Print Array after it is right rotated K times where K can be large or negative

Last Updated : 21 Sep, 2022

Given an array arr[] of size N and a value K (-10^5<K<10^5), the task is to print the array rotated by K times to the right.

Examples:

Input: arr = {1, 3, 5, 7, 9}, K = 2
Output: 7 9 1 3 5
Explanation:
Rotating array 1 time right: 9, 1, 3, 5, 7
Rotating array 2 time right: 7, 9, 1, 3, 5

Input: arr = {1, 2, 3, 4, 5}, K = -2
Output: 3 4 5 1 2
Explanation:
Rotating array -1 time right: 2, 3, 4, 5, 1
Rotating array -2 time right: 3, 4, 5, 1, 2

Naive Approach: The brute force approach to solve this problem is to use a temporary array to rotate the array K or -K times.

Time Complexity: O(N)
Auxiliary Space: O(N)

Efficient Approach: The given problem can be solved by breaking the problem into the following parts:

Round up the value of K in range [0, N), using below steps:
- If K is negative, first change it into positive, find the modulo with N, and then again change it to negative
- If K is positive, just find the modulo with N
Handle the case when K is negative. If K is negative, it means we need to rotate the array K times left, or -K times right.
Next we can simply rotate the array K times by reversing subarrays. Below steps can be followed to solve the problem:
- Reverse all the array elements from 1 to N -1
- Reverse the array elements from 1 to K – 1
- Reverse the array elements from K to N -1

C++

// C++ implementation for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to rotate the array
// to the right, K times
void RightRotate(vector<int>& nums, int K)
{
    int n = nums.size();

    // Case when K > N or K < -N
    K = K < 0 ? ((K * -1) % n) * -1 : K % n;

    // Case when K is negative
    K = K < 0 ? (n - (K * -1)) : K;

    // Reverse all the array elements
    reverse(nums.begin(), nums.end());

    // Reverse the first k elements
    reverse(nums.begin(), nums.begin() + K);

    // Reverse the elements from K
    // till the end of the array
    reverse(nums.begin() + K, nums.end());
}

// Driver code
int main()
{

    // Initialize the array
    vector<int> Array = { 1, 2, 3, 4, 5 };

    // Find the size of the array
    int N = Array.size();

    // Initialize K
    int K = -2;

    // Call the function and
    // print the answer
    RightRotate(Array, K);

    // Print the array after rotation
    for (int i = 0; i < N; i++) {

        cout << Array[i] << " ";
    }

    cout << endl;
    return 0;
}

Java

// Java implementation for the above approach
import java.util.*;

class GFG{

    // Initialize the array
   static int[] Array = { 1, 2, 3, 4, 5 };
   
   static void reverse( int start, int end) {

       // Temporary variable to store character 
       int temp;
       while (start <= end)
       {
         
           // Swapping the first and last character 
           temp = Array[start];
           Array[start] = Array[end];
           Array[end] = temp;
           start++;
           end--;
       }
   }
  
// Function to rotate the array
// to the right, K times
static void RightRotate( int K)
{
    int n = Array.length;

    // Case when K > N or K < -N
    K = K < 0 ? ((K * -1) % n) * -1 : K % n;

    // Case when K is negative
    K = K < 0 ? (n - (K * -1)) : K;

    // Reverse all the array elements
    reverse(0, n-1);

    // Reverse the first k elements
    reverse(0, n - K);

    // Reverse the elements from K
    // till the end of the array
    reverse( K, n-1);
}

// Driver code
public static void main(String[] args)
{


    // Find the size of the array
    int N = Array.length;

    // Initialize K
    int K = -2;

    // Call the function and
    // print the answer
    RightRotate(K);

    // Print the array after rotation
    for (int i = 0; i < N; i++) {

        System.out.print(Array[i]+ " ");
    }

    System.out.println();
}
}

// This code is contributed by Rajput-Ji

Python3

# Python code for the above approach 

# Function to rotate the array
# to the right, K times
def RightRotate(nums, K) :
    n = len(nums)

    # Case when K > N or K < -N
    K = ((K * -1) % n) * -1 if K < 0 else K % n;

    # Case when K is negative
    K = (n - (K * -1)) if K < 0 else K;

    # Reverse all the array elements
    nums.reverse();

    # Reverse the first k elements
    p1 = nums[0:K]
    p1.reverse();

    # Reverse the elements from K
    # till the end of the array
    p2 = nums[K:]
    p2.reverse();
    arr = p1 + p2

    return arr;

# Driver code

# Initialize the array
Array = [1, 2, 3, 4, 5];

# Find the size of the array
N = len(Array)

# Initialize K
K = -2;

# Call the function and
# print the answer
Array = RightRotate(Array, K);

# Print the array after rotation
for i in Array:
    print(i, end=" ")

# This code is contributed by Saurabh jaiswal

// C# implementation for the above approach
using System;
public class GFG {

  // Initialize the array
  static int[] Array = { 1, 2, 3, 4, 5 };
  static void reverse(int start, int end) 
  {

    // Temporary variable to store character
    int temp;
    while (start <= end) {

      // Swapping the first and last character
      temp = Array[start];
      Array[start] = Array[end];
      Array[end] = temp;
      start++;
      end--;
    }
  }

  // Function to rotate the array
  // to the right, K times
  static void RightRotate(int K) {
    int n = Array.Length;

    // Case when K > N or K < -N
    K = K < 0 ? ((K * -1) % n) * -1 : K % n;

    // Case when K is negative
    K = K < 0 ? (n - (K * -1)) : K;

    // Reverse all the array elements
    reverse(0, n - 1);

    // Reverse the first k elements
    reverse(0, n - K);

    // Reverse the elements from K
    // till the end of the array
    reverse(K, n - 1);
  }

  // Driver code
  public static void Main(String[] args) {

    // Find the size of the array
    int N = Array.Length;

    // Initialize K
    int K = -2;

    // Call the function and
    // print the answer
    RightRotate(K);

    // Print the array after rotation
    for (int i = 0; i < N; i++) {

      Console.Write(Array[i] + " ");
    }

    Console.WriteLine();
  }
}

// This code is contributed by Rajput-Ji

JavaScript

 <script>
        // JavaScript code for the above approach 

        // Function to rotate the array
        // to the right, K times
        function RightRotate(nums, K) 
        {
            let n = nums.length;

            // Case when K > N or K < -N
            K = K < 0 ? ((K * -1) % n) * -1 : K % n;

            // Case when K is negative
            K = K < 0 ? (n - (K * -1)) : K;

            // Reverse all the array elements
            nums = nums.reverse();

            // Reverse the first k elements
            let p1 = nums.slice(0, K)
            p1 = p1.reverse();

            // Reverse the elements from K
            // till the end of the array
            let p2 = nums.slice(K)
            p2 = p2.reverse();


            let arr = p1.concat(p2);

            return arr;
        }

        // Driver code


        // Initialize the array
        let Array = [1, 2, 3, 4, 5];

        // Find the size of the array
        let N = Array.length;

        // Initialize K
        let K = -2;

        // Call the function and
        // print the answer
        Array = RightRotate(Array, K);

        // Print the array after rotation
        for (let i = 0; i < N; i++) {

            document.write(Array[i] + " ");
        }
        document.write('<br>')

       // This code is contributed by Potta Lokesh
    </script>

Output

3 4 5 1 2

Time Complexity: O(N)
Auxiliary Space: O(1)

Javascript Program to Find Mth element after K Right Rotations of an Array

code_r

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Print Array after it is right rotated K times where K can be large or negative

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