Neat code , thanks Hitscotty !!

hmm.. I'm surprised that worked for you. This one worked for me:

str = ''

length_of_array.times do |i|
  new_index = (i + no_of_left_rotation) % length_of_array
  str += "#{array[new_index]} "
end

puts str.strip

I guess the logic fails if the input is as follows: 5-n 6-d 1 2 3 4 5

The question asks to shift a fully formed array, not to shift elements to their position as they're read in. Start with a fully formed array, then this solution does not work.

I had the same idea! Just find the starting point of the array with the shift and count on from there, taking modulo and the size of the array into account.

(i + shift) % lenght Should be enough

modular arithmetic is cool. I solved that way too

for idx in range(0, _size):
  indexes[(idx - shift + _size) % _size] = _list[idx]

what if newLocation becomes negative

What if lengthOfArray < shiftAmount? I think you should use abs value

Aweosme !!!

pretty simple in js:

a.splice(k).concat(a.slice(0, k)).join(' ')

nice one

i think that the new location = ((i + Shift)) % lenght of array

godlike solution.

this doesnt work if you shift by 3.

Spoiler! You can do it even simpler: rotated[i] = a[(i + k) % n]. Also spoilers should be removed from the discussion or the discussion should only be available after solving. I will complain about this until its changed :P

Very nice.

@hitscotty what brought you to modular arithmetic for this solution?

what if shiftAmount is greater than length of array

your solution is cool but if you have an array as input then you are in trouble bcoz in that case you have space complexity of O(n) as you need an another array to store element in new place.. think..

It is not following the instructions. Sorry.

Great Logic!! Thank you so much!!

can u pls expalin what is shiftAmount???

Hey @Hitscotty! What will be the newLocation if you wish to do a right rotation?

Hey, guys
Here is a solution based on modular arithmetic for the case when k > n:

new_index = (n + i - abs(k-n)) % n

(note: n - abs(k-n) can be collapsed to a single number)

This only work when you read directly from input. But in the question they have asked to do rotation operation only on Array, so your solution cannot be used for this problem.

I understand how this works, but how did you think of it?

Worked like a charm! Thanks!

This will also fail when my shiftAmount = 7 and lengthOfArray = 3, in short lengthOfArray is less than shiftAmount. In this case we can use Math.abs(). for(int a_i=0; a_i < n; a_i++){ int new_index = Math.abs((a_i + (lengthOfArray - shiftAmount))) % lengthOfArray ; a[new_index] = in.nextInt(); }

Superb man.....

That's nice but its kinda cheating :)

Well done! I did it by finding the correct number for the index, rather than the new position of a given number:

for (int i =0; i < n; i++){
    int num = a[((n + k) % n + i) % n];
    System.out.print((num) + " ");
}

very good, thanks!!!

can u please elaborate some more about your code as i dont have much knowledge about modular maths

the requirement is to take an array and left rotate the array d times. Your solution returns the correct result, but takes an integer one at a time.

what is in.nextInt() here.

Thanks for sharing this code it really helpped. I felt the constraints were to be includes by ifstatements but after viewing your code I was able get it. I have a small suggestion, would it improve the code if one were to seperate the (LengthOfArray - ShiftAmount) part into a variable and then reuse it since its kind of a constant value. Once again kudos.

what is the use of in.nextInt(),will u plz explain me.

Neat code, the only possible optimization is extra space used. even for a single rotation on say 1 million elements, the algo is greedy on space and will create 1 million auxilary elements.

Brilliant

what is in.nextInt() which language is that did you create another scanner object of in can you be more specific?

Very celever! thanks.

reverse it

arr1=a[0:d]

arr2=a[d:]

return arr2+arr1

It's easy when you directly read them from system input. Try to make it work on already stored array. That's what problem statement says. It gets tricky and interesting after that to solve it in o(n) without extra memory.

i.e. // Complete roLeft function

My solution

private static int getIncomingIndex(int index, int rotations, int length) {
    if(index < (length - rotations)) {
        return index + rotations;
    }
    return index + rotations - length;
}

// Complete the rotLeft function below.
    static int[] rotLeft(int[] a, int d) { 
           int rotations = d % a.length;

    if(a.length == 0 || a.length == 1 || rotations == 0) {
        return a;
    }

    if( a.length % 2 == 0 && a.length / 2 == rotations) {
        for(int i =0 ; i < a.length / 2 ; i++) {
            swap(a, i, i + rotations);
        }
    } else {
        int count = 0;
        int i = 0;
        while(true) {
            int dstIndex = getIncomingIndex(i, rotations, a.length);
            swap(a, i, dstIndex);
            i = dstIndex;
            count++;
            if(count == a.length - 1) {
                break;
            }
        }
    }
    return a;
}

nice code tq

The part I'm missing here is why use a loop (O(n)). Can't you take the array and find the effective rotation based on the shift amount (using the same modular arithemetic you're doing? (Which is now O(1) since the length of the array is a property)

function rotLeft(a, d) {

  //calculate effective rotation (d % a.length)
  let effectiveRotation = d % a.length;

  // split a at index of effective rotation into left and right
  let leftPortion = a.slice(0, effectiveRotation);
  let rightPortion = a.slice(effectiveRotation); 
	
  // concat left to right
  return rightPortion.concat(leftPortion)
}

vera level..!

Why would you loop for every element when in essence the rotation operation is nothing but just a rearrangement of the array elements in a specified fashion?

helps a lot

Verry Good!

I saw this answer in stackoverflow too.. Please be kind enough to explain this.
Thanks

Tried a different approach

def rotLeft(a, d): return reversed(list(reversed(a[:d])) + list(reversed(a[d:])))

But isn't the whole point that you are not placing them as they come, the array is pre-populated and then rotate it. My solution is O(dn), not sure if there is anything better. Clearly I am not an algorithm guy (anymore)!

for (int i = 0; i < d; i++) { int pop=a[0]; //shift left for (int j = 1; j < a.length; j++) { a[j-1] = a[j]; } //push a[a.length-1]=pop; }

Excellent !!! I am new to problem solving. I had solved it via normal shifting using one for loop and one while loop. How did you arrive at this kind of solution?? Little bit of explanation as what you thought while solving this would help a lot.

Thanks.

How can I do this in C++?

a[newLocation] = in.nextInt();

If the number of rotations are greater than array length (I know it's less than array length which is given in the question, let us assume), then how would this formula change? BTW That's a great way to get the array indices without having to traverse the whole array

Neat Solution. A nice add would be shiftAmount = shiftAmount % lengthofArray; in the cases where shiftAmount exceeds the array length.

Yeah index slicing FTW, here was my 1-liner in definition, lol:

def array_left_rotation(a, n, k):
    return a[k:]+a[:k]

n, k = map(int, input().strip().split(' '))
a = list(map(int, input().strip().split(' ')))
answer = array_left_rotation(a, n, k);
print(*answer, sep=' ')

Pretty smart, but are you sure you are not copying the kth eleement, in ruby it would be:

def array_left_rotation(a, k)
  a[k..-1] + a[0...k]
end

In ruby, ... means excluding the right number.

Haha that's what I did, got to love Python !

@qzhang63 isnt a already a list when it is passed? why convert it to a list a second time?

thanks. makes more sense

Thanks ....... Its working and simple to learn

I think 'k' should be allowed to be greater than 'n'':

def array_left_rotation(a, n, k):
    d = k % n
    ret_arr = a[d:]
    ret_arr.extend(a[:d])
    return ret_arr

def array_left_rotation(a, k):
    d = k % len(a)
    return a[d:] + a[:d]

What about if 'k' is greater than 'n'? You should use modular arithmetic to get actual rotate count.

actual_rotate_count = k % n

Then your solution would work for every k values.

This won't work if the number of rotations is greater than the length of the array. The test cases must have all had d < n I guess This is a modification that I did so it satisfies the above case too.

• def rotLeft(a, d,n):

• rem = d%n
  

• new = a[rem:]+a[:rem]
  

• return new
  

return a[k:] + a[:k]

Great answer, you just need %size_of_list for cases the k is greater than the size of the list

where did the k parameter come from?

This is a very good answer. But if the constraint k

The test case is a large array with a large amount of shits. If your algorithim is not effecient than it will time out.

Array size: 73642 Left shifts: 60581

If u are using javascript, this will work var b = a.splice(0, d); a = a.concat(b);

That's the simplest solution in Python!

instead of if-else statement temp[i] = a[(i+k)%n]; would be enough.

Also this solution would take up extra memory(for temp).

why u have used that command of else please explain me

I used the System.arraycopy() method which was used in the video tutorial. I'm wondering if this solution is more efficient or mine?

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        int a[] = new int[n];
        for(int a_i=0; a_i < n; a_i++){
            a[a_i] = in.nextInt();
        }
        a = leftRotation(n, k, a);
        for (int i=0; i<a.length; i++) {
            System.out.print(a[i]+" ");
        }
    }
    public static int[] leftRotation(int n, int k, int[] a){
        int[] copy = new int[n];
        System.arraycopy(a, k, copy, 0, (n - k));
        System.arraycopy(a, 0, copy, (n - k), (n - (n - k)));
        return copy;
    }
}

Is calculation really required ?

for(int i=k; i < a.length; i++){
    System.out.print(a[i] + " ");
}
for(int i=0; i < k; i++){
    System.out.print(a[i] + " ");
}