Rotation of the array means that each element is shifted right by one index, and the last element of the array is also moved to the first place

Question

For example, the rotation of array A = [3, 8, 9, 7, 6] is [6, 3, 8, 9, 7]. The goal is to rotate array A K times; that is, each element of A will be shifted to the right by K indexes.

For example, given array A = [3, 8, 9, 7, 6] and K = 3, the function should return [9, 7, 6, 3, 8].

I want this in java. I have tried this.

public static int[] rotation(int[] a,int k) {

    int[] newArray = new int[a.length];
    for(int i = 0 ; i < a.length ; i++) {
        int newPosition = (i + k)%a.length;
        newArray[newPosition] = a[i];
    }
    return newArray;
}

duplicate of http://stackoverflow.com/questions/4457277/algorithm-to-rotate-an-array-in-linear-time — Arjit, Dec 30 '16 at 07:04
What is the problem? What do you expect and what actually happens? — mm759, Dec 30 '16 at 07:07
This is working! Please do some test first and some research on how to print an array in console. — AxelH, Dec 30 '16 at 07:08
Try `Arrays.stream(resultArray).forEach(System.out::println);` — Kishore Bandi, Dec 30 '16 at 07:09

Iryna Prokopenko · Answer 1 · 2019-07-16T09:21:20.357

5

I didnt get what exactly wants topic starter but here my code for this task

class Solution {
public int[] solution(int[] arr, int k) {
    int[] newArr = new int[arr.length];

    if (arr.length == 0) return arr;
       k = k%arr.length;
    for (int i=0; i<arr.length; i++) {

      newArr[i] = arr[(i + (arr.length - k)) % (arr.length)];

    }

    return newArr;
}

}

edited Jul 16 '19 at 09:21

answered Jul 16 '19 at 08:02

Iryna Prokopenko

831
8
6

Great solution! – GTRekter Jun 30 '20 at 13:16

score 2 · Accepted Answer · edited May 09 '19 at 10:25

You can also use A = B.clone();

public int[] solution(int[] A, int K) {
    // write your code in Java SE 8

    int [] B =new int [A.length];

    for(int l=0;K>l;K--){
        int j=0;
        for(int i=0;i<A.length;i++){
            if(i==0){
                B[j]=A[A.length-1];
                j++;
            }
            else{
                B[j]=A[i-1];
                j++;
            }
        }

        //below part
        /*for(int i= 0;i<A.length;i++){
            A[i]=B[i];
        }*/
        A = B.clone();

    }
    return B;
}

If you like :D

mm759 · Answer 3 · 2016-12-30T07:11:46.870

1

You can print the result using Arrays.toString. For example:

System.out.println(Arrays.toString(rotation(new int[] { 3, 8, 9, 7, 6}, 3)));

edited Dec 30 '16 at 07:11

answered Dec 30 '16 at 07:08

mm759

1,404
1
9
7

1

you can make a comment for such small thing. – Wasi Ahmad Dec 30 '16 at 07:09
It answers the question as explained in comment of Dhaval Mistry. A small question gets a small answer. I added an example. – mm759 Dec 30 '16 at 07:13

score 0 · Answer 4 · edited Jan 17 '18 at 14:12

0

public int[] solution(int[] A, int K) {
    // write your code in Java SE 8

    int [] B =new int [A.length];

    for(int l=0;K>l;K--){
        int j=0;
        for(int i=0;i<A.length;i++){
            if(i==0){
                B[j]=A[A.length-1];
                j++;
            }
            else{
                B[j]=A[i-1];
                j++;
            }
        }
        for(int i= 0;i<A.length;i++){
            A[i]=B[i];
        }

    }
    return B;
}

edited Jan 17 '18 at 14:12

Keyur Potdar

7,158
6
25
40

answered Jan 17 '18 at 13:50

user9229864

1

1

Please add some description to explain answer for benefit of other. – jjj Jan 17 '18 at 14:10

Shailendra Singh · Answer 5 · 2019-05-27T02:29:58.653

  function solution(A, K) {

        function shiftArray(arrayToShift, newArray=[] ){
            newArray[0] = arrayToShift[arrayToShift.length-1] ; 
            for (var i=1; i<arrayToShift.length; i++){
                newArray[i] = arrayToShift[i-1];   
            }
            // console.log("arrayToShift");
            // console.log(newArray);
            return newArray;
        }


        var newArray = A;
        for(var i=0; i<K; i++){
            newArray =  shiftArray(newArray);
        }
        return newArray
    }


console.log(solution([3, 8, 9, 7, 6], 3));

score 0 · Answer 6 · answered Oct 25 '19 at 11:48

Achieved 100% correctness

public int[] solution(int[] A, int K) {
    // write your code in Java SE 8
    if(K == A.length || A.length == 0)
        return A;
    if(K > A.length) {
        K = K%A.length;
    }
    int[] arr1 = Arrays.copyOfRange(A, A.length-K, A.length);
    int[] arr2 = Arrays.copyOfRange(A, 0, A.length-K);
    int aLen = arr1.length;
    int bLen = arr2.length;
    int[] result = new int[aLen + bLen];
    System.arraycopy(arr1, 0, result, 0, aLen);
    System.arraycopy(arr2, 0, result, aLen, bLen);
    return result;
}

score 0 · Answer 7 · answered Jan 11 '20 at 19:33

Here's my solution using JavaScript, tested 100%.

function solution(A, K) {
    for(let i=0; i<K; i++) {
        let lastIndex = A.length - 1;
        let lastItem = A[lastIndex];
        for(let j=(A.length-1); j>-1; j--) {
            if(j>0) {
                A[j] = A[j-1];
            } else {
                A[j] = lastItem;
            }
        }
    }
    return A;
}

score 0 · Answer 8 · answered Jan 16 '20 at 11:07

class Solution { public int[] solution(int[] A, int K){

    if((A.length == 0) || (K == 0)){
        return A;
    }
    int [] B = new int[A.length];
    int c = K;
    while(c != 0){
        for(int i = 1; i< A.length; i++){
            B[i] = A[i-1];
        }
        c--;
        B[0] = A[A.length-1];
        System.arraycopy(B, 0, A, 0, A.length);
    }
    return A;
}

}

score 0 · Answer 9 · answered May 16 '20 at 07:27

In Kotlin:

fun flipList(list: List<Int>, times: Int): List<Int> {
    val flippedList = list.toMutableList()
    val referenceList = list.toMutableList()
    repeat(times) {
        var position = 0
        referenceList.forEach {
            position++
            if (position < list.size)
                flippedList[position] = referenceList[position -1]
            else
                flippedList[0] = referenceList.last()
        }
        referenceList.clear()
        referenceList.addAll(flippedList)
    }
    return flippedList
}

Unit Teste

class FlipListUnitTest {
        private val referenceListMock = listOf(1,2,3,4)
        private val referenceListOneTimeFlipResultMock = listOf(4,1,2,3)
        private val referenceListFourTimesFlipResultMock = listOf(1,2,3,4)
        @Test
        fun `check test api is working`(){
            assertTrue(true)
        }

        @Test
        fun `check list flip 1 time`(){
            assertTrue(flipList(referenceListMock, 1) == referenceListOneTimeFlipResultMock)
        }

        @Test
        fun `check list flip 4 time`(){
            assertTrue(flipList(referenceListMock, 4) == referenceListFourTimesFlipResultMock)
        }
}

score 0 · Answer 10 · answered May 24 '20 at 08:49

Here in my code

public static int[] solution(int[] A, int K){
    if(A.length > 0) {
        for(int i = 0; i < K ; i++){
            int temp = A[A.length - 1];
            for(int j = A.length - 2; j >= 0; j--){
                A[j + 1] = A[j];
            }
            A[0] = temp;
        }
    }
    return A;
}

score 0 · Answer 11 · answered Jan 25 '21 at 09:22

public static int[] solution(int[] A, int K) {
    if(A.length<1)
        return A;
    int[] newArray = new int[A.length];

    while (K>0){
        newArray[0]=A[A.length-1];
        for(int i=0; i<A.length-1; i++){
            newArray[i+1]=A[i];
        }
        for (int i=0; i<newArray.length; i++) {
            A[i]=newArray[i];
        }
        K--;
    }
    return A;
}

score 0 · Answer 12 · answered Jun 15 '21 at 15:46

An array A consisting of N integers is given. Rotation of the array means that each element is shifted right by one index, and the last element of the array is moved to the first place.

Hi everyone here is another simple solution for this problem in JAVA, 100% working.

class Solution {
    public int[] solution(int[] A, int K) {
        
        // Corner cases to save resources
         if(K == 0 || A.length <= 0 || A.length == K){
             return A;
         }
         
         // Loop to traverse K times
         for(int i=0; i<K; i++){
             int last = A[A.length - 1]; // Last digit
             
             // Loop to traverse A.Length times in swing order,
             // so that first element can be set
             for(int j=A.length-1; j>0; j--){
              A[j] = A[j-1];   
             }
             
             // Set last element
             A[0] = last;
         }
         
         // Return result
         return A;
    }
}

score 0 · Answer 13 · answered Feb 11 '22 at 04:43

Here is my answer in JavaScript

function solution(A, K){
let arr = [];
let lastIndex = A.length -1;
        let rotation = K-1;
    for(let i = 0; i < K; i++){
        arr[rotation] = A[lastIndex];
        --lastIndex; 
        --rotation;
    }

    for(let j = 0; j <= lastIndex; j++){
        arr.push(A[j]);
    }
    return arr;
}

score 0 · Answer 14 · answered May 05 '22 at 00:23

Here is my answer in Python without loop and using string substitution

def solution(A, K):
    K = K % len(A) if K > len(A) else K
    if (K == 0) or (K == len(A)) or (len(A) in [0, 1]):
        return A
    
    first_index = len(A) - K
    return A[first_index:] + A[:first_index]

score 0 · Answer 15 · answered Jul 20 '22 at 19:12

0

function solution(A, K) {
   let tmpA = [];
   for (let i = 1; i <= K; i++){
       tmpA.unshift(A.pop());
   }
   for (let i = 1; i <= K; i++){
       A.unshift(tmpA.pop());
   }
   return A;   
}

answered Jul 20 '22 at 19:12

Chukwuemeka Joseph Onwukwe

1
2

This answer appears to be Javascript code. The question requested Java code and gave an attempt in Java, not Javascript. – user9712582 Jul 24 '22 at 01:20

Mohneesh · Answer 16 · 2022-09-23T14:15:37.697

This is the code written in PHP but logic can be used in any language.

function solution($A, $K) {
// write your code in PHP7.0      
    $cnt = count($A);
    for($i=1; $i<=$K; $i++){ 
        foreach($A as $key=>$val){
            $key++;
            if($key==$cnt){
                $A[0] = $val;
            }
            else
            {   
                $A[$key] = $val;
            }
        }
    }
    return $A;
}

score 0 · Answer 17 · edited Oct 13 '22 at 07:42

0

public static int[] rotateArrayKTimes(int[] A, int K) {

        if (A.length == 0 || A.length > 1000 || K > 100) {
            return new int[]{};
        }
        for (int index = 0; index < K; index++) {
            int temp = A[A.length - 1];
            System.arraycopy(A, 0, A, 1, A.length - 1);
            A[0] = temp;
        }
        return A;
    }

edited Oct 13 '22 at 07:42

Procrastinator

2,526
30
27
36

answered Oct 08 '22 at 05:55

RAKESH YEDLAPPALLI

3
4

Simão Rocha · Answer 18 · 2022-10-10T15:56:28.663

0

javascript | score: 100%

function solution(A, K) {

for(let i=0 ; i<K ; i++){
    let lastIndex = A.length-1;
    let lastNumber = A[lastIndex];
    A.unshift(lastNumber);
    A.pop();
}
return A;

}

edited Oct 10 '22 at 15:56

answered Oct 10 '22 at 15:55

Simão Rocha

1
1

Your post is mostly code, please add some more explanation of how it works – André Walker Oct 14 '22 at 14:49
1

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Oct 14 '22 at 14:50

score 0 · Answer 19 · answered Oct 23 '22 at 07:45

JAVA solution -

class Solution2 {
    public int[] solution(int[] arr, int k) {
        int temp;
        int startindex = 0;
        if (arr.length != 0) {
            for (int i = 1; i <= k; i++) {
                temp = arr[arr.length - 1];
                for (int j = arr.length - 1; j >= startindex; j--) {

                    if (j != (startindex)) {
                        arr[j] = arr[j - 1];
                    } else
                        arr[startindex] = temp;
                }
            }
        } else
            System.out.println("This is empty array");
        return arr;
    }

}

Rotation of the array means that each element is shifted right by one index, and the last element of the array is also moved to the first place

19 Answers19