Quickly find multiple left rotations of an array

Given an array of size n and multiple values around which we need to left rotate the array. How to quickly find multiple left rotations?

Examples:

Input : arr[] = {1, 3, 5, 7, 9}
        k1 = 1
        k2 = 3
        k3 = 4
        k4 = 6
Output : 3 5 7 9 1
         7 9 1 3 5
         9 1 3 5 7
         3 5 7 9 1

Input : arr[] = {1, 3, 5, 7, 9}
        k1 = 14 
Output : 9 1 3 5 7

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Simple Approach: We have already discussed different approaches in below posts.

The best of above approaches take O(n) time and O(1) extra space.

Efficient Approach:
The above approaches work well when there is single rotation required. The approaches also modify the original array. To handle multiple queries of array rotation, we use a temp array of size 2n and quickly handle rotations.

Step 1 : Copy the entire array two times in temp[0..2n-1] array.
Step 2 : Starting position of array after k rotations in temp[] will be k % n. We do k
Step 3 : Print temp[] array from k % n to k % n + n.

C++

// CPP implementation of left rotation of 
// an array K number of times 
#include<bits/stdc++.h> 
using namespace std; 
  
// Fills temp[] with two copies of arr[] 
void preprocess(int arr[], int n, int temp[]) 
{ 
    // Store arr[] elements at i and i + n 
    for (int i = 0; i<n; i++) 
         temp[i] = temp[i + n] = arr[i]; 
} 
  
// Function to left rotate an array k times 
void leftRotate(int arr[], int n, int k, int temp[]) 
{ 
    // Starting position of array after k 
    // rotations in temp[] will be k % n 
    int start = k % n; 
  
    // Print array after k rotations 
    for (int i = start; i < start + n; i++) 
         cout << temp[i] << " "; 
  
    cout << endl; 
} 
  
// Driver program 
int main() 
{ 
    int arr[] = {1, 3, 5, 7, 9}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    int temp[2*n]; 
    preprocess(arr, n, temp); 
  
    int k = 2; 
    leftRotate(arr, n, k, temp); 
  
    k = 3; 
    leftRotate(arr, n, k, temp); 
  
    k = 4; 
    leftRotate(arr, n, k, temp); 
  
    return 0; 
} 

Java

// Java implementation of left rotation of 
// an array K number of times 
class LeftRotate 
{ 
    // Fills temp[] with two copies of arr[] 
    static void preprocess(int arr[], int n, 
                                   int temp[]) 
    { 
        // Store arr[] elements at i and i + n 
        for (int i = 0; i<n; i++) 
             temp[i] = temp[i + n] = arr[i]; 
    } 
   
    // Function to left rotate an array k time 
    static void leftRotate(int arr[], int n, int k,  
                                    int temp[]) 
    { 
        // Starting position of array after k 
        // rotations in temp[] will be k % n 
        int start = k % n; 
       
        // Print array after k rotations 
        for (int i = start; i < start + n; i++) 
            System.out.print(temp[i] + " "); 
       
        System.out.print("\n"); 
    } 
  
    // Driver program 
    public static void main (String[] args) 
    { 
        int arr[] = {1, 3, 5, 7, 9}; 
        int n = arr.length; 
       
        int temp[] = new int[2*n]; 
        preprocess(arr, n, temp); 
       
        int k = 2; 
        leftRotate(arr, n, k, temp); 
       
        k = 3; 
        leftRotate(arr, n, k, temp); 
       
        k = 4; 
        leftRotate(arr, n, k, temp); 
    } 
} 
/*This code is contributed by Prakriti Gupta*/

Python3

# Python3 implementation of left rotation 
# of an array K number of times 
  
# Fills temp with two copies of arr 
def preprocess(arr, n): 
    temp = [None] * (2 * n) 
      
    # Store arr elements at i and i + n 
    for i in range(n): 
        temp[i] = temp[i + n] = arr[i] 
    return temp 
  
# Function to left rotate an array k times 
def leftRotate(arr, n, k, temp): 
      
    # Starting position of array after k 
    # rotations in temp will be k % n 
    start = k % n 
      
    # Print array after k rotations 
    for i in range(start, start + n): 
        print(temp[i], end = " ") 
    print("") 
  
# Driver program 
arr = [1, 3, 5, 7, 9] 
n = len(arr) 
temp = preprocess(arr, n) 
  
k = 2
leftRotate(arr, n, k, temp) 
        
k = 3
leftRotate(arr, n, k, temp) 
        
k = 4
leftRotate(arr, n, k, temp) 
  
# This code is contributed by Sanghamitra Mishra  

C#

// Java implementation of left rotation of 
// an array K number of times 
using System; 
class LeftRotate 
{ 
    // Fills temp[] with two copies of arr[] 
    static void preprocess(int []arr, int n, 
                                int[] temp) 
    { 
        // Store arr[] elements at i and i + n 
        for (int i = 0; i<n; i++) 
            temp[i] = temp[i + n] = arr[i]; 
    } 
  
    // Function to left rotate an array k time 
    static void leftRotate(int []arr, int n, int k,  
                                    int []temp) 
    { 
        // Starting position of array after k 
        // rotations in temp[] will be k % n 
        int start = k % n; 
      
        // Print array after k rotations 
        for (int i = start; i < start + n; i++) 
        Console.Write(temp[i] + " "); 
        Console.WriteLine(); 
    } 
  
    // Driver program 
    public static void Main () 
    { 
        int []arr = {1, 3, 5, 7, 9}; 
        int n = arr.Length; 
      
        int []temp = new int[2*n]; 
        preprocess(arr, n, temp); 
      
        int k = 2; 
        leftRotate(arr, n, k, temp); 
      
        k = 3; 
        leftRotate(arr, n, k, temp); 
      
        k = 4; 
        leftRotate(arr, n, k, temp); 
    } 
} 
//This code is contributed by vt_m. 

PHP

<?php  
// PHP implementation of  
// left rotation of an  
// array K number of times 
  
// Fills $temp with 
// two copies of $arr 
function preprocess(&$arr, $n,  
                    &$temp) 
{ 
    // Store $arr elements 
    // at i and i + n 
    for ($i = 0; $i < $n; $i++) 
        $temp[$i] = $temp[$i + $n] = $arr[$i]; 
} 
  
// Function to left rotate 
// an array k times 
function leftRotate(&$arr, $n, 
                     $k, &$temp) 
{ 
    // Starting position of  
    // array after k rotations 
    // in temp[] will be k % n 
    $start = $k % $n; 
  
    // Print array after 
    // k rotations 
    for ($i = $start;  
         $i < $start + $n; $i++) 
        echo $temp[$i] . " "; 
  
    echo "\n"; 
} 
  
// Driver Code 
$arr = array(1, 3, 5, 7, 9); 
$n = sizeof($arr); 
  
$temp[2 * $n] = array(); 
preprocess($arr, $n, $temp); 
  
$k = 2; 
leftRotate($arr, $n, $k, $temp); 
  
$k = 3; 
leftRotate($arr, $n, $k, $temp); 
  
$k = 4; 
leftRotate($arr, $n, $k, $temp); 
  
// This code is contributed 
// by ChitraNayal 
?> 

Output:

5 7 9 1 3 
7 9 1 3 5 
9 1 3 5 7

Note that the task to find starting address of rotation takes O(1) time. It is printing the elements that takes O(n) time.

Space optimized Approach : The above method takes extra space. Below given is the space optimized solution. Thanks to frenzy77 for suggesting this approach.

C++

// CPP implementation of left rotation of 
// an array K number of times 
#include<bits/stdc++.h> 
using namespace std; 
  
// Function to left rotate an array k times 
void leftRotate(int arr[], int n, int k) 
{ 
    // Print array after k rotations 
    for (int i = k; i < k + n; i++) 
        cout << arr[i%n] << " "; 
} 
  
// Driver program 
int main() 
{ 
    int arr[] = {1, 3, 5, 7, 9}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    int k = 2; 
    leftRotate(arr, n, k); 
    cout << endl; 
  
    k = 3; 
    leftRotate(arr, n, k); 
    cout << endl; 
  
    k = 4; 
    leftRotate(arr, n, k); 
    cout << endl; 
  
    return 0; 
} 

Java

// Java implementation of  
// left rotation of an  
// array K number of times 
  
import java.io.*; 
  
class GFG 
{ 
  
// Function to left rotate 
// an array k times 
static void leftRotate(int arr[],  
                       int n, int k) 
{ 
    // Print array after 
    // k rotations 
    for (int i = k; i < k + n; i++) 
        System.out.print(arr[i % n] + " "); 
} 
  
// Driver Code 
public static void main (String[] args) 
{ 
    int arr[] = {1, 3, 5, 7, 9}; 
    int n = arr.length; 
      
    int k = 2; 
    leftRotate(arr, n, k); 
    System.out.println(); 
      
    k = 3; 
    leftRotate(arr, n, k); 
    System.out.println(); 
      
    k = 4; 
    leftRotate(arr, n, k); 
    System.out.println();  
} 
} 
  
// This code is contributed by ajit 

Python 3

# Python3 implementation of  
# left rotation of an array  
# K number of times  
  
# Function to left rotate 
# an array k times 
def leftRotate(arr, n, k): 
      
    # Print array  
    # after k rotations 
    for i in range(k, k + n): 
        print(str(arr[i % n]),  
                   end = " ") 
  
# Driver Code 
arr = [1, 3, 5, 7, 9] 
n = len(arr) 
k = 2; 
leftRotate(arr, n, k) 
print() 
  
k = 3; 
leftRotate(arr, n, k) 
print() 
  
k = 4
leftRotate(arr, n, k) 
print() 
  
# This code is contributed  
# by ChitraNayal 

C#

// C# implementation of  
// left rotation of an  
// array K number of times 
using System; 
  
class GFG 
{ 
  
// Function to left rotate 
// an array k times 
static void leftRotate(int []arr,  
                       int n, int k) 
{ 
    // Print array after 
    // k rotations 
    for (int i = k; i < k + n; i++) 
        Console.Write(arr[i % n] + " "); 
} 
  
// Driver Code 
static public void Main () 
{ 
int []arr = {1, 3, 5, 7, 9}; 
int n = arr.Length; 
  
int k = 2; 
leftRotate(arr, n, k); 
Console.WriteLine(); 
  
k = 3; 
leftRotate(arr, n, k); 
Console.WriteLine(); 
  
k = 4; 
leftRotate(arr, n, k); 
Console.WriteLine();  
} 
} 
  
// This code is contributed  
// by akt_mit 

PHP

<?php 
  
// PHP implementation of left rotation of 
// an array K number of times 
  
// Function to left rotate an array k times 
function leftRotate($arr, $n, $k) 
{ 
      
    // Print array after k rotations 
    for ($i = $k; $i < $k + $n; $i++) 
        echo $arr[$i % $n] ," "; 
} 
  
// Driver program 
$arr = array (1, 3, 5, 7, 9); 
$n = sizeof($arr); 
  
$k = 2; 
leftRotate($arr, $n, $k); 
echo "\n"; 
  
$k = 3; 
leftRotate($arr, $n, $k); 
echo "\n"; 
  
$k = 4; 
leftRotate($arr, $n, $k); 
echo "\n"; 
  
// This code is contributed by aj_36 
?> 

Output:

5 7 9 1 3 
7 9 1 3 5 
9 1 3 5 7

This article is contributed by nuclode and Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Sep	OCT	Nov
	22
2019	2020	2021

Quickly find multiple left rotations of an array | Set 1

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python 3

C#

PHP

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