C++ Program to Left Rotate an Array by d Elements.

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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 : 

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

// Function to rotate array
void Rotate(int arr[], int d, int n)
{
    // Storing rotated version of array
    int temp[n];

    // Keeping track of the current index
    // of temp[]
    int k = 0;

    // Storing the n - d elements of
    // array arr[] to the front of temp[]
    for (int i = d; i < n; i++) {
        temp[k] = arr[i];
        k++;
    }

    // Storing the first d elements of array arr[]
    //  into temp
    for (int i = 0; i < d; i++) {
        temp[k] = arr[i];
        k++;
    }

    // Copying the elements of temp[] in arr[]
    // to get the final rotated array
    for (int i = 0; i < n; i++) {
        arr[i] = temp[i];
    }
}

// Function to print elements of array
void PrintTheArray(int arr[], int n)
{
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int d = 2;

    // Function calling
    Rotate(arr, d, N);
    PrintTheArray(arr, N);

    return 0;
}

3 4 5 6 7 1 2 

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++ program to rotate an array by
// d elements
#include <bits/stdc++.h>
using namespace std;

/*Function to left rotate arr[] of size n by d*/
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++;
    }
}

// Function to print an array
void printArray(int arr[], int size)
{
    for (int i = 0; i < size; i++)
        cout << arr[i] << " ";
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int d = 2;
  
    // Function calling
    Rotate(arr, d, N);
    printArray(arr, N);

    return 0;
}

3 4 5 6 7 1 2 

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 i^th set).

Below is the implementation of the above approach :

// C++ program to rotate an array by
// d elements
#include <bits/stdc++.h>
using namespace std;

/*Function to get gcd of a and b*/
int gcd(int a, int b)
{
    if (b == 0)
        return a;

    else
        return gcd(b, a % b);
}

/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
    /* To handle if d >= n */
    d = d % n;
    int g_c_d = gcd(d, n);
    for (int i = 0; i < g_c_d; i++) {
        /* move i-th values of blocks */
        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;
    }
}

// Function to print an array
void printArray(int arr[], int size)
{
    for (int i = 0; i < size; i++)
        cout << arr[i] << " ";
}

/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function calling
    leftRotate(arr, 2, n);
    printArray(arr, n);

    return 0;
}

3 4 5 6 7 1 2 

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

Approach 4 :

(Using Collections module [python][/python])

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.

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

int main() {
    deque<int> dq {1, 2, 3, 4, 5, 6, 7};
    int d = 2;
    
    // Rotate the deque left by d elements
    for(int i=0; i<d; i++) {
        int temp = dq.front();
        dq.pop_front();
        dq.push_back(temp);
    }
    
    // Print the rotated deque
    for(auto it=dq.begin(); it!=dq.end(); it++) {
        cout << *it << " ";
    }
    return 0;
}

deque([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.