Given an array of integers arr[] of size N and an integer, the task is to rotate the array elements to the left by d positions.
Examples:
Input:
arr[] = {1, 2, 3, 4, 5, 6, 7}, d = 2
Output: 3 4 5 6 7 1 2
Input: arr[] = {3, 4, 5, 6, 7, 1, 2}, d=2
Output: 5 6 7 1 2 3 4
Approach 1 (Using temp array): This problem can be solved using the below idea:
After rotating d positions to the left, the first d elements become the last d elements of the array
- First store the elements from index d to N-1 into the temp array.
- Then store the first d elements of the original array into the temp array.
- Copy back the elements of the temp array into the original array
Illustration:
Suppose the give array is arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2.
First Step:
=> Store the elements from 2nd index to the last.
=> temp[] = [3, 4, 5, 6, 7]
Second Step:
=> Now store the first 2 elements into the temp[] array.
=> temp[] = [3, 4, 5, 6, 7, 1, 2]
Third Steps:
=> Copy the elements of the temp[] array into the original array.
=> arr[] = temp[] So arr[] = [3, 4, 5, 6, 7, 1, 2]
Follow the steps below to solve the given problem.
- Initialize a temporary array(temp[n]) of length same as the original array
- Initialize an integer(k) to keep a track of the current index
- Store the elements from the position d to n-1 in the temporary array
- Now, store 0 to d-1 elements of the original array in the temporary array
- Lastly, copy back the temporary array to the original array
Below is the implementation of the above approach :
C++
#include <bits/stdc++.h>
using namespace std;
void Rotate(int arr[], int d, int n)
{
int temp[n];
int k = 0;
for (int i = d; i < n; i++) {
temp[k] = arr[i];
k++;
}
for (int i = 0; i < d; i++) {
temp[k] = arr[i];
k++;
}
for (int i = 0; i < n; i++) {
arr[i] = temp[i];
}
}
void PrintTheArray(int arr[], int n)
{
for (int i = 0; i < n; i++) {
cout << arr[i] << " ";
}
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int N = sizeof(arr) / sizeof(arr[0]);
int d = 2;
Rotate(arr, d, N);
PrintTheArray(arr, N);
return 0;
}
|
Java
import java.io.*;
class GFG {
static void Rotate(int arr[], int d, int n)
{
int temp[] = new int[n];
int k = 0;
for (int i = d; i < n; i++) {
temp[k] = arr[i];
k++;
}
for (int i = 0; i < d; i++) {
temp[k] = arr[i];
k++;
}
for (int i = 0; i < n; i++) {
arr[i] = temp[i];
}
}
static void PrintTheArray(int arr[], int n)
{
for (int i = 0; i < n; i++) {
System.out.print(arr[i]+" ");
}
}
public static void main (String[] args) {
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int N = arr.length;
int d = 2;
Rotate(arr, d, N);
PrintTheArray(arr, N);
}
}
|
Python3
def rotate(L, d, n):
k = L.index(d)
new_lis = []
new_lis = L[k+1:]+L[0:k+1]
return new_lis
if __name__ == '__main__':
arr = [1, 2, 3, 4, 5, 6, 7]
d = 2
N = len(arr)
arr = rotate(arr, d, N)
for i in arr:
print(i, end=" ")
|
C#
using System;
public class GFG
{
public static void Rotate(int[] arr, int d, int n)
{
int[] temp = new int[n];
var k = 0;
for (int i = d; i < n; i++)
{
temp[k] = arr[i];
k++;
}
for (int i = 0; i < d; i++)
{
temp[k] = arr[i];
k++;
}
for (int i = 0; i < n; i++)
{
arr[i] = temp[i];
}
}
public static void PrintTheArray(int[] arr, int n)
{
for (int i = 0; i < n; i++)
{
Console.Write(arr[i].ToString() + " ");
}
}
public static void Main(String[] args)
{
int[] arr = {1, 2, 3, 4, 5, 6, 7};
var N = arr.Length;
var d = 2;
GFG.Rotate(arr, d, N);
GFG.PrintTheArray(arr, N);
}
}
|
Javascript
function Rotate_and_Print(arr,d,n)
{
var temp=new Array(n);
let k = 0;
for (let i = d; i < n; i++) {
temp[k] = arr[i];
k++;
}
for (let i = 0; i < d; i++) {
temp[k] = arr[i];
k++;
}
for (let i = 0; i < n; i++) {
console.log(temp[i]+" ");
}
}
let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
let n = arr.length;
let d = 2;
Rotate_and_Print(arr, d, n);
|
Time complexity: O(N)
Auxiliary Space: O(N)
Approach 2 (Rotate one by one): This problem can be solved using the below idea:
- At each iteration, shift the elements by one position to the left circularly (i.e., first element becomes the last).
- Perform this operation d times to rotate the elements to the left by d position.
Illustration:
Let us take arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2.
First Step:
=> Rotate to left by one position.
=> arr[] = {2, 3, 4, 5, 6, 7, 1}
Second Step:
=> Rotate again to left by one position
=> arr[] = {3, 4, 5, 6, 7, 1, 2}
Rotation is done by 2 times.
So the array becomes arr[] = {3, 4, 5, 6, 7, 1, 2}
Follow the steps below to solve the given problem.
- Rotate the array to left by one position. For that do the following:
- Store the first element of the array in a temporary variable.
- Shift the rest of the elements in the original array by one place.
- Update the last index of the array with the temporary variable.
- Repeat the above steps for the number of left rotations required.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void Rotate(int arr[], int d, int n)
{
int p = 1;
while (p <= d) {
int last = arr[0];
for (int i = 0; i < n - 1; i++) {
arr[i] = arr[i + 1];
}
arr[n - 1] = last;
p++;
}
}
void printArray(int arr[], int size)
{
for (int i = 0; i < size; i++)
cout << arr[i] << " ";
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int N = sizeof(arr) / sizeof(arr[0]);
int d = 2;
Rotate(arr, d, N);
printArray(arr, N);
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void rotate(int arr[], int d, int n)
{
int p = 1;
while (p <= d) {
int last = arr[0];
for (int i = 0; i < n - 1; i++) {
arr[i] = arr[i + 1];
}
arr[n - 1] = last;
p++;
}
for (int i = 0; i < n; i++) {
System.out.print(arr[i] + " ");
}
}
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int N = arr.length;
int d = 2;
rotate(arr, d, N);
}
}
|
Python3
def Rotate(arr, d, n):
p = 1
while(p <= d):
last = arr[0]
for i in range (n - 1):
arr[i] = arr[i + 1]
arr[n - 1] = last
p = p + 1
def printArray(arr, size):
for i in range (size):
print(arr[i] ,end = " ")
arr = [1, 2, 3, 4, 5, 6, 7]
N = len(arr)
d = 2
Rotate(arr, d, N)
printArray(arr, N)
|
C#
using System;
public class GFG
{
public static void rotate(int[] arr, int d, int n)
{
var p = 1;
while (p <= d)
{
var last = arr[0];
for (int i = 0; i < n - 1; i++)
{
arr[i] = arr[i + 1];
}
arr[n - 1] = last;
p++;
}
for (int i = 0; i < n; i++)
{
Console.Write(arr[i].ToString() + " ");
}
}
public static void Main(String[] args)
{
int[] arr = {1, 2, 3, 4, 5, 6, 7};
var N = arr.Length;
var d = 2;
GFG.rotate(arr, d, N);
}
}
|
Javascript
function printArray(arr,n,d)
{
let p = 1;
while (p <= d) {
let last = arr[0];
for (let i = 0; i < n - 1; i++) {
arr[i] = arr[i + 1];
}
arr[n - 1] = last;
p++;
}
for (let i = 0; i < n; i++) {
console.log(arr[i] + " ");
}
}
let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
let n = arr.length;
let d=2;
printArray(arr, n,d);
|
Time Complexity: O(N * d)
Auxiliary Space: O(1)
Approach 3 (A Juggling Algorithm): This is an extension of method 2.
Instead of moving one by one, divide the array into different sets where the number of sets is equal to the GCD of N and d (say X. So the elements which are X distance apart are part of a set) and rotate the elements within sets by 1 position to the left.
- Calculate the GCD between the length and the distance to be moved.
- The elements are only shifted within the sets.
- We start with temp = arr[0] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.
Follow the below illustration for a better understanding
Illustration:
Each steps looks like following:
Let arr[] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} and d = 10
First step:
=> First set is {0, 5, 10}.
=> Rotate this set by d position in cyclic order
=> arr[0] = arr[0+10]
=> arr[10] = arr[(10+10)%15]
=> arr[5] = arr[0]
=> This set becomes {10,0,5}
=> Array arr[] = {10, 1, 2, 3, 4, 0, 6, 7, 8, 9, 5, 11, 12, 13, 14}
Second step:
=> Second set is {1, 6, 11}.
=> Rotate this set by d position in cyclic order.
=> This set becomes {11, 1, 6}
=> Array arr[] = {10, 11, 2, 3, 4, 0, 1, 7, 8, 9, 5, 6, 12, 13, 14}
Third step:
=> Second set is {2, 7, 12}.
=> Rotate this set by d position in cyclic order.
=> This set becomes {12, 2, 7}
=> Array arr[] = {10, 11, 12, 3, 4, 0, 1, 2, 8, 9, 5, 6, 7, 13, 14}
Fourth step:
=> Second set is {3, 8, 13}.
=> Rotate this set by d position in cyclic order.
=> This set becomes {13, 3, 8}
=> Array arr[] = {10, 11, 12, 13, 4, 0, 1, 2, 3, 9, 5, 6, 7, 8, 14}
Fifth step:
=> Second set is {4, 9, 14}.
=> Rotate this set by d position in cyclic order.
=> This set becomes {14, 4, 9}
=> Array arr[] = {10, 11, 12, 13, 14, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Follow the steps below to solve the given problem.
- Perform d%n in order to keep the value of d within the range of the array where d is the number of times the array is rotated and N is the size of the array.
- Calculate the GCD(N, d) to divide the array into sets.
- Run a for loop from 0 to the value obtained from GCD.
- Store the value of arr[i] in a temporary variable (the value of i denotes the set number).
- Run a while loop to update the values according to the set.
- After exiting the while loop assign the value of arr[j] as the value of the temporary variable (the value of j denotes the last element of the ith set).
Below is the implementation of the above approach :
C++
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
void leftRotate(int arr[], int d, int n)
{
d = d % n;
int g_c_d = gcd(d, n);
for (int i = 0; i < g_c_d; i++) {
int temp = arr[i];
int j = i;
while (1) {
int k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
void printArray(int arr[], int size)
{
for (int i = 0; i < size; i++)
cout << arr[i] << " ";
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
leftRotate(arr, 2, n);
printArray(arr, n);
return 0;
}
|
C
#include <stdio.h>
void printArray(int arr[], int size);
int gcd(int a, int b);
void leftRotate(int arr[], int d, int n)
{
int i, j, k, temp;
d = d % n;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
temp = arr[i];
j = i;
while (1) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
void printArray(int arr[], int n)
{
int i;
for (i = 0; i < n; i++)
printf("%d ", arr[i]);
}
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
getchar();
return 0;
}
|
Java
import java.io.*;
class RotateArray {
void leftRotate(int arr[], int d, int n)
{
d = d % n;
int i, j, k, temp;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
temp = arr[i];
j = i;
while (true) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
void printArray(int arr[], int size)
{
int i;
for (i = 0; i < size; i++)
System.out.print(arr[i] + " ");
}
int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
public static void main(String[] args)
{
RotateArray rotate = new RotateArray();
int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
rotate.leftRotate(arr, 2, 7);
rotate.printArray(arr, 7);
}
}
|
Python3
def leftRotate(arr, d, n):
d = d % n
g_c_d = gcd(d, n)
for i in range(g_c_d):
temp = arr[i]
j = i
while 1:
k = j + d
if k >= n:
k = k - n
if k == i:
break
arr[j] = arr[k]
j = k
arr[j] = temp
def printArray(arr, size):
for i in range(size):
print("% d" % arr[i], end=" ")
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
arr = [1, 2, 3, 4, 5, 6, 7]
n = len(arr)
d = 2
leftRotate(arr, d, n)
printArray(arr, n)
|
C#
using System;
class GFG {
static void leftRotate(int[] arr, int d, int n)
{
int i, j, k, temp;
d = d % n;
int g_c_d = gcd(d, n);
for (i = 0; i < g_c_d; i++) {
temp = arr[i];
j = i;
while (true) {
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
static void printArray(int[] arr, int size)
{
for (int i = 0; i < size; i++)
Console.Write(arr[i] + " ");
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
public static void Main()
{
int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
leftRotate(arr, 2, 7);
printArray(arr, 7);
}
}
|
Javascript
<script>
function gcd( a, b){
if (b == 0)
return a;
else
return gcd(b, a % b);
}
function leftRotate(arr, d, n){
d = d % n;
let g_c_d = gcd(d, n);
for (let i = 0; i < g_c_d; i++) {
let temp = arr[i];
let j = i;
while (1) {
let k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
function printArray(arr, size){
for (let i = 0; i < size; i++)
document.write(arr[i] +" ");
}
let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
let n = arr.length;
leftRotate(arr, 2, n);
printArray(arr, n);
</script>
|
Time complexity : O(N)
Auxiliary Space : O(1)
Approach 4 :
(Using Collections module )
Python module have module named “collections” which provides various data structures. One of them is “deque“.
Deque is also known as double ended queue. Module also provides different in-built methods. One of them is “rotate”.
To know more about DEQUE, click here.
C++14
#include <bits/stdc++.h>
#include <deque>
using namespace std;
int main() {
deque<int> dq {1, 2, 3, 4, 5, 6, 7};
int d = 2;
for(int i=0; i<d; i++) {
int temp = dq.front();
dq.pop_front();
dq.push_back(temp);
}
for(auto it=dq.begin(); it!=dq.end(); it++) {
cout << *it << " ";
}
return 0;
}
|
Java
import java.util.ArrayDeque;
import java.util.Deque;
public class Main {
public static void main(String[] args) {
int[] inp = {1, 2, 3, 4, 5, 6, 7};
int d = 2;
Deque<Integer> deq = new ArrayDeque<>();
for (int i : inp) {
deq.add(i);
}
for (int i = 0; i < d; i++) {
int temp = deq.remove();
deq.addLast(temp);
}
System.out.println(deq);
}
}
|
Python3
from collections import deque
inp=[1,2,3,4,5,6,7]
d=2
deq = deque(inp)
U = type(deq)
deq.rotate(-d)
print(deq)
|
Javascript
let inp = [1, 2, 3, 4, 5, 6, 7];
let d = 2;
let deque = [...inp];
for (let i = 0; i < d; i++) {
let element = deque.shift();
deque.push(element);
}
console.log(deque);
|
C#
using System;
using System.Collections.Generic;
class GFG {
static void Main(string[] args)
{
List<int> inp
= new List<int>{ 1, 2, 3, 4, 5, 6, 7 };
int d = 2;
Queue<int> deq = new Queue<int>(inp);
for (int i = 0; i < d; i++) {
int temp = deq.Dequeue();
deq.Enqueue(temp);
}
Console.WriteLine(string.Join(" ", deq));
}
}
|
Outputdeque([3, 4, 5, 6, 7, 1, 2])
Time complexity: The time complexity of the code is O(d*n), where d is the number of rotations and n is the size of the deque.
The auxiliary space is O(n), where n is the size of the deque.
Please see the following posts for other methods of array rotation:
Block swap algorithm for array rotation
Reversal algorithm for array rotation
Please write comments if you find any bugs in the above programs/algorithms.