errors is generating

public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int d = in.nextInt(); int[] array = new int[n]; int c=0,i,temp; for(i=0; i

} }

I used the same logic but timeout error is coming for the last case.

what is the issue if i write this way.

Awesome , this really saves time .Thank You..!!

I was thining on the same lines after I got the timeout error. But didn't reach anywhere, I was still using the same number of For loops some how! Thanks a lot!

So without using an extra array u did it. Basically u dont have an input array.. Great

is the array[(i+n-d)%n] is some standard algo?

suppose if i enter value of n=5;d=2; then it will accept five numbers not more than that but if i wil not hit the Enter key then it will have been accepting the many values i will hav to take only 5 values not more than that

Awesome...

amazing!!

¿This works for you? In my analysis, this does a right rotation instead of a left rotation. Isn't it? You should implement the logic for translate right to left rotations or to invert the direction.

Nice

Hi akueisara39 : i appreciate your code as it appears to be out of box solution . but i feel it throws an exception when the number of shift operations is greater than the size(d>n) of the array so just added few lines of code to it .

int[] array = new int[n]; if(d>n) d=d%n; for(int i=0; i array[(i+n-d)%n] = scan.nextInt();

}

Instead of rotating each element we can just cut the string upto that index and paste(append) it at the end of the string.

A solution for python would be

beautiful solution!!!

but this will not work when d > len(a)

Correct, I added this line first to account for cases when d > len(a).

true, but the constraints say 1 <= d <= n :)

Here's my solution in Kotlin. I did it using circular LinkedList. data class Node(val value: Int, var nextNode: Node?)

fun main(args: Array) { val (n, d) = readLine()!!.split(" ").map(String::toInt) val list = readLine()!!.split(" ").map(String::toInt) var head = Node(list[0], null) var current = head var rotatedHead = head for (i in 1 until n) { val node = Node(list[i], null) current.nextNode = node current = node if (i == d % n) rotatedHead = node } current.nextNode = head

}

genius!

I liked your approach.But I would like to pin point one thing here as you have been already given an array and N number of rotation.ON that array you have to perform operations.You are changing it while reading from console. please have a look at this and let me know if this can be optimized.

static int[] leftRotation(int[] a, int d) { // Complete this function

how did you do that it is amazing...how can you think like that.....

nice explanation genius

Brilliant

very cool way of doing!

Your approach to this problem was great and one correction if d > n then in those cases you should recompute d as d = d%n;

cool! i mean hats off

If the number of rotations is greater than the number of elements, than this can still throw an IndexOutOfBounds as (i+n-d) is negative.

To solve this one can use a true modulus function like Math.floorMod(i-d, n) which rather than returning the remainder returns a modulus. https://stackoverflow.com/questions/5385024/mod-in-java-produces-negative-numbers

Yes this code is useful.

wow bro you are great

How did you come up with this idea? I was puzzled with this task for two days and still didn't succeed in it...

hey, your logic is cool . but i dont understand it . like i am not able to figure out how it works (in the logical sense ) . why do you add n in(i+n-d)? It would be really nice if you explain . thank you

what a piece of outstanding code........well done

Nice....

Just calculate the effective number of rotations d % n and then simply change the list accordingly. Simple Pythonic solution with O(1) complexity:

Thank you, GOOD EXPLANATION