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 2Input: 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;// Function to rotate arrayvoid 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 arrayvoid PrintTheArray(int arr[], int n){ for (int i = 0; i < n; i++) { cout << arr[i] << " "; }}// Driver codeint 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;} |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { // Function to rotate arraystatic void Rotate(int arr[], int d, int n){ // Storing rotated version of array int temp[] = new int[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 arraystatic 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; // Function calling Rotate(arr, d, N); PrintTheArray(arr, N); }}// This code is contributed by ishankhandelwals. |
Python3
def rotate(L, d, n): k = L.index(d) new_lis = [] new_lis = L[k+1:]+L[0:k+1] return new_lisif __name__ == '__main__': arr = [1, 2, 3, 4, 5, 6, 7] d = 2 N = len(arr) # Function call arr = rotate(arr, d, N) for i in arr: print(i, end=" ") |
C#
// Include namespace systemusing System;public class GFG{ // Function to rotate array public static void Rotate(int[] arr, int d, int n) { // Storing rotated version of array int[] temp = new int[n]; // Keeping track of the current index // of temp[] var 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 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; // Function calling GFG.Rotate(arr, d, N); GFG.PrintTheArray(arr, N); }}// This code is contributed by ishankhandelwals. |
Javascript
function Rotate_and_Print(arr,d,n) { //Initializing array temp with size n var temp=new Array(n); let k = 0; // Storing the n - d elements of // array arr[] to the front of temp[] for (let i = d; i < n; i++) { temp[k] = arr[i]; k++; } // Storing the first d elements of array arr[] // into temp for (let i = 0; i < d; i++) { temp[k] = arr[i]; k++; } //Printing the temp array which stores the result 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; //number of times rotating the arrayRotate_and_Print(arr, d, n);//contributed by keerthikarathan123 |
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++
// 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 arrayvoid printArray(int arr[], int size){ for (int i = 0; i < size; i++) cout << arr[i] << " ";}// Driver codeint 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;} |
Java
/*package whatever //do not write package name here */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; // Rotate 2 times int d = 2; // Function call rotate(arr, d, N); }}// contributed by keerthikarathan123 |
Python3
# Python program to rotate an array by d elements# Function to left rotate arr[] of size n by ddef 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 # Function to print an arraydef printArray(arr, size): for i in range (size): print(arr[i] ,end = " ") # Driver codearr = [1, 2, 3, 4, 5, 6, 7]N = len(arr)d = 2# Function callingRotate(arr, d, N)printArray(arr, N)# This code is contributed by Atul_kumar_Shrivastava |
C#
// Include namespace systemusing 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; // Rotate 2 times var d = 2; // Function call 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; //number of times rotating the array// Function callingprintArray(arr, n,d);//contributed by keerthikarathan123 |
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 ith set).
Below is the implementation of the above approach :
C++
// 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 arrayvoid 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;} |
C
// C program to rotate an array by// d elements#include <stdio.h>/* function to print an array */void printArray(int arr[], int size);/*Function to get gcd of a and b*/int gcd(int a, int b);/*Function to left rotate arr[] of size n by d*/void leftRotate(int arr[], int d, int n){ int i, j, k, temp; /* To handle if d >= n */ d = d % n; int g_c_d = gcd(d, n); for (i = 0; i < g_c_d; i++) { /* move i-th values of blocks */ 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; }}/*UTILITY FUNCTIONS*//* function to print an array */void printArray(int arr[], int n){ int i; for (i = 0; i < n; i++) printf("%d ", arr[i]);}/*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);}/* Driver program to test above functions */int main(){ int arr[] = { 1, 2, 3, 4, 5, 6, 7 }; leftRotate(arr, 2, 7); printArray(arr, 7); getchar(); return 0;} |
Java
// Java program to rotate an array by// d elementsimport java.io.*;class RotateArray { /*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 i, j, k, temp; int g_c_d = gcd(d, n); for (i = 0; i < g_c_d; i++) { /* move i-th values of blocks */ 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; } } /*UTILITY FUNCTIONS*/ /* function to print an array */ void printArray(int arr[], int size) { int i; for (i = 0; i < size; i++) System.out.print(arr[i] + " "); } /*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); } // Driver program to test above functions 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); }}// This code has been contributed by Mayank Jaiswal |
Python3
# Python3 program to rotate an array by# d elements# Function to left rotate arr[] of size n by ddef leftRotate(arr, d, n): d = d % n g_c_d = gcd(d, n) for i in range(g_c_d): # move i-th values of blocks 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# UTILITY FUNCTIONS# function to print an arraydef printArray(arr, size): for i in range(size): print("% d" % arr[i], end=" ")# Function to get gcd of a and bdef gcd(a, b): if b == 0: return a else: return gcd(b, a % b)# Driver program to test above functionsarr = [1, 2, 3, 4, 5, 6, 7]n = len(arr)d = 2leftRotate(arr, d, n)printArray(arr, n)# This code is contributed by Shreyanshi Arun |
C#
// C# program for array rotationusing System;class GFG { /* Function to left rotate arr[] of size n by d*/ static void leftRotate(int[] arr, int d, int n) { int i, j, k, temp; /* To handle if d >= n */ d = d % n; int g_c_d = gcd(d, n); for (i = 0; i < g_c_d; i++) { /* move i-th values of blocks */ 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; } } /*UTILITY FUNCTIONS*/ /* Function to print an array */ static void printArray(int[] arr, int size) { for (int i = 0; i < size; i++) Console.Write(arr[i] + " "); } /* Function to get gcd of a and b*/ static int gcd(int a, int b) { if (b == 0) return a; else return gcd(b, a % b); } // Driver code public static void Main() { int[] arr = { 1, 2, 3, 4, 5, 6, 7 }; leftRotate(arr, 2, 7); printArray(arr, 7); }}// This code is contributed by Sam007 |
Javascript
<script>// JavaScript program to rotate an array by// d elements/*Function to get gcd of a and b*/function gcd( a, b){ if (b == 0) return a; else return gcd(b, a % b);}/*Function to left rotate arr[] of size n by d*/function leftRotate(arr, d, n){ /* To handle if d >= n */ d = d % n; let g_c_d = gcd(d, n); for (let i = 0; i < g_c_d; i++) { /* move i-th values of blocks */ 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 to print an arrayfunction printArray(arr, size){ for (let i = 0; i < size; i++) document.write(arr[i] +" ");}/* Driver program to test above functions */let arr = [ 1, 2, 3, 4, 5, 6, 7 ];let n = arr.length;// Function callingleftRotate(arr, 2, n);printArray(arr, n);</script> |
3 4 5 6 7 1 2
Time complexity : O(N)
Auxiliary Space : O(1)
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.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
