Print left rotation of array in O(n) time and O(1) space
Last Updated :
05 Dec, 2024
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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
Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
We have discussed a solution in the below post.
Quickly find multiple left rotations of an array | Set 1
Method I: The solution discussed above requires extra space. In this post, an optimized solution is discussed that doesn’t require extra space.
Implementation:
// C++ implementation of left rotation of
// an array K number of times
#include <bits/stdc++.h>
using namespace std;
// Function to leftRotate array multiple times
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++)
cout << (arr[(mod + i) % n]) << " ";
cout << "\n";
}
// Driver Code
int 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;
}
#include <stdio.h>
// Function to leftRotate array multiple times
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++)
printf("%d ", arr[(mod + i) % n]);
printf("\n");
}
int 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 times
import 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 code
arr = [1, 3, 5, 7, 9]
n = len(arr)
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 Sachin Bisht
// C# implementation of left
// rotation of an array K
// number of times
using 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
<script>
// JavaScript implementation of left rotation of
// an array K number of times
// Function to leftRotate array multiple times
function 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 Code
let arr = [ 1, 3, 5, 7, 9 ];
let n = arr.length;
let k = 2;
// Function Call
leftRotate(arr, n, k);
document.write("<br>");
k = 3;
// Function Call
leftRotate(arr, n, k);
document.write("<br>");
k = 4;
// Function Call
leftRotate(arr, n, k);
</script>
<?php
// PHP implementation of
// left rotation of an
// array K number of times
// Function to leftRotate
// array multiple times
function 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);
$k = 2;
// Function Call
leftRotate($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
?>
Output
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.
Implementation:
// 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 Rotation
void 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 program
int 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;
}
#include <stdio.h>
// Function For k Times Left Rotation
void leftRotate(int arr[], int k, int n)
{
int i, temp;
// Perform k left rotations
for (i = 0; i < k; i++) {
// Store the first element of the array
temp = arr[0];
// Shift all elements one position to the left
for (int j = 0; j < n - 1; j++) {
arr[j] = arr[j + 1];
}
// Place the first element at the end
arr[n - 1] = temp;
}
// Print the rotated array
for (i = 0; i < n; i++) {
printf("%d ", arr[i]);
}
printf("\n");
}
// Driver program
int 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 times
import java.io.*;
import java.util.*;
class GFG{
// Function for the k times left rotation
static 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 code
public 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 times
from collections import deque
# Function For The k Times Left Rotation
def 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 Code
if __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 approach
using 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 Code
public 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.
let arr = [4, 3, 7, 6, 2, 1, 5,8];
arr.sort((a, b) => a - b); // Sort the array
let k = 6;
let n = arr.length
for (let i = 0; i < k; i++) {
for (let j = 0; j < n-1; j++) {
[arr[j + 1] , arr[j]] = [arr[j] , arr[j + 1]];
}
}
console.log("Final array:", arr);
Output
5 7 9 1 3
Note: the array itself gets updated after the rotation.
Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.
Method III(Using Reversal):
To left rotate an array by “k” units we will perform 3 simple reversals-
- Reverse the first “k” elements
- Reverse the last “n-k” elements where n is the size of the array
- Reverse the whole array
Code-
// 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 Rotation
void leftRotate(int arr[], int k, int n)
{
// if k>n , k%n will bring k back in range
k = (k%n);
reverse(arr,arr+k);
reverse(arr+k,arr+n);
reverse(arr,arr+n);
// Print the rotated array from start position
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
cout << "\n";
}
// Driver program
int 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;
}
#include <stdio.h>
// Function For k Times Left Rotation
void leftRotate(int arr[], int k, int n)
{
// if k > n, k % n will bring k back in range
k = (k % n);
// Reverse the first part of the array (0 to k-1)
for (int i = 0, j = k - 1; i < j; i++, j--) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
// Reverse the second part of the array (k to n-1)
for (int i = k, j = n - 1; i < j; i++, j--) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
// Reverse the entire array
for (int i = 0, j = n - 1; i < j; i++, j--) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
// Print the rotated array
for (int i = 0; i < n; i++) {
printf("%d ", arr[i]);
}
printf("\n");
}
// Driver program
int 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;
}
import java.util.*;
public class Main {
// Function for k times left rotation
public static void leftRotate(int[] arr, int k)
{
// if k>arr.length,k%arr.length will bring k back to range
k%=arr.length;
// Reverse the first k elements
reverseArray(arr, 0, k - 1);
// Reverse the remaining n-k elements
reverseArray(arr, k, arr.length - 1);
// Reverse the entire array
reverseArray(arr, 0, arr.length - 1);
// Print the rotated array from start position
String result = Arrays.toString(arr).replaceAll("\\[|\\]|,|\\s", " ");
System.out.println(result);
}
// Helper function to reverse a section of an array from start to end (inclusive)
public static void reverseArray(int[] arr, int start, int end) {
while (start < end) {
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
// Driver code
public static void main(String[] args) {
int[] arr = {1, 3, 5, 7, 9};
int k = 2;
// Function Call
leftRotate(arr, k);
}
}
# Function for k times left rotation
def leftRotate(arr, k):
# if k>len(arr) , k%=len(arr) bring k back to range
k %= len(arr)
# Reverse the first k elements
arr = reverseArray(arr, 0, k - 1)
# Reverse the remaining n-k elements
arr = reverseArray(arr, k, len(arr) - 1)
# Reverse the entire array
arr = reverseArray(arr, 0, len(arr) - 1)
# Print the rotated array from start position
print(" ".join(map(str,arr)))
# Helper function to reverse a section of an array from start to end (inclusive)
def reverseArray(arr, start, end):
while start < end:
temp = arr[start]
arr[start] = arr[end]
arr[end] = temp
start += 1
end -= 1
return arr
# Driver code
arr = [1, 3, 5, 7, 9]
k = 2
# Function Call
leftRotate(arr, k)
// C# Implementation For Print Left Rotation Of Any Array K
// Times
using System;
using System.Collections.Generic;
class Program
{
// Driver program
static void Main(string[] args)
{
int[] arr = { 1, 3, 5, 7, 9 };
int n = arr.Length;
int k = 2;
leftRotate(arr, k, n);
Console.ReadKey();
}
// Function For The k Times Left Rotation
static void leftRotate(int[] arr, int k, int n)
{
k%=n;
Array.Reverse(arr, 0, k);
Array.Reverse(arr, k, n - k);
Array.Reverse(arr, 0, n);
// Print the rotated array from start position
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
Console.WriteLine();
}
}
// This code is contributed by Tapesh(tapeshdua420)
// Function for k times left rotation
function leftRotate(arr, k) {
k%=arr.length
// Reverse the first k elements
arr = reverseArray(arr, 0, k - 1);
// Reverse the remaining n-k elements
arr = reverseArray(arr, k, arr.length - 1);
// Reverse the entire array
arr = reverseArray(arr, 0, arr.length - 1);
// Print the rotated array from start position
console.log(arr.join(" "));
}
// Helper function to reverse a section of an array from start to end (inclusive)
function reverseArray(arr, start, end) {
while (start < end) {
let temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
return arr;
}
// Driver code
let arr = [1, 3, 5, 7, 9 ];
let n = arr.length;
let k = 2;
// Function Call
leftRotate(arr, k, n);
fun leftRotate(arr: IntArray, k: Int, n: Int) {
// If k > n, k % n will bring k back in range
val rotation = k % n
// Reverse the first part of the array (0 to rotation-1)
for (i in 0 until rotation / 2) {
val temp = arr[i]
arr[i] = arr[rotation - i - 1]
arr[rotation - i - 1] = temp
}
// Reverse the second part of the array (rotation to n-1)
for (i in 0 until (n - rotation) / 2) {
val temp = arr[rotation + i]
arr[rotation + i] = arr[n - i - 1]
arr[n - i - 1] = temp
}
// Reverse the entire array
for (i in 0 until n / 2) {
val temp = arr[i]
arr[i] = arr[n - i - 1]
arr[n - i - 1] = temp
}
// Print the rotated array
for (i in arr) {
print("$i ")
}
println()
}
fun main() {
val arr = intArrayOf(1, 3, 5, 7, 9)
val n = arr.size
val k = 2
// Function Call
leftRotate(arr, k, n)
}
fun leftRotate(arr: IntArray, k: Int, n: Int) {
// If k > n, k % n will bring k back in range
val rotation = k % n
// Reverse the first part of the array (0 to rotation-1)
for (i in 0 until rotation / 2) {
val temp = arr[i]
arr[i] = arr[rotation - i - 1]
arr[rotation - i - 1] = temp
}
// Reverse the second part of the array (rotation to n-1)
for (i in 0 until (n - rotation) / 2) {
val temp = arr[rotation + i]
arr[rotation + i] = arr[n - i - 1]
arr[n - i - 1] = temp
}
// Reverse the entire array
for (i in 0 until n / 2) {
val temp = arr[i]
arr[i] = arr[n - i - 1]
arr[n - i - 1] = temp
}
// Print the rotated array
for (i in arr) {
print("$i ")
}
println()
}
fun main() {
val arr = intArrayOf(1, 3, 5, 7, 9)
val n = arr.size
val k = 2
// Function Call
leftRotate(arr, k, n)
}
Output
5 7 9 1 3
Note: the array itself gets updated after the rotation.
Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.
