Given an array, cyclically rotate the array clockwise by one.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}
Output: arr[] = {5, 1, 2, 3, 4}
Following are steps.
1) Store last element in a variable say x.
2) Shift all elements one position ahead.
3) Replace first element of array with x.
C++
// C++ code for program // to cyclically rotate // an array by one # include <iostream> using namespace std; void rotate(int arr[], int n) { int x = arr[n - 1], i; for (i = n - 1; i > 0; i--) arr[i] = arr[i - 1]; arr[0] = x; } // Driver code int main() { int arr[] = {1, 2, 3, 4, 5}, i; int n = sizeof(arr) / sizeof(arr[0]); cout << "Given array is "; for (i = 0; i < n; i++) cout << arr[i] << ' '; rotate(arr, n); cout << " Rotated array is "; for (i = 0; i < n; i++) cout << arr[i] << ' '; return 0; } // This code is contributed by jit_t |
Given array is 1 2 3 4 5 Rotated array is 5 1 2 3 4
Time Complexity: O(n) As we need to iterate through all the elements
Auxiliary Space: O(1)
The above question can also be solved by using reversal algorithm.
Another approach:
We can use two pointers, say i and j which point to first and last element of array respectively. As we know in cyclic rotation we will bring last element to first and shift rest in forward direction, so start swapping arr[i] and arr[j] and keep j fixed and i moving towards j. Repeat till i is not equal to j.
C++
#include <iostream> using namespace std; void rotate(int arr[], int n) { int i = 0, j = n-1; // i and j pointing to first and last element respectively while(i != j){ swap(arr[i], arr[j]); i++; } } // Driver code int main() { int arr[] = {1, 2, 3, 4, 5}, i; int n = sizeof(arr) / sizeof(arr[0]); cout << "Given array is \n"; for (i = 0; i < n; i++) cout << arr[i] << " "; rotate(arr, n); cout << "\nRotated array is\n"; for (i = 0; i < n; i++) cout << arr[i] << " "; return 0; } |
Given array is 1 2 3 4 5 Rotated array is 5 1 2 3 4
Time Complexity: O(n)
Auxiliary Space: O(1)
Please refer complete article on Program to cyclically rotate an array by one for more details!
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