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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

## Left Rotation

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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...????

%.....?

It means to divide the numbers and return the remainder

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.

Hey @qianyonan

You have forgot to initialize the empty array i.e. array variable .

that's why little bit hard to understand to newbie.

One more thing , Don't you think you are using extra space bcz you can solve this problem using existing array too.