Hitscotty With my solution I used modular arithmetic to calculate the position of the each element and placed them as I read from input.
for(int i = 0; i < lengthOfArray; i++){ int newLocation = (i + (lengthOfArray - shiftAmount)) % lengthOfArray; a[newLocation] = in.nextInt(); }
antrikshverma2 Neat code , thanks Hitscotty !!
manishdas 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
darwin57721 what is the starting value of your i? (i dont know ruby). d=2, n = 10. Because if it is 0, it would be (0+2)%10 = 2. What am I getting wrong?
manishdas The starting value of the i is 0. Looks like correct calculation to me. What result are you expecting?
darwin57721 ha, yeah i wasn't understanding right! I made it this way, that's why I was confused. rotated[(n+i-d)%n] = a[i]. Which is analogous to yours, but calculating the index in destination. Yours is more clear I think. Thanks!
Usernamer89 are you a mathematician? because i came out with a bit similar answer
dev1ender me too
jambekardhanash1 why do we need i? Can you please explain?
manishdas Based on current index (i), you need to generate new index. For example: let's say array = [1, 2, 3, 4] and k = 2, then after 2 left rotation it should be [3, 4, 1, 2] => 3 4 1 2 (space separated string output)
Now let's walk through my algorithm:
# Initial assignments: # array = [1, 2, 3, 4] # length_of_array = array.length = 4 # no_of_left_rotation = k = 2 # new_arr = Arra.new(length_of_array) # new_arr: [nil, nil, nil, nil] # NOTE: # length_of_array.times do |i| # is equivalent to # for(i = 0; i < length_of_array; i++) # Algorithm to calculate new index and update new array for each index (i): # new_index = (i + no_of_left_rotation) % length_of_array # new_arr[i] = array[new_index] # LOOP1: # i = 0 # new_index = (0 + 2) % 4 = 2 # new_arr[i = 0] = array[new_index = 2] = 3 # new_arr: [3, nil, nil, nil] # LOOP2: # i = 1 # new_index = (1 + 2) % 4 = 3 # new_arr[i = 1] = array[new_index = 3] = 4 # new_arr: [3, 4, nil, nil] # LOOP3: # i = 2 # new_index = (2 + 2) % 4 = 0 # new_arr[i = 2] = array[new_index = 0] = 1 # new_arr: [3, 4, 1, nil] # LOOP4: # i = 3 # new_index = (3 + 2) % 4 = 1 # new_arr[i = 3] = array[new_index = 1] = 2 # new_arr: [3, 4, 1, 2] # After final loop our new roated array is [3, 4, 1, 2] # You can return the output: # new_arr.join(' ') => 3 4 1 2
Hope that's clear.
glezin Crystal.
sarikaaa nice algo
bhavikpatel576 Great explanation! Thanks.
MobilityWins I am trying to understand this, but this is the first time I have seen value assignments that involve a val= val= anotherVal
I am not quite understanding how that is supposed to work, also what is "nil" and its purpose for an array
jenish9599 Cool....
manavmehra96 Excellent Description!
mzancanella if the length of the array is = 3 then it seems it won't work.
p_callebat new_index = (i + no_of_left_rotation) % length_of_array;
seems incorrect. You will see the problem if you test, for example [1,2,3,4,5] and k = 2 .
I guess would be better:
new_index = (i + (lengthOfArray - no_of_left_rotation)) % lengthOfArray;
sfelde21 why?
sruthi_1401227 y ? its not working
supertrens Seems like this algorith only works for small number because when the array is big enough due to long looping period u will have system "timeout"
2017A7PS0931G I was facing the same problem.I gave several attempts but the issue couldn't be solved. Can you please tell me how to define a loop for a set of array with so many elements as such... :)
pawel_jozkow In java8 the problem was in String; You have to use more efficient StringBuilder instead; And of couse use only one loop to iterate over array;
here is my code snippet:
StringBuilder output = new StringBuilder(); for(int i = 0; i<n; i++) { b[i] = a[(i+k) % n]; output = output.append(b[i]).append(" "); }
__raviraj__ include
using namespace std; int main() { long int a[1000000],n,d,i,f; cin>>n>>d; for(i=0;i>a[i];
for(int j=0;j<d;j++) { f=a[0]; for(i=0;i<n;i++) { a[i]=a[i+1]; } a[n-1]=f; } for(i=0;i<n;i++) cout<<a[i]<<" ";} //this is my code and im getting time out could u please solve
nsaikaly12 its because your solution is O(n^2) with the inner loop. Try and find an O(xn) solution and iterate over the whole array only once.
__raviraj__ [deleted]__raviraj__ i didnt get u
joelvanpatten I was facing the same issue in PHP. My solution worked for 9 out of 10 test cases but timed out on one of them every time. You have to re-write the solution to be less memory intensive. In my case I was using array_shift() which re-indexes the arrays, so for large arrays it uses too much memory. My solution was to use array_reverse() and then array_pop() instead, because those methods don't re-index.
haroon_1993 This Does not suits for all entries if you make the rotation to more than 4 its fails
waseefakhtar Brilliant explanation!
mchandravadhana good explanation..working fine.
[deleted] nice algo...easy to understand ...thank u soo much
to_vishalsachde1 Adbsolute geeky, How did it appeared to your mind ?
lakshman1055 How to think like this ? Once the code is there I know its easy to understand.I want to know how did you know to use modulous and how did you come up thinking that logic ?
thanks in advance.
amrelbehairy88 Have you ever heard about Data Structure ? because if you do , you would probably heard about circular array.
I was able to solve the question because I'm knew about circular arrays , we use % + size of array to create a cirural array , then all you need to do is to complete the puzzle to solve the problem.
check this video, https://www.youtube.com/watch?v=okr-XE8yTO8&t=773s
tcjcastillo This is super helpful, thanks so much for sharing!
KivenL96 cool
hendradedisupri1 really run ?
sasuke_10 Great solution. Any tips on how to know if you need to use modulus in your algorithm? I solved this problem using 2 for loops...
priteshbonde Nice explanation. helped me a lot. Thanks You
anilkumaryadav91 Awesome Explanation....Tq
anisharya16 thankyou so much, it helped a lot. but can you please tell how did you think about the new index position. what did you think?
zodiac_797 Great explanation!Thanks
g_panwar great way of explaining.big thank!
Narendr_Gadde Nice algo..
polaki Good Solution.
rakeshreddy5566 simple is peace
return arr[d:] +arr[0:d]
igor_atf Greatest explanation so far. Thanks!
itshiteshverma Well Explained !!
morrisontech i is a variable used to iterate through the loop, it generally represents the index of the array that is being referenced on a particular iteration of the loop.
abhash24oct Your code if for right rotation, and the explanation gave you right answer as the size was 4 and k =2 , so no matters you do left/right you will get same. For left it will be int newLoc= (n +(i-k))%n;
ang_7jd Array = {1,2,3,4,5} d = 4 Dosen't work for me
My Code :
for (int i = 0; i < a.Length; i++) { position = Math.Abs((i + (a.Length - d))% a.Length); newArray[position] = a[i]; }
ayushasthana025 nice algorithm manish
himaang I guess the logic fails if the input is as follows: 5-n 6-d 1 2 3 4 5
manishdas [deleted]phaniyelisi According to given constraint : 1 <= d <=n, d will not exceed n
hacknroll Even if there was no constraint, you could just do: k % n, and then apply the same logic.
zenmasterchris 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.
cmshiyas007 thats what me too thinking of..was wondering why the logic writte here was arranging the array on read...
andritogv That's exactly the point of the exercise. You have to rotate an already existing array.
avi_roychow Correct!
Turings_Ghost I noticed that right away. If the point was to produce printed output, then this is fine (and a lot of analysis works backward from output). But, as stated, one is supposed to shift an array, so this missed it.
aubreylolandt this could easily be modified though by creating another array of the same size:
vector b(n); for(int i = 0; i < n; i++) { b[i] = a[(i+k) % n]; } return b;
buzzaldrin 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.
denis_ariel (i + shift) % lenght Should be enough
robertgbjones Except that describes a right shift, and specification says a left shift. You might consider left shift to be negative shift, in which case you are correct mathematically, but I'd feel much more comfortable keeping the whole calculation in positive territory.
chrislucas modular arithmetic is cool. I solved that way too
for idx in range(0, _size): indexes[(idx - shift + _size) % _size] = _list[idx]
marwinko19 Can you please explain how that works?
jericogantuangc1 Hello, where did this solution from? what should I study to be able to come up with solutions like this?
jattilah Looks a lot like my C# solution:
static int[] Rotate(int[] a, int n) { n %= a.Length; var ret = new int[a.Length]; for(int i = 0; i < a.Length; ++i) { ret[i] = a[(i + n) % a.Length]; } return ret; }
rodrigoramirez93 Perfect solution. Thanks for sharing
Aggasti Nice solution. You dont need:
n %= a.Length;
purshottamV This line usefull when n >= a.Length
ronaldabellano Nice
sidnext2none I agree modular arithmetic is awesome. But, simple list slicing as follows solves too ;)
def rotLeft(a, d): return a[d:]+a[:d]
jarvian what if newLocation becomes negative
diegobonales then you send it to the end of the array
Turings_Ghost The modulus operation always returns positive. If, as in Java, it really does remainder, rather than the mathematical modulus, it can return negative. So, depends on which language.
marakhakim What if lengthOfArray < shiftAmount? I think you should use abs value
digibluh there is a constraint that says there wont be negatives so it should be fine. if it wasn't given then use abs.
jattilah You deal with lengthOfArray < shiftAmount by using:
shiftAmount = shiftAmount % lengthOfArray;
If the array length is 4, and you're shifting 6, then you really just want to shift 2.
The constraints say that shiftAmount will always be >= 1, so you don't have to worry about negative numbers.
iAmBipinPaul Aweosme !!!
vovchuck_bogdan pretty simple in js:
a.splice(k).concat(a.slice(0, k)).join(' ')
cpbx12 [deleted]HackDaddy even simpler is a.splice(k).concat(a).join(' ')
ajkhatibi what does a represent?
flyinghorsedanc1 The array of initial values.
rags1981 [deleted]amezolma Did something similar in C#..
using System; using System.Collections.Generic; using System.IO; using System.Linq; class Solution { static string rotate(int rot, int[] arr) { string left = string.Join( " ", arr.Take(rot).ToArray() ); string right = string.Join( " ", arr.Skip(rot).ToArray() ); return right + ' ' + left; } static void Main(String[] args) { string[] tokens_n = Console.ReadLine().Split(' '); int n = Convert.ToInt32(tokens_n[0]); int k = Convert.ToInt32(tokens_n[1]); string[] a_temp = Console.ReadLine().Split(' '); int[] a = Array.ConvertAll(a_temp,Int32.Parse); // rotate and return as string string result = Solution.rotate(k, a); // print result Console.WriteLine(result); } }
merkman Or you can one line it with LINQ
Console.Write(string.Join(" ", a.Skip(k).Concat(a.Take(k)).ToArray()));
rahulbhansali While it is definitely elegant looking with a single line of code, how many times will this iterate over the array when performing 'skip', 'take' and 'concating' them? In other words, what's the complexity of this algorithm?
mihir7759 O(n)
avi_roychow Can any one please tell me why the below code is timing out for large data set:
for(int j=0;j<k;j++) { for(int current=n-1;current>=0;current--) { if(current!=0) { if(temp!=0) { a[current-1]= a[current-1]+temp; temp= a[current-1]-temp; a[current-1]=a[current-1]-temp; } else { temp=a[current-1]; a[current-1]=a[current];//for the first time } } else//when current reaches the first element { a[n-1]=temp; } } } Console.WriteLine(string.Join(" ",a));
rishabh10 mine is also a brute force approach but it worked check it out if it helps you
import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static int[] arrayLeftRotation(int[] a, int n, int k) { int temp,i,j; for(i=0;i<k;i++){ temp=a[0]; for(j=1;j<n;j++){ a[j-1]=a[j]; } a[n-1]=temp; } return a; } 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(); } int[] output = new int[n]; output = arrayLeftRotation(a, n, k); for(int i = 0; i < n; i++) System.out.print(output[i] + " "); System.out.println(); } }
pateldeep18 int main(){ int n; int k; int temp1, temp2; scanf("%d %d",&n,&k); int *a = malloc(sizeof(int) * n); for(int a_i = 0; a_i < n; a_i++){ scanf("%d",&a[a_i]); } k = k %n; for(int a_i = 0; a_i < k; a_i++){ temp1 = a[0]; for(int i = 1; i < n; i++){ a[i-1] = a[i]; } a[n-1] = temp1; } for(int a_i = 0; a_i < n; a_i++){ printf("%d ", a[a_i]); } return 0; }
my code is the same as yours but i still time in test case 8, why is that?
not_nigel You're not wrong but this solution is inefficient. You're solving it in O(((n-1) * k) + 2n). The solution below is in O(2n).
private static void solution(int size, int shift, int[] arr) { int count = 0; for (int i = shift; i < size; i++) { System.out.print(arr[i]); System.out.print(" "); count++; } count = 0; for (int i = size - shift; i < size; i++) { System.out.print(arr[count]); if (i != size - 1) System.out.print(" "); count++; } }
ash_jo4444 I got a timeout error for TC#8 and #9 for the same logic in Python :(
Muthukumar_T i got time out for tc#8 in c why??????
val3rian Iterate over array only once
russelljuma No loops. Just split and reconnect. def rotLeft(a, d): b = [] b = a[d:len(a)] + a[0:d] return b
gdahis Because it is O(n*k), if you have a big n and a big k, it could timeout. See if you can think of an algorithm that would visit each array element only once and make it o(n). Also, is there any optimization you can make? For example: if k is bigger than n, then you don't need to do k rotations you just need to do k % n rotations and k will be much smaller, smaller than n. Example:
[ 1, 2, 3, 4, 5 ]
K=2, K=7=(1*5)+2, K=12=(2*5)+2, they are all equivant, leading the array to be:
[3, 4, 5, 1, 2]
mihir7759 [deleted]
Nitin304 My Solution :
public static int[] arrayLeftRotation(int[] a, int n, int k) { int[] b = new int[n]; for(int i=0;i<n-k;i++){ b[i] = a[k+i]; } int l = 0; for(int i=n-k;i<n;i++){ b[i] = a[l++]; } return b; }
hatem_ali64 [deleted]amit_feb06 with one for loop i have subitted the code
donald_thai Nice one! Didn't even think of that
thefonso in an actual interview they will ask you not to use splice or slice. had that happen ti me.
akimy pretty elegant solution
amandakoster [deleted]
_e_popov indeed, forgot that
endgoes through the end of a sequence, so here is my solutionfunction rotLeft(a, d) { const index = d % a.length; return [...a.slice(index), ...a.slice(0, index)]; }
vladimirvuj function rotLeft(a, d) { return [...a.slice(d), ...a.slice(0, d)] }
visbs58 nice one
douglascvas [deleted]jenish9599 i think that the new location = ((i + Shift)) % lenght of array
sidch95 godlike solution.
parthshorey this doesnt work if you shift by 3.
Paul_Denton 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
rishabh10 wouldn't then space complexity be O(n)
jigarsnaik Very nice.
wjmillon @hitscotty what brought you to modular arithmetic for this solution?
chatpras what if shiftAmount is greater than length of array
gurdeeps158 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..
[DELETED] [deleted]rishabh10 why do you use this approach if in a realtime env. this wouldn't be correct as you are just given the function to complete.
praveshjamgade rishabh10 can u please explain ??
dtoxecko It is not following the instructions. Sorry.
anushadandu Great Logic!! Thank you so much!!
rajeswariramasa1 can u pls expalin what is shiftAmount???
theshishodia Hey @Hitscotty! What will be the newLocation if you wish to do a right rotation?
sabujp right rotation : b[i] = a[(i-k)%n]
alexzaitsev 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)
ViNii 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.
sfelde21 I understand how this works, but how did you think of it?
ciceromoura13 [deleted]
ro63rtD Worked like a charm! Thanks!
milindmehtamsc 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(); }
AshrafShekh Superb man.....
codextj That's nice but its kinda cheating :)
mihir7759 It's not cheating exactly. Using the same method you can even rotate the array, instead of printing the array just give the values of the array to a new array.
codextj I was nitpicking, I thought of the same soln at first but then changed my mind;
As the question was GIVEN an array ..so if this was an interview there is this constraint that your array is already populated with the elements.
btw r u 14 ? its great to see my young indian frnds indulging in programing
pul_k3333 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) + " "); }
gptran96 very good, thanks!!!
96rishu_nidhi can u please elaborate some more about your code as i dont have much knowledge about modular maths
mihir7759 Have I used modulo in my code? I really forgot what have I written if you could show me the code then I can elaborate :)
greengalaxy2016 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.
venkataramanago1 what is in.nextInt() here.
c00301223 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.
rinshukr3 what is the use of in.nextInt(),will u plz explain me.
t_hacker 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.
moazashraf007 Brilliant
riyaz_rayyan07 what is in.nextInt() which language is that did you create another scanner object of in can you be more specific?
golzar_mohammad Very celever! thanks.
rakeshreddy5566 reverse it
arr1=a[0:d]
arr2=a[d:]
return arr2+arr1
qzhang63 Python 3
It is way easier if you choose python since you can play with indices.
def array_left_rotation(a, n, k): alist = list(a) b = alist[k:]+alist[:k] return b
kevinmathis08 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=' ')
Lord_krishna is that scala?
domar It's Python 3
aniket_vartak you dont need to pass n to your function, right..
michael_bubb I agree - I ended up not using 'n' (Python):
def left_shift(n,k,a): for _ in range(k): a.append(a.pop(0)) print(*a)
unitraxx Obviously this solves the problem, but is a terrible solution. Pop is an O(N) operation. So your solution becomes O(K*N). This should be done in O(N) total time complexity. You do have the space requirement of O(1) correct. All the standard solutions have a O(N) space complexity.
asfaltboy True. However, it becomes an elegant solution if we use collections.dequeue instead of list. Double ended queues have a popleft method, which is an O(1) operation:
def array_left_rotation(a, n, k): for _ in range(k): a.append(a.popleft()) return a
More info: https://wiki.python.org/moin/TimeComplexity#collections.deque https://docs.python.org/3/library/collections.html#collections.deque
AffineStructure They have rotate built into the deque
def array_left_rotation(a, n, k): a = deque(a) for i in range(k): a.rotate(-1) return a
josegabriel_st You have O(n) in:
a = deque(a)
In order to avoid this, you should use a deque from the beginning like:
from collections import deque def array_left_rotation(a, n, k): a.rotate(-k) n, k = map(int, input().strip().split(' ')) a = deque(map(int, input().strip().split(' '))) array_left_rotation(a, n, k); print(*a, sep=' ')
And array_left_rotation only takes O(k) instead of O(n).
Note that a is pased by reference, so there is no need to return anything, but this could be an issue for some user cases, for this particular problem, it works.
navi_srm def array_left_rotation(a, n, k): return (a[k:] + a[:k])
ansimionescu k -> k%nfrostje k%n handles any range of shifts, but the problem specifies n < k so you can just use k in this case
loks30 This logic fails when k>n.
madhuvasu7 for that, do:
k = k%n
chrislucas Nice, only one line was added
marwinko19 Perfection. Short, simple, & clean.
ahmettortumlu25 hmm nice....
aliie62 It's so smart! well done:)
toaa_kama2354 Can you please explain how (a[k:]+a[:k]) is working?
jae_duk_seo Damn this is super impressive.
todor_90 [deleted]domar 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.kevinmathis08 Yes both qzangs and my answer is correct. In python index slicing (indices[start:stop:step]), works like so...
We will begin with the index specified at start and traverse to the next index by our step amount (i.e. if step = 2, then we jump over every other element, if step = 3, we jump over 2 elements at a time). If step is not specified it is defaulted to 1. We continue steping from our start point until we come to or exeed our stop point. We do NOT get the stop point, it simply represents the point at which we stop.
I love Python :)
domar Ok, so it's exclusive by default in Python, interesting.
JXSan Haha that's what I did, got to love Python !
pandasagar thanks. makes more sense
RVK5233 Thanks ....... Its working and simple to learn
crassus88 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]
MaxNRG def array_left_rotation(a, n, k): return a[k:] + a[:k]
burakozdemir32 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.
sdasara95 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%nnew = a[rem:]+a[:rem]return new
jhaaditya14 I am getting request timeout for test case 8... anyone with same problem?? or anyone knows the solution??
vabanagas 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
nie_cyril I agree that it timesout if the algorithm is not efficient enough. Instead of relying on rudimental loops, try using a linked list which simplifies its complexity significantly to O(dn).
chenyu_zhu86 Python index slicing makes this trivial :D
def array_left_rotation(a, n, k): return a[k:] + a[:k]
varellap That's the simplest solution in Python!
darkOverLord I did it this way, in Java
public static int[] arrayLeftRotation(int[] a, int n, int k) { if (k >= n) { k = k % n; } if (k == 0) return a; int[] temp = new int[n]; for (int i = 0; i < n; i++) { if (i + k < n) { temp[i] = a[i + k]; } else { temp[i] = a[(i + k) - n]; } } return temp; }
vinaysh instead of if-else statement temp[i] = a[(i+k)%n]; would be enough.
Also this solution would take up extra memory(for temp).
shortcut2alireza Would you mind sharing your solution that does it in-place? Thanks
jacob0306 in case you need it! hope it help!
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[((n- (k%n))+a_i)%n] = in.nextInt(); } for(int i = 0; i < n; i++) System.out.print(a[i] + " "); System.out.println(); } }
ngokli This is a good reason to start looping on the first element of the input, rather than the output.
But the problem statement does say you are "Given an array" (not a stream).
hackerrank_com23 Here's my in-place function implementation:
public static int[] arrayLeftRotation(int[] a, int n, int k) { // Rotate in-place int[] temp = new int[k]; System.arraycopy(a, 0, temp, 0, k); System.arraycopy(a, k, a, 0, n - k); System.arraycopy(temp, 0, a, n - k, k); return a; }
Angi29 [deleted]Angi29 idem, just a little variation : kifs
static int[] rotLeft(int[] a, int d) { int length = a.length; int[] b = new int[length]; System.arraycopy(a, d, b, 0, length - d); System.arraycopy(a, 0, b, length - d, d); return b; }
denis631 How is this in-place when you allocate a new array of size k.
mohitgautam2k16 [deleted]123shuklaprasha1 why u have used that command of else please explain me
cc_insp 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; } }
yash_97373 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] + " "); }
danyim 2 lines of JS:
while(k-- !== 0) a.push(a.shift()) console.log(a.join(' '));
HeinousTugboat One line of JS, no looping:
console.log(a.concat(a.splice(0,k).join(' ')));
akimy I've the same solution. gj
main = () => { let [integersCount, rotations] = readLine().split(' ').map(Number) let array = readLine().split(' ').map(Number) while(rotations--) array.push(array.shift()) console.log(array.join(' ')) }
noahlbw There is no need to do copying of data. The question is just to print out the values. The following code suffices, I think.
int *b; int b_i; b = a + k; for(b_i = 0; b_i < n; b_i++) { if((int)(b - a) >= n) { b = a; } printf("%d ",*b++); }devesh23 I also thought the same. Even if the rotations are made subsequent we can have two pointers x and y take care of the start of new array after rotations and end of new array after rotations. Then we can print from x to y.(Only in case of multiple query)
marochm I don't think that's a valid assumption. The question is pretty clear about that: "perform left rotations ON THE ARRAY. THEN print THE UPDATED ARRAY as a single line of space-separated integers."
MohammedKadir Not sure about that. The instructions are two fold, and they distinguish between rotations and printing. I think it's expecting the values to physically shift.
"perform left rotations on the array. Then print the updated array as a single line of space-separated integers."
aishwaryagaud17 [deleted]
mmicael In PHP:
$list = array_merge(array_slice($a, $k), array_slice($a, 0, $k)); echo implode(" ", $list);
quliyev_rustam Hi!
You are in right way, but in wrong code.
You need use this:
a, a)-a, 0, $k));
kuttumiah Hi,
For cases including shifts more than the array size this should work.
$actual_shift = $d % count($a); $list = array_merge(array_slice($a, $actual_shift), array_slice($a, 0, $actual_shift));
quliyev_rustam It is not necessary. By the hypothesis of the task shifts cant bo more than array size
kuttumiah Yeah, Thanks for the response. I missed that hypothesis.
In that case this shouldn't this be just fine?
array_merge(array_slice($a, $k), array_slice($a, 0, $k));quliyev_rustam In this test, for "Tester 8", you need to use third argument for array_slice. Otherwise, an error is returned.
This is because for "Tester 8" uses a large array of transmitted values
kuttumiah According to the PHP documentation of
array_slicethird argument denotes the length of the slice.array array_slice ( array $array , int $offset [, int $length = NULL [, bool $preserve_keys = FALSE ]] )If I do not pass third argument then it will be
NULLby default. As a result the slice will be created until the end of the array. I believe there is nothing to do with the volume of the array. Moreover, without passing the third argument like I denoted in my previous comment I am having all the test cases passed.quliyev_rustam Moreover, without passing the third argument like I denoted in my previous comment I am having all the test cases passed.
It's very strange. But, yesterday, when I tested, it didnt work.
Yes, you are right! It have to be like you write!
pavangopal449 #include <math.h> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <assert.h> #include <limits.h> #include <stdbool.h> int main(){ int n; int k,p=0; scanf("%d %d",&n,&k); int *a = malloc(sizeof(int) * n); for(int a_i = 0; a_i < n; a_i++){ scanf("%d",&a[a_i]); } for(int i=0;i<n;i++) { if(i+k>n-1) { printf("%d ",a[p]); p++; } else printf("%d ",a[i+k]); } return 0; }
cbsrk yaa,.. okk got it man !!!!!!!!!!!
yildirim84 for(int i = 0; i < n; i++){ b[i] = a[(i+k)%n]; }
I guess this is the most primitive solution to understand.
Greydon for i in range(k): a.append(a.pop(0))
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