FOR FREE YEAR SOLVED
Right Rotate by 2 of an Array
Back to ProgrammingDescription
The elements are taken as the input. The array is rotated right twice. That means, the element of index 0 is stored in index 2, the element of index 1 is stored in index 3 and so on. And the element of index n-1 is stored in index 1, the element of index n-2 is stored in index 0.
Let the array be:
14 | 52 | 41 | 63 | 74 |
0 1 2 3 4
First the element of index n-1 (here, 74) and n-2 (here, 63) will be stored in a temporary variable and at last the values will be stored at index 1 and 0 respectively.
14 | 52 | 41 | 63 | 74 |
0 1 2 3 4
After shifting the array will be:
63 | 74 | 14 | 52 | 41 |
0 1 2 3 4
Algorithm
INPUT: The elements of the array
OUTPUT: Array after rotating it right two times
PROCESS:
Step 1: [Taking the input]
Read n [number of elements]
For i=0 to n-1 repeat
Read a[i]
[End of ‘for’ loop]
Step 2: [Right Rotating the array]
Set t1<-a[n-1]
Set t2<-a[n-2]
For i=n-1 to 2 decremented by 1 repeat
Set a[i]<-a[i-2]
[End of ‘for’ loop]
Set a[1]<-t1
Set a[0]<-t2
[Printing the array after rotation]
Print "The array after rotation: "
For i=0 to n-1 repeat
Print a[i]
[End of ‘for’ loop]
Step 3: Stop.
Code
TIME COMPLEXITY:
for(i=n-1;i>1;i--)---------------------------------O(n)
a[i]=a[i-2];
The time complexity of this program is O(n) where ‘n’ is the number of elements of the array.
SPACE COMPLEXITY:
The space complexity of the program is O(n) where ‘n’ is the number of elements of the array.
