Print left rotation of array in O(n) time and O(1) space

27 May, 2022
Given an array of size n and multiple values around which we need to left rotate the array. How to quickly print 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
We have discussed a solution in the below post. Quickly find multiple left rotations of an array | Set 1Method I:
The solution discussed above requires extra space. In this post, an optimized solution is discussed that doesn’t require extra space.
C++
Java
Python
C#
PHP
Javascript
// C++ implementation of left rotation of// an array K number of times#include <bits/stdc++.h>using namespace std; // Function to leftRotate array multiple timesvoid leftRotate(int arr[], int n, int k){    /* To get the starting point of rotated array */    int mod = k % n;     // Prints the rotated array from start position    for (int i = 0; i < n; i++)        cout << (arr[(mod + i) % n]) << " ";     cout << "\n";} // Driver Codeint main(){    int arr[] = { 1, 3, 5, 7, 9 };    int n = sizeof(arr) / sizeof(arr[0]);     int k = 2;         // Function Call    leftRotate(arr, n, k);     k = 3;         // Function Call    leftRotate(arr, n, k);     k = 4;         // Function Call    leftRotate(arr, n, k);     return 0;}
// JAVA implementation of left rotation// of an array K number of timesimport java.util.*;import java.lang.*;import java.io.*; class arr_rot {    // Function to leftRotate array multiple    // times    static void leftRotate(int arr[], int n, int k)    {        /* To get the starting point of        rotated array */        int mod = k % n;         // Prints the rotated array from        // start position        for (int i = 0; i < n; ++i)            System.out.print(arr[(i + mod) % n] + " ");         System.out.println();    }     // Driver code    public static void main(String[] args)    {        int arr[] = { 1, 3, 5, 7, 9 };        int n = arr.length;         int k = 2;                 // Function Call        leftRotate(arr, n, k);         k = 3;                 // Function Call        leftRotate(arr, n, k);         k = 4;                 // Function Call        leftRotate(arr, n, k);    }} // This code is contributed by Sanjal
# Python implementation of left rotation of# an array K number of times # Function to leftRotate array multiple times  def leftRotate(arr, n, k):     # To get the starting point of rotated array    mod = k % n    s = ""     # Prints the rotated array from start position    for i in range(n):        print str(arr[(mod + i) % n]),    print    return  # Driver codearr = [1, 3, 5, 7, 9]n = len(arr)k = 2 # Function CallleftRotate(arr, n, k) k = 3 # Function CallleftRotate(arr, n, k) k = 4 # Function CallleftRotate(arr, n, k) # This code is contributed by Sachin Bisht
// C# implementation of left// rotation of an array K// number of timesusing System; class GFG {     // Function to leftRotate    // array multiple times    static void leftRotate(int[] arr, int n, int k)    {        // To get the starting        // point of rotated array        int mod = k % n;         // Prints the rotated array        // from start position        for (int i = 0; i < n; ++i)            Console.Write(arr[(i + mod) % n] + " ");         Console.WriteLine();    }     // Driver Code    static public void Main()    {        int[] arr = { 1, 3, 5, 7, 9 };        int n = arr.Length;         int k = 2;                 // Function Call        leftRotate(arr, n, k);         k = 3;                 // Function Call        leftRotate(arr, n, k);         k = 4;                 // Function Call        leftRotate(arr, n, k);    }} // This code is contributed by m_kit
<?php// PHP implementation of// left rotation of an// array K number of times // Function to leftRotate// array multiple timesfunction leftRotate($arr, $n, $k){    // To get the starting    // point of rotated array    $mod = $k % $n;     // Prints the rotated array    // from start position    for ($i = 0; $i < $n; $i++)        echo ($arr[($mod +                    $i) % $n]) , " ";     echo "\n";} // Driver Code$arr = array(1, 3, 5, 7, 9);$n = sizeof($arr); $k = 2; // Function CallleftRotate($arr, $n, $k); $k = 3; // Function CallleftRotate($arr, $n, $k); $k = 4; // Function CallleftRotate($arr, $n, $k); // This code is contributed by m_kit?>
<script>// JavaScript implementation of left rotation of// an array K number of times // Function to leftRotate array multiple timesfunction leftRotate(arr, n, k){    /* To get the starting point of rotated array */    let mod = k % n;     // Prints the rotated array from start position    for (let i = 0; i < n; i++)        document.write((arr[(mod + i) % n]) + " ");     document.write("\n");} // Driver Codelet arr = [ 1, 3, 5, 7, 9 ];let n = arr.length; let k = 2;// Function CallleftRotate(arr, n, k);document.write("<br>"); k = 3;// Function CallleftRotate(arr, n, k);document.write("<br>"); k = 4;// Function CallleftRotate(arr, n, k); </script>
5 7 9 1 3 
7 9 1 3 5 
9 1 3 5 7 
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
Method II:  
In the below implementation we will use Standard Template Library (STL) which will be making the solution more optimize and easy to Implement.
C++
Java
Python3
C#
Javascript
// C++ Implementation For Print Left Rotation Of Any Array K// Times #include <bits/stdc++.h>#include <iostream>using namespace std;// Function For The k Times Left Rotationvoid leftRotate(int arr[], int k, int n){     // Stl function rotates takes three parameters - the    // beginning,the position by which it should be rotated    // ,the end address of the array      // The below function will be rotating the array left        // in linear time (k%arraySize) times    rotate(arr, arr + (k % n), arr + n);           // Print the rotated array from start position    for (int i = 0; i < n; i++)        cout << arr[i] << " ";    cout << "\n";}// Driver programint main(){    int arr[] = { 1, 3, 5, 7, 9 };    int n = sizeof(arr) / sizeof(arr[0]);    int k = 2;         // Function Call    leftRotate(arr, k, n);      return 0;}
// Java implementation for print left// rotation of any array K timesimport java.io.*;import java.util.*; class GFG{     // Function for the k times left rotationstatic void leftRotate(Integer arr[], int k,                                      int n){        // In Collection class rotate function     // takes two parameters - the name of     // array and the position by which it     // should be rotated     // The below function will be rotating     // the array left  in linear time           // Collections.rotate()rotate the     // array from right hence n-k    Collections.rotate(Arrays.asList(arr), n - k);         // Print the rotated array from start position    for(int i = 0; i < n; i++)        System.out.print(arr[i] + " ");} // Driver codepublic static void main(String[] args){    Integer arr[] = { 1, 3, 5, 7, 9 };    int n = arr.length;    int k = 2;         // Function call    leftRotate(arr, k, n);}} // This code is contributed by chahattekwani71
# Python3 implementation to print left# rotation of any array K timesfrom collections import deque # Function For The k Times Left Rotationdef leftRotate(arr, k, n):         # The collections module has deque class    # which provides the rotate(), which is    # inbuilt function to allow rotation    arr = deque(arr)         # using rotate() to left rotate by k    arr.rotate(-k)    arr = list(arr)         # Print the rotated array from    # start position    for i in range(n):        print(arr[i], end = " ") # Driver Codeif __name__ == '__main__':           arr = [ 1, 3, 5, 7, 9 ]    n = len(arr)    k = 2       # Function Call    leftRotate(arr, k, n) # This code is contributed by math_lover
// C# program for the above approachusing System; class GFG{    static void leftRotate(int[] arr, int d,                           int n)    {        for (int i = 0; i < d; i++)            leftRotatebyOne(arr, n);    }      static void leftRotatebyOne(int[] arr, int n)    {        int i, temp = arr[0];        for (i = 0; i < n - 1; i++)            arr[i] = arr[i + 1];          arr[n - 1] = temp;    }      /* utility function to print an array */    static void printArray(int[] arr, int size)    {        for (int i = 0; i < size; i++)            Console.Write(arr[i] + " ");    } // Driver Codepublic static void Main(){    int[] arr = { 1, 3, 5, 7, 9 };    int n = arr.Length;    int k = 2;         // Function call    leftRotate(arr, k, n);    printArray(arr, n);}} // This code is contributed by avijitmondal1998.
<script>// Javascript program for the above approachfunction leftRotate(arr, d, n){    for (let i = 0; i < d; i++)        leftRotatebyOne(arr, n);}  function leftRotatebyOne(arr, n){    let i, temp = arr[0];    for (i = 0; i < n - 1; i++)        arr[i] = arr[i + 1];      arr[n - 1] = temp;}  /* utility function to print an array */function printArray(arr, size){    for (let i = 0; i < size; i++)        document.write(arr[i] + " ");} // Driver Codelet arr = [ 1, 3, 5, 7, 9 ];let n = arr.length;let k = 2;     // Function callleftRotate(arr, k, n);printArray(arr, n); // This code is contributed by Samim Hossain Mondal.</script>
5 7 9 1 3 
Note: the array itself gets updated after the rotation.
Time Complexity: O(n)
Auxiliary Space: O(1)
This article is contributed by Sridhar Babu and improved by Geetansh Sahni. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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