Given an array, the task is to cyclically right-rotate the array by one.
Examples:
Input: arr[] = [1, 2, 3, 4, 5]
Output: [5, 1, 2, 3, 4]
Input: arr[] = [2, 3, 4, 5, 1]
Output: [1, 2, 3, 4, 5]
Shifting Each Element - O(n) Time and O(1) Space
The basic idea is to store the last element in a temp variable and shift every other element one position ahead. After shifting, update the first element with value stored in temp.
C++14 #include <iostream> #include<vector> using namespace std; void rotate(vector<int> &arr) { int n = arr.size(); // store the last element in a variable int lastElement = arr[n-1]; // assign every value by its predecessor for (int i = n - 1; i > 0; i--) { arr[i] = arr[i - 1]; } // first element will be assigned by last element arr[0] = lastElement; } int main() { vector<int> arr = {1, 2, 3, 4, 5 }; rotate(arr); for (int i = 0; i < arr.size(); i++) cout << arr[i] << ' '; return 0; } #include <stdio.h> #include <stdlib.h> void rotate(int arr[], int n) { // store the last element in a variable int lastElement = arr[n-1]; // assign every value by its predecessor for (int i = n - 1; i > 0; i--) { arr[i] = arr[i - 1]; } // first element will be assigned by last element arr[0] = lastElement; } int main() { int arr[] = {1, 2, 3, 4, 5}; int n = sizeof(arr) / sizeof(arr[0]); rotate(arr, n); for (int i = 0; i < n; i++) printf("%d ", arr[i]); return 0; } import java.util.Arrays; class GfG { static void rotate(int[] arr) { // store the last element in a variable int lastElement = arr[arr.length - 1]; // assign every value by its predecessor for (int i = arr.length - 1; i > 0; i--) { arr[i] = arr[i - 1]; } // first element will be assigned by last element arr[0] = lastElement; } public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 5}; rotate(arr); System.out.println(Arrays.toString(arr)); } } def rotate(arr): # store the last element in a variable lastElement = arr[-1] # assign every value by its predecessor for i in range(len(arr) - 1, 0, -1): arr[i] = arr[i - 1] # first element will be assigned by last element arr[0] = lastElement if __name__ == "__main__": arr = [1, 2, 3, 4, 5] rotate(arr) for i in range(0, len(arr)): print(arr[i], end=' ')
using System; class GfG { static void Rotate(int[] arr) { // store the last element in a variable int lastElement = arr[arr.Length - 1]; // assign every value by its predecessor for (int i = arr.Length - 1; i > 0; i--) { arr[i] = arr[i - 1]; } // first element will be assigned by last element arr[0] = lastElement; } static void Main() { int[] arr = {1, 2, 3, 4, 5}; Rotate(arr); Console.WriteLine(string.Join(" ", arr)); } } function rotate(arr) { // store the last element in a variable const lastElement = arr[arr.length - 1]; // assign every value by its predecessor for (let i = arr.length - 1; i > 0; i--) { arr[i] = arr[i - 1]; } // first element will be assigned by last element arr[0] = lastElement; } // Driver Code const arr = [1, 2, 3, 4, 5]; rotate(arr); console.log(arr); Time Complexity: O(n), as we need to iterate through all the elements. Where n is the number of elements in the array.
Auxiliary Space: O(1), as we are using constant space.
Using Two Pointers - O(n) Time and O(1) Space
We can use two pointers, As we know in cyclic rotation we will bring last element to first and shift rest in forward direction, we can do this by swapping every element with last element till we get to the last point.
C++14 #include <iostream> #include<vector> using namespace std; void rotate(vector<int> &arr) { int n=arr.size(); // i and j pointing to first and last // element respectively int i = 0, j = n - 1; while (i != j) { swap(arr[i], arr[j]); i++; } } int main() { vector<int> arr = {1, 2, 3, 4, 5 }; int i; rotate(arr); for (i = 0; i < arr.size(); i++) cout << arr[i] << " "; return 0; } #include <stdio.h> #include <stdlib.h> void rotate(int *arr, int n) { // i and j pointing to first and last // element respectively int i = 0, j = n - 1; while (i != j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i++; } } int main() { int arr[] = {1, 2, 3, 4, 5}; int n = sizeof(arr) / sizeof(arr[0]); rotate(arr, n); for (int i = 0; i < n; i++) printf("%d ", arr[i]); return 0; } import java.util.Arrays; class GfG { static void rotate(int[] arr) { // i and j pointing to first and last // element respectively int i = 0, j = arr.length - 1; while (i != j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i++; } } public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 5}; rotate(arr); System.out.println(Arrays.toString(arr)); } } def rotate(arr): # i and j pointing to first and last # element respectively i, j = 0, len(arr) - 1 while i != j: arr[i], arr[j] = arr[j], arr[i] i += 1 if __name__ == "__main__": arr = [1, 2, 3, 4, 5] rotate(arr) for i in range(0, len(arr)): print(arr[i], end=' ')
using System; class GfG { static void Rotate(int[] arr) { // i and j pointing to first and last // element respectively int i = 0, j = arr.Length - 1; while (i != j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i++; } } static void Main() { int[] arr = {1, 2, 3, 4, 5}; Rotate(arr); Console.WriteLine(string.Join(" ", arr)); } } function rotate(arr) { // i and j pointing to first and last // element respectively let i = 0, j = arr.length - 1; while (i !== j) { [arr[i], arr[j]] = [arr[j], arr[i]]; i++; } } // Driver Code let arr = [1, 2, 3, 4, 5]; rotate(arr); console.log(arr); Time Complexity: O(n), as we need to iterate through all the elements. Where n is the number of elements in the array.
Auxiliary Space: O(1), as we are using constant space.
Reversing the Array - O(n) Time and O(1) Space
We can use Reversal Algorithm also, reverse first n-1 elements and then whole array which will result into one right rotation.
C++14 #include <iostream> #include<vector> using namespace std; void rotate(vector<int> &arr) { int n = arr.size(); // Reverse the first n-1 terms int i, j; for (i = 0, j = n - 2; i < j; i++, j--) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } // Reverse the entire array for (i = 0, j = n - 1; i < j; i++, j--) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } int main() { vector<int> arr = {1, 2, 3, 4, 5 }; int i, j; rotate(arr); for (i = 0; i < arr.size(); i++) cout << arr[i] << " "; return 0; } #include <stdio.h> #include <stdlib.h> void rotate(int *arr, int n) { // Reverse the first n-1 terms int i, j; for (i = 0, j = n - 2; i < j; i++, j--) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } // Reverse the entire array for (i = 0, j = n - 1; i < j; i++, j--) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } int main() { int arr[] = {1, 2, 3, 4, 5}; int n = sizeof(arr) / sizeof(arr[0]); rotate(arr, n); for (int i = 0; i < n; i++) printf("%d ", arr[i]); return 0; } import java.util.Arrays; class GfG { static void rotate(int[] arr) { int n = arr.length; // Reverse the first n-1 terms for (int i = 0, j = n - 2; 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; } } public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 5}; rotate(arr); System.out.println(Arrays.toString(arr)); } } def rotate(arr): n = len(arr) # Reverse the first n-1 terms i, j = 0, n - 2 while i < j: arr[i], arr[j] = arr[j], arr[i] i += 1 j -= 1 # Reverse the entire array i, j = 0, n - 1 while i < j: arr[i], arr[j] = arr[j], arr[i] i += 1 j -= 1 if __name__ == "__main__": arr = [1, 2, 3, 4, 5] rotate(arr) for i in range(0, len(arr)): print(arr[i], end=' ')
using System; class GfG { static void Rotate(int[] arr) { int n = arr.Length; // Reverse the first n-1 terms for (int i = 0, j = n - 2; 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; } } static void Main() { int[] arr = {1, 2, 3, 4, 5}; Rotate(arr); Console.WriteLine(string.Join(" ", arr)); } } function rotate(arr) { const n = arr.length; // Reverse the first n-1 terms for (let i = 0, j = n - 2; i < j; i++, j--) { [arr[i], arr[j]] = [arr[j], arr[i]]; } // Reverse the entire array for (let i = 0, j = n - 1; i < j; i++, j--) { [arr[i], arr[j]] = [arr[j], arr[i]]; } } // Driver Code const arr = [1, 2, 3, 4, 5]; rotate(arr); console.log(arr); Time Complexity: O(n), as we are reversing the array. Where n is the number of elements in the array.
Auxiliary Space: O(1), as we are using constant space.
Similar Reads
Rotate bits of a number Given a 32-bit integer n and an integer d, rotate the binary representation of n by d positions in both left and right directions. After each rotation, convert the result back to its decimal representation and return both values in an array as [left rotation, right rotation].Note: A rotation (or cir
7 min read
Rotate an Array by d - Counterclockwise or Left Given an array of integers arr[] of size n, the task is to rotate the array elements to the left by d positions.Examples:Input: arr[] = {1, 2, 3, 4, 5, 6}, d = 2Output: {3, 4, 5, 6, 1, 2}Explanation: After first left rotation, arr[] becomes {2, 3, 4, 5, 6, 1} and after the second rotation, arr[] bec
15+ min read
Rotate an Array - Clockwise or Right Rotations in the array is defined as the process of rearranging the elements in an array by shifting each element to a new position. This is mostly done by rotating the elements of the array clockwise or counterclockwise.Table of ContentTypes of Rotations in ArrayHow to implement rotations in an arr
15+ min read
Rotate Matrix Clockwise by 1 Given a square matrix, the task is to rotate its elements clockwise by one step.Examples:Input 1 2 34 5 6 7 8 9Output: 4 1 27 5 3 8 9 6Input: 1 2 3 4 5 6 7 8 9 10 11 1213 14 15 16 Output: 5 1 2 3 9 10 6 4 13 11 7 8 14 15 16 12The idea is to use nested loops to move elements in four directions (right
9 min read
Rotate a Matrix by 180 degree Given a square matrix, the task is to turn it by 180 degrees. Note that when we rotate a matrix by 180 degree, clockwise and anticlockwise both give same results. Examples: Input: mat[][] = [[1, 2, 3] [4, 5, 6] [7, 8, 9]]Output: [9, 8, 7] [6, 5, 4] [3, 2, 1]Input: mat[][] = [[1, 2, 3, 4], [5, 6, 7,
13 min read